Laxmi

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions

Additional Questions and Answers

Exercise 4.1

Question 1.
“The sum of any two angles of a triangle is always greater than the third angle”. Is this statement true. Justify your answer.
Solution:
No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be 90°, 45°, 45°.
Here sum of two angles 45° + 45° = 90°.

Question 2.
The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the angles of the triangle be x, 2x, x.
Using the angle sum property, we have
x + 2x + x = 180°
4x = 180°
x = \(\frac{180^{\circ}}{4}\)
x = 45°
2x = 2 × 45° = 90°
Thus the three angles of the triangle are 45°, 90°, 45°.
Its two angles are equal. It is an isoscales triangle. Its one angle is 90°.
∴ It is a right angled triangle.

Question 3.
Find the values of the unknown x and y in the following figures

Solution:
(i) Since angles y and 120° form a linear pair.
y + 120° = 180°
y = 180° – 120°
y = 60°
Now using the angle sum property of a triangle, we have
x + y + 50° = 180°
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° – 110°= 70°
x = 70°
y = 60

(ii) Using the angle sum property of triangle, we have
50° + 60° + y = 180°
110° + y = 180°
y = 180° – 110°
y = 70°
Again x and y form a linear pair
∴ x + y = 180°
x + 70° = 180°
x = 180° – 70°= 110°
∴ x = 110°; y = 70°

Question 4.
Two angles of a triangle are 30° and 80°. Find the third angle.
Solution:
Let the third angle be x.
Using the angle sum property of a triangle we have,
30° + 80° + x = 180°
x + 110° = 180°
x = 180° – 110° = 70°
Third angle = 70°.

Exercise 4.2

Question 1.
In an isoscleles ∆ABC, AB = AC. Show that angles opposite to the equal sides are equal.
Solution:

Given: ∆ABC in which \(\overline{A B}\) = \(\overline{A C}\).
To Prove: ∠B = ∠K.
Construction: Draw AD ⊥ BC.
Proof: In right ∆ADB and right ∆ADC.
we have side AD = side AD (common)
AB = AC (Hypoteneous) (given)
∆ADB = ADC (RHS criterion]
∴ Their corresponding parts are equal. ∠B = ∠C.

Question 2.
ABC is an isosceles triangle having side \(\overline{A B}\) = side \(\overline{A C}\). If AD is perpendicular to BC, prove that D is the mid-point of \(\overline{B C}\).
Solution:
In ∆ABD and ∆ACD, we have
∠ADB = ∠ADC [∵ AD ⊥ BC]
Side \(\overline{A D}\) = Side \(\overline{A D}\) [Common]
Side \(\overline{A B}\) = Side \(\overline{A C}\) [Common]
Using RHS congruency, we get
∆ABD ≅ ∆ACDc
Their corresponding parts are equal

∴ BD = CD
∴ O is the mid point of BC.

Question 3.
In the figure PL ⊥ OB and PM ⊥ OA such that PL = PM.
Prove that ∆PLO ≅ ∆PMO.
Solution:

In ∆PLO and ∆PMO, we have
∠PLO = ∠PMO = 90° [Given]
\(\overline{O P}\) = \(\overline{O P}\) [Hypotenuse]
PL = PM
Using RHS congruency, we get
∆PLO ≅ ∆PMO

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Intext Questions

Exercise 3.1
Try This (Text book Page No. 60)

Question 1.
How will you prove ∆ ABC ~ AD AC?
Proof:
In ∆ ABC & ∆ DAC, ∠C is common and ∠BAC = ∠ADC = 90°
Therefore ∆ ABC ~ AD AC (AA similarity)

Try This (Text book Page No. 61)

Question 1.
Check whether the following are Pythagorean triplets

1. 57, 176, 185
2. 264, 265, 23
3. 8, 41, 40

Solution:
1. For Pythagorean triplet, the sum of the squares of 2 sides is equivalent to square of 3rd side (hypotenuse)
Let us check of
57² + 176² whether = 185² or not
57² = 3249
176² = 30976
185² = 34225
57² + 176² = 3249 + 30976 = 34225 = 185²
∴ 57, 176 & 185 are Pythagorean triplet.

2. 23, 264, 265
23² = 529
264² = 69696
23² + 264² = 70225
265² = 70225
∴ 23² + 264² = 265²
∴ Pythagorean triplet

3. 8, 41, 40
8² = 64
40² = 1600
8² + 40² = 1664
41² = 1681
8² + 40² ≠ 41² = 1
∴ They are not Pythagorean triplet

Activity – 1 (Text book Page No. 62)

We can construct sets of Pythagorean triplets as follows. Let m and n be any two positive integers (m > n):
(a, b, c) is a Pythagorean triple if a = m² – n², b = 2mn and c = m² + n² (Think, why?) Complete the table.

Solution:

Activity – 2 (Text book Page No. 65)

Question 1.
Find all integer-sided right angled triangles with hypotenuse 85
Solution:
(x + y)² – 2xy = 85²

Pythagorean triplets with hypotenuse 85.

Exercise 3.3
Try These (Text book Page No. 68)

Question 1.
The area of the trepezium is ……..
Answer:
\(\frac{1}{2}\) x h x (a + b) sq. units!

Question 2.
The distance between the parallel sides of a trapezium is called as ………
Answer:
its height

Question 3.
If the height and parallel sides of a trapezium are 5 cm, 7 cm and 5 cm respectively, then its area is ……..
Answer:
30 sq cm
Hint:
= \(\frac{1}{2}\) x h x (a + b) sq. units
= \(\frac{1}{2}\) x 5 x (7 + 5)
= \(\frac{1}{2}\) x 5 x 12 = 30 sq.cm

Question 4.
In an isosceles trapezium, the non-parallel sides are ……….. in length.
Answer:
Equal.

Question 5.
To construct a trapezium, ………… measurements are enough.
Answer:
Four.

Question 6.
If the area and sum of the parallel sides are 60 cm2 and 12 cm, its height is ………..
Answer:
10 cm
Hint:
Area of the trapezium = \(\frac{1}{2}\) x h (a + b)
60 = \(\frac{1}{2}\) x h x (12)
h = \(\frac{60×2}{12}\) = 10 cm

Exercise 3.4
Think (Text book Page No. 74)

Question 1.
Can a rhombus, a square or a rectangle be called as a parallelogram? Justify your answer.
Solution:
Yes, a rhombus, a square or a rectangle can be called as parallelogram as the opposite sides are equal and parallel and diagonals bisect each other in this figures.

Try These (Text book Page No. 74)

Question 1.
In a parallelogram, the opposite sides are …….. and ……….
Answer:
equal, parallel

Question 2.
If ∠A of a parallelogram ABCD is 100° then, find ∠B, ∠C and ∠D .
Answer:

∠B = 80°
∠C = 100°
∠D = 80°

Question 3.
Diagonals of a parallelogram each other.
Answer:
Bisect

Question 4.
If the base and height of the parallelogram are 20 cm and 5 cm then, its area is ………
Answer:
100 sq.cm

Question 5.
Find the unknown values in the given parallelograms and write the property Used to find them.

Solution:

Students can Download Maths Chapter 4 Geometry Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Intext Questions

Exercise 4.1

Try These (Text book Page No. 66)

Answer the following questions.

Question 1.
Triangle is formed by joining three ______ points.
Answer::
Non collinear

Question 2.
A triangle has ______ vertices and ______ sides.
Answer:
three, three

Question 3.
A point where two sides of a triangle meet is known as ______ of a triangle.
Answer:
vertese

Question 4.
Each angle of an equilateral triangle is of measure.
Answer:
same

Question 5.
A triangle has angle measurements of 29°, 65° and 86°. Then it is ______ triangle.
(i) an acute angled
(ii) a right angled
(iii) an obtuse angled
(iv) a scalene
Answer:
(i) an acute angled

Question 6.
A triangle has angle measurements of 30°, 30° and 120°. Then it is ______ triangle.
(i) an acute angled
(ii) scalene
(iii) obtuse angled
(iv) right angled
Answer:
obtuse angled

Question 7.
Which of the following can be the sides of a triangle?
(i) 5.9.14
(ii) 7,7,15
(iii) 1,2,4
(iv) 3, 6, 8
Answer:
(iv) 3, 6, 8
Solution:
(i) Here 5 + 9 = 14 = the measure of the third side.
In a triangle the sum of the measures of any two sides must be greater than the third side.
∴ 5, 9, 14 cannot be the sides of a triangle.

(ii) 7.7.15
Here sum of two sides 7 + 7 = 14 < the measures of the thrid side.
So 1,1, 15 cannot be the sides of a triangles.

(iii) 1,2,4
Here sum of two sides 1 + 2 = 3 < the measure of the third side.
∴ 1, 2, 4 cannot be the sides of a triangle.

(iv) 3, 6, 8
Sum of two sides 3 + 6 = 9 > the third side.
∴ 3, 6, 8 can be the sides of a triangle.

Question 8.
Ezhil wants to fence his triangular garden. If two of the sides measure 8 feet and 14 feet then the length of the third side is ______
(i) 11 ft
(ii) 6 ft
(iii) 5 ft
(iv) 22 ft
Answer:
(i) 11 ft

Question 9.
Can we have more than one right angle in a triangle?
Solution:
No, we cannot have more than one right angle in a triangle.
Because the sum of three angles of a triangle is 180°.
But if two angles are right angles then their sum itself become 180°.

Question 10.
How many obtuse angles are possible in a triangle?
Solution:
Only one.

Question 11.
In a right triangle, what will be the sum of other two angles?
Solution:
Sum of three angles of a triangle = 180°
If one angle is right angle (i.e. 90°) .
Sum of other two sides = 180° – 90° = 90°

Question 12.
Is it possible to form an isosceles right angled triangle? Explain.
Solution:
Yes, it is possible.
If one angle is right angle, then the other two angles will be 45° and 45°.

Exercise 4.2

Try These (Text book Page No. 76)

Question 1.
Measure and group the pair of congruent line segments.

Solution:
\(\overline{A B}\) = 3 cm
\(\overline{C D}\) = 4.8 cm
\(\overline{I J}\) = 4.8 cm
\(\overline{P Q}\) = 3 cm
\(\overline{R S}\) = 1.7 cm
\(\overline{X Y}\) = 1.7 cm
From the above measurement S, we can conclude that
(i) \(\overline{A B}\) ≅ \(\overline{P Q}\)
(ii) \(\overline{C D}\) ≅ \(\overline{I J}\)
(iii) \(\overline{R S}\) ≅ \(\overline{X Y}\)

Try These (Text book Page No. 77)

Question 1.
Find the pairs of congruent angles either by superposition method or by measuring them.

Solution:
From the given figures
∠ABC = 50°
∠EFG = 120°
∠HIJ = 120°
∠KLH = 90°
∠PON = 50°
∠RST = 90°
From the above measures, we can conclude that
(i) ∠ABC = ∠PON
(ii) ∠EFG = ∠HIJ
(iii) ∠KLH ≅ ∠RST

Try These (Text book Page No. 83)

Question 1.
If ∆ABC ≅ ∆XYZ then list the corresponding sides and corresponding angles.

Solution:
If ∆ABC ≅ ∆XYZ
\(\overline{A B}\) ≅ \(\overline{X Y}\) – \(\overline{B C}\) ≅ \(\overline{Y Z}\)
\(\overline{A C}\) ≅ \(\overline{X Z}\)
And also
∠A ≅ ∠X – ∠B ≅ ∠Y
∠C ≅ ∠Z

Question 2.
Given triangles are congruent. Identify the corresponding parts and write the congruent statement.

Solution:
Given the set of triangles are congruent. Also we observe from the triangles that the corresponding sides.
\(\overline{A B}\) = \(\overline{A C}\)
\(\overline{B C}\) = \(\overline{Y Z}\)
\(\overline{A C}\) = \(\overline{X Z}\)
Here three sides of ∆ABC are equal to the corresponding sides of ∆XYZ.
This criterion of congruency is side – side – side.

Question 3.
Mention the conditions needed to conclude the congruency of the triangles with reference to the above said criterions. Give reasons for your answer.

Solution:
(i) In ∆ABC and ∆XYZ
if \(\overline{A B}\) = \(\overline{X Y}\)
\(\overline{B C}\) = \(\overline{Y Z}\)
\(\overline{A C}\) = \(\overline{X Z}\)
then ∆ABC ≅ ∆XYZ. By the Side – Side -Side Congruency Criterion.

then ∆AB ≅ ∆XYZ.
By Side – Angle – Side Criterion.

then ∆ABC ≅ ∆XYZ.
By Angle – Side – Angle Congruency Critirion.

then by RHS criterion.
∆ABC ≅ ∆XYZ

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
In an isoscales triangle one angle is 76°. If the other two angles are equal, find them.
Solution:
In an isoscales triangle, angle opposite to equal sides are equal. Let the equal angles be x° and x°.
In a triangle the sum of the three angles is 180°.
x° + x° + 76° = 180°
x° (1 + 1) = 180° – 76° = 104°
2x = 104°
x = \(\frac{104^{\circ}}{2}\) = 52°
x = 52°
∴ Other two angles are 52° and 52°.

Question 2.
If two angles of a triangle are 46° each, how can you classify the triangle?
Solution:
Given two angles of the triangle are same and is equal to 46°. If two angles are equal the sides opposite to equal angles are equal. Therefore it will be an isoscales triangle.

Question 3.
If an angle of a triangle is equal to the sum of the other two angles, find the type of the triangle.
Solution:
Let ∠B is the greater angle then by the given condition ∠B = ∠A + ∠C.
Sum of three angle of a triangle = 180°.

∠A + ∠B + ∠C = 180°.
∠A + (∠A + ∠C) + ∠C) = 180°.
2∠A + 2∠C = 180°
2(∠A + ∠C) = 180°
∠A + ∠C = \(\frac{180^{\circ}}{2}\)
∠B = 90°
∴ One of the angle of the triangle = 90°
It will be a right angled triangle.

Question 4.
If the exterior angle of a triangle is 140° and its interior opposite angles are equal, find all the interior angles of the triangle.
Solution:
Given the exterior angle = 140°
Interior opposite angle are equal.

Let one of the interior opposite angle be x.
Then x + x = 140°.
[∵ Exterior angle = sum of interior opposite angles]
2x = 140°
x = \(\frac{140^{\circ}}{2}\) = 70°
x = 70°
Interior opposite angle = 70°, 70°.
Sum of the three angles of a triangle = 180°.
70° + 70° + Third angle = 180°
140° + Third angle = 180°
Third angle = 180° – 140° = 40°
∴ Interior angle are 40°, 70°, 70°.

Question 5.
In ∆JKL, if ∠J = 60° and ∠K = 40°, then find the value of exterior angle formed by extending the side KL.
Solution:
When extending the side KL, the exterior angle formed in equal
to the sum of the interior opposite angles.

∠JLX = ∠LJK + ∠LKJ
= 60°+ 40° =100°
Exterior angle formed = 100°

Question 6.
Find the value of ‘x’ in the given figure.

Solution:
Given ∠DCB = 1000 and ∠DBA = 128°
In the given figure
∠CBD + ∠DBA = 180°
∠CBD + 128° = 180°
∠CBD = 52°
Now exterior angle x = Sum of interior opposite angles.
x = ∠DCB + ∠CBD = 100° + 52° = 152°
x = 152°

Question 7.
If ∆MNO ≅ ∆DEF, ∠M = 60° and ∠E = 45° then find the value of ∠O.
Solution:
Given ∆MNO ≅ ∆DEF

∴ Corresponding parts of conqruent triangle are congruent.
∠M = ∠D = 60° [given ∠M = 60°]
∠N = ∠E = 45° [given ∠E = 45°]
∠O = ∠F
In triangle MNO, sum of the three angle – 180°.
∠M + ∠N + ∠O = 180°
60° + 45° + ∠O = 180°
105° + ∠O = 180°
∠O = 180° – 105° = 75°
Value of ∠O = 75°

Question 8.
In the given figure ray AZ bisects ∠BAD and ∠DCB, prove that
(i) ∆BAC ≅ ∆DAC
(ii) AB = AD

Solution:
(i) In ∆BAC and ∆DAC
∠BAC = ∠DAC [Given \(\overline{A Z}\) bisects ∠BAD]
∠BCA = ∠DCA[\(\overline{A Z}\) bisects ∠DCB]
AC = AC [∵ common side]
∴ Here AC is the included side of the angles. By ASA criterior, ∆BAC ≅ ∆DAC.

(ii) By (i) ∆BAC ≅ ∆DAC
BA = DA [By CPCTC]
i.e., AB = AD

Question 9.
In the given figure FG = FI and H is midpoint of GI, prove that ∆FGH ≅ ∆FHI
Solution:
In ∆FGH and ∆FHI
Given FG = HI
Also, GH = HI [∵ H is the midpoint of GI]

FH = FH [Common]
∴ By S.SS congruency criteria, ∆FGH ≅ ∆FIH. Hence proved.

Question 10.
Using the given figure, prove that the triangles are congruent. Can you conclude that AC is parallel to DE.
Solution:
In ∆ABC and ∆EBD,

AB = EB
BC = BD
∠ABC = ∠EBD [∵ Vertically opposite angles]
By SAS congruency criteria. ∆ABC ≅ ∆EBD.
We know that corresponding parts of congruent triangles are congruent.
∴ ∠BCA ≅ ∠BDE
and ∠BAC ≅ ∠BED
∠BCA ≅ ∠BDE means that alternate interior angles are equal if CD is the transversal to lines AC and DE.
Similarly, if AE is the transversal to AC and DE, we have ∠BAC ≅ ∠BED
Again interior opposite angles are equal.
We can conclude that AC is parallel to DE.

Challenge Problems

Question 11.
In given figure BD = BC, find the value of x.

Solution:
Given that BD = BC
∆BDC is on isoscales triangle.
In isoscales triangle, angles opposite to equal sides are equal.
∠BDC = ∠BCD ……(1)
Also ∠BCD + ∠BCX = 180° [∵ Liner Pair]
∠BCD + 115° = 180°
∠BCD = 180° – 115°
∠BCD = 65° [By (1)]
In ∆ADB
∠BAD + ∠ADB = ∠BDC
[∵ BDC is the exterior angle and ∠BAD and ∠ABD are interior opposite angles]
35° + x = 65°
x = 65° – 35°
x = 30°

Question 12.
In the given figure find the value of x.

Solution:
For ∆LNM, ∠LMK is the exterior angle at M.
Exterior angle = sum of opposite interior angles
∠LMK = ∠MLN + ∠LNM = 26° + 30° = 56°
∠JMK = 56° [∵ ∠LMK = ∠JMK]
x is the exterior angle at J for ∆JKM.
∴ x = ∠JKM + ∠KMJ [∵ Sum of interior opposite angles]
x = 58° + 56° [∵ ∠JMK = 56°]
x = 114°

Question 13.
In the given figure find the values of x and y.

Solution:
In ∆BCA, ∠BAX = 62° is the exterior angle at A.
Exterior angle = sum of interior opposite angles.
∠ABC + ∠ACB = ∠BAX
28°+ x = 62°
x = 62° – 28° = 34°
Also ∠BAC + ∠BAX = 180° [∵ Linear pair]
y + 62° = 180°
y = 180° – 62° = 118°
x = 34°
y = 118°

Question 14.
In ∆DEF, ∠F = 48°, ∠E = 68° and bisector of ∠D meets FE at G. Find ∠FGD.

Solution:
Given ∠F = 48°
∠E = 68°
In ∆DEF,
∠D + ∠F + ∠E = 180° [By angle sum property]
∠D + 68° + 68° = 180°
∠D + 116° = 180°
∠D = 180° – 116° = 64°
Since DG is the angular bisector of ∠D.
∠FDG = ∠GDE
Also ∠FDG + ∠GDE = ∠D
2 ∠FDG = 64°
2 ∠FDG = 64°
∠FDG = \(\frac{64^{\circ}}{2}\) = 32°
∠FDG = 32°
In ∆FDG,
∠FDG + ∠GFD = 180° [By angle sum property of triangles]
32° + ∠FDG + 48° = 180°
∠FDG + 80° = 180°
∠FDG = 180° – 80°
∠FDG = 100°

Question 15.
In the figure find the value of x.

Solution:
Exterior angle is equal to the sum of opposite interior angles.
in ∆TSP ∠TSP + ∠SPT = ∠UTP
75° + ∠SPT = 105°
∠SPT = 105° – 75°
∠SPT = 30° ……(1)
∠SPT + ∠TPR + ∠RPQ = 180° [∵ Sum of angles at a point on a line is 180°]
30° + 90° + ∠RPQ = 180°
120° + ∠RPQ = 180°
∠RPQ = 180° – 120°
∠RPQ = 60° …… (2)
∠VRQ + ∠QRP = 180° [∵ linear pair]
145° + ∠QRP = 180°
∠QRP = 180° – 145°
∠QRP = 35°
Now in ∆ PQR
∠QRP + ∠RPQ = x [∵ x in the exterior angle]
35° + 60° = x
95° = x

Question 16.
From the given figure find the value of y.

Solution:
From the figure,
∠ACB = ∠XCY [Vertically opposite angles]
∠ACB = 48° …(1)
In ∆ABC, ∠CBD is the exterior angle at B.
Exterior angle = Sum of interior opposite angles.
∠CBD = ∠BAC + ∠ACB
∠CBE + ∠EBD = 57° + 48°
65° + ∠EBD = 105°
∠EBD = 105° + 65° = 40° ……… (2)
In ∆EBD, y is the exterior angle at D.
y = ∠EBD + ∠BED
[∵ Exterior angle = Sum of opposite interior angles]
y = 40° + 97° [∵ From (2)]
y = 137°

Students can Download Maths Chapter 3 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Additional Questions

Question 1.
State Pythagoras theorem.
Solution:
In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Question 2.
Write two uses of Pythagoras theorem in our daily life.
Solution:
The Pythagoras theorem is useful in finding the distance and the heights of objects.

Question 3.
Prove Pythagoras theorem.
Solution:
Proof of the Pythagoras theorem using similarity of triangles.

Given: ∠BAC = 90°
To Prove BC² = AB² + AC²
Construction: Draw AD ⊥ BC
Proof:

In ∆ ABC and ∆ DBA
∠B is common and ∠BAC = ∠ADB = 90°
Therefore, ∆ ABC ~ ∆ DBA, (AA similarity)
Hence, the ratio of corresponding sides are equal.
⇒ \(\frac{AB}{DB}\) = \(\frac{BC}{BA}\)
⇒ AB² = BC x DB
Similarity, ∆ ABC ~ ∆ DAC,
⇒ \(\frac{AC}{DC}\) = \(\frac{BC}{AC}\)
⇒ AC² = BC x DB
Adding (1) and (2), we get
AB² + AC² = BC x DB + BC x DC
= BC x (DB + DC) = BC x BC
⇒ AB² + AC² = BC² and hence the proof of the theorem

Question 4.
State the converse of Pythagoras theorem.
Solution:
If in a triangle, the square on the greatest side is equal to the sum of squares on the other two sides, then the triangle is right angled triangle.

Question 5.
Give 2 examples for Pythagorean triplet.
Solution:
(21, 28, 35), (24, 32, 40)

Question 6.
Check whether (2, 3, 5) is Pythagorean triplet.
Solution:

2² + 3² = 4 + 9 = 13
5² = 25
13 ≠ 25
∴ It is not a Pythagorean triplet.

Question 7.
Can the sides that measure 15 cm, 20 cm and 25 cm make a right triangle?
Solution:

20² + 15² = 400 + 225 = 625
25² = 625
20² + 15² = 25²
∴ It is Pythagorean triplet.
∴ The given sides make a right angled triangle.

Question 8.
In the figure, find the value of x.

Solution:
In ABC
12² + 9² = 144 + 81 =225 = 15²
∴ x = 15

Question 9.
In the figure find the value of a and b and verify ∆ ABD is a right angled triangle.
Solution:
Now by altitude – on – hypotenuse theorem,
AB² = BC x BD gives
10² = a x 26
a = \(\frac{100}{26}\) = \(\frac{50}{13}\)

and AD² = CD x BD
24² = b x 26
b = \(\frac{576}{26}\) = \(\frac{288}{13}\)
and in ∆ ABD;
AB² + AD² = BD²
10² + 24² = 576 = 26²
BD = 26
Therefore ∆ ABD is a right angled triangle.

Question 9.
State the altitude-on-Hypotenuse theorem.
Solution:
If an altitude is drawn to the hypotenuse of an right angled triangle then (i) the two triangles are similar to the given triangle and also to each other.

That is ∆ PRQ ~ ∆ PSR ~ ∆ RSQ

1. h² = xy
2. p² = yr and
3. q² = xr, where r = x + y

You can Download Samacheer Kalvi 9th English Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th English Grammar Active Voice, Passive Voice

Read the following sentences and analyze the difference.
The team leader presented the report.
The report was presented by the team leader.

1. In the first sentence, the verb shows that the subject is the doer of the action. Therefore, the sentence is in active voice.
2. In the second sentence, the verb shows that the subject is not the doer of the action. Therefore, the sentence is in passive voice.

We use the Passive voice when –

• the focus is on the action rather than the doer of the action.
  (e.g.), About 50 percent of the graduates are employed in IT-related sectors.
• we do not know who the doer is.
  (e.g.) My bike was stolen yesterday.
• we talk of a system or a process.
  (e.g.) The vegetables are washed well. Then, they are cut into cubes.
• we write newspaper headlines and notices at public places, (‘be’ verb is omitted as the language has to be concise)
  (e.g.) 20 sportsmen felicitated by PM.
• we describe changes that have taken place.
  (e.g.) Our school looks completely different. The whole place has been painted.

Look at the table below. It shows the changes in tense while changing sentences from active voice in to passive voice.

When using the active voice, the subjects are the ones performing the action.
God loves all men.
Birds build nests.
Dogs eats bones.

In these three sentences the subject does the action. Hence they are in the active voice. In the active voice, the verb takes an object.
All men are loved by God.
Nests are built by birds.
Bones are eaten by dogs.

Rules for changing Active Voice into Passive Voice:

• Identify the subject, the verb and the object: SVO.
• Change the object into subject.
• Put the suitable helping verb or auxiliary verb according to the tense given.
• Change the verb into past participle of the verb.
• Add the preposition “by”
• Change the subject into object.

How to form passive forms of verbs?

Changes of Pronouns:

Type 1: Statements:

Type 2: Imperative Sentence :
If the given sentence in the active voice is in the imperative, to get the passive voice use ‘Let’.
Passive Voice = Let + Object + be + Past Participle

• We can begin the passive sentence with ‘you’ if we want to put emphasis on the person addressed to.

Type 3 : Interrogative Questions in the Passive :
If the question in the Active Voice begins with a Helping verb, the Passive voice must also begin with a suitable helping verb. Supposing the question begins with ‘Wh or How’ form (what, when, how …) the Passive Voice must begin with the same.

If a sentence contains two objects namely Indirect Object and Direct Object in the Active Voice, two forms of Passive Voice can be formed.

• She brought me a cup of coffee. (AV)
  I was brought a cup of coffee by her. (PV) (or)
  A cup of coffee was brought to me by her. (PV)
• The teacher teaches us grammar. (AV)
  We are taught grammar by the teacher. (PV) (or)
  Grammar is taught [to] us by the teacher. (PV)

Infinitive and Gerund :

• I want to shoot the tiger. (AV)
  I want the tiger to be shot. (PV)
• I remember my father taking me to the theatre. (AV)
  I remember being taken to the theatre by my father. (PV)

Passive to Active form:
While changing Passive Voice into Active Voice, we must keep in mind all the rules of the Active Voice in the reverse order. We come across sentences in the Passive Voice without subject or agent. In this case, supply the appropriate subject.

Changing Passive Voice to Active Voice.

• Last year, the Swach Bharat scheme was announced by the Government.
• Rare plants are found in Silent Valley.

In the first sentence, the doer/agent is explicitly mentioned because the doer is important in that sentence. But in the second sentence, it is not so, because either the agent or doer of the action is too obvious or unknown.

The passive construction is quite common in scientific/technical/ business writing. In these types of objective writing, the emphasis is usually on the action or process or thing that is described. So the ‘by’ phrase is generally omitted in these writings. It is called Impersonal Passive.

• They say that might is right.
  It is said that might is right.
• One finds mosquitoes everywhere.
  Mosquitoes are found everywhere.
• He gave us a cheque.
  A cheque was given to us.

Exercises
(i) Choose the correct passive form for the following sentences in active voice.

1. I did not beat her.
(a) She is not beaten by me.
(b) She has not beaten by me.
(c) She was not beaten by me.
Answer:
(c) She was not beaten by me.

2. I will never forget this experience.
(a) This experience is not forgotten by me.
(b) This experience would never be forgotten by me.
(c) This experience will never be forgotten by me.
Answer:
(c) This experience will never be forgotten by me.

3. Mother made a cake yesterday.
(a) A cake made by mother yesterday.
(b) A cake is made by mother yesterday,
(c) A cake was made by mother yesterday.
Answer:
(c) A cake was made by mother yesterday.

4. The boy congratulated the girl.
(a) The girl was congratulated by the boy.
(b) The girl had congratulated by the boy.
(c) The girl congratulated the boy.
Answer:
(a) The girl was congratulated by the boy.

5. Did she do her duty?
(a) Was she done her duty?
(b) Was her duty done by her?
(c) Had her duty done by her?
Answer:
(b) Was her duty done by her?

6. The tiger was chasing the deer.
(a) The deer was chased by the tiger.
(b) The deer was being chased by the tiger,
(c) The deer had chased by the tiger.
Answer:
(b) The deer was being closed by the tiger

7. She has written a novel.
(a) A novel has written by her.
(b) A novel has been written by her.
(c) A novel had written by her.
Answer:
(b) A novel has been written by her.

8. She has learned her lessons.
(a) Her lessons has learned by her.
(b) Her lessons have been learned by her.
(c) Her lessons had been learned by her.
Answer:
(b) Her lessons have been learned by her.

9. Have you finished the report?
(a) Has the report finished by you?
(b) Has the report been finished by you?
(c) Had the report been finished by you?
Answer:
(b) Has the report been finished by you?

10. The police have caught the thief.
(a) The thief has been caught by the police.
(b) The thief was caught by the police.
(c) The thief had been caught by the police.
Answer:
(a) The thief has been caught b– police,

(ii) Rewrite the following sentences into active voice.

1. We are taught grammar by Mr. Harsha.
Answer:
Mr. Harsha teaches us grammar.

2. He was praised by the teacher.
Answer:
The teacher praised him.

3. The injured were taken to the hospital by the firemen.
Answer:
The firemen took the injured to the hospital.

4. The town was destroyed by an earthquake.
Answer:
An earthquake destroyed the town.

5. The teacher was pleased with the boy’s work.
Answer:
The boy’s work pleased the teacher.

6. The building was damaged by the fire.
Answer:
The fire damaged die building.

7. By whom were you taught Hindi?
Answer:
Who taught you Hindi? ,

8. You will be given a ticket by the manager.
Answer:
The manager will give you a ticket.

9. The streets were thronged with spectators.
Answer:
Spectators thronged the streets.

10. We will be blamed by everyone.
Answer:
Everyone will blame us.

11. The trees were blown down by the wind.
Answer:
The wind blew down the trees.

12. The thieves were caught by the police.
Answer:
The police caught the thieves.

13. The letter was posted by Alice.
Answer:
Alice posted the letter.

14. We were received by the hostess.
Answer:
The hostess received us.

15. The snake was killed with a stick.
Answer:
They / somebody killed the snake with a stick.

16. The minister was welcomed by the people.
Answer:
The people welcomed the minister.

17. He was found guilty of murder.
Answer:
They found him guilty of murder.

18. This house was built by Afsar.
Answer:
After wilting this house.

(iii) Rewrite the following sentences in the passive voice.

1. She is writing a poem.
Answer:
A poem is being written by her.

2. The team has won the series.
Answer:
The series has been won by the team.

3. Can you break the door?
Answer:
Can the door be broken by you?

4. Will she sing a song?
Answer:
Will a song be sung by her?

5. Is he speaking English?
Answer:
Is English being spoken by him?

6. Are you eating a banana?
Answer:
Is a banana being eaten by you?

7. Why are you washing the car?
Answer:
Why is the car being washed by you ?

8. When will he give the money?
Answer:
When will the money be given by him?

9. Where will he meet you?
Answer:
Where will you be met by him?

10. How do you make a cake?
Answer:
How is a cake made by you?

11. Who did you tell the story?
Answer:
To whom was the story told by you?

12. Lighting struck him.
Answer:
He was struck by lighting.

13. The language that he used quite shocked me.
Answer:
I was quite shocked by the language that he used.

14. We were both filled with horror by the sight of that event.
Answer:
The sight of the event filled both of us with horror.

15. I was much hurt by his voice and manner.
Answer:
His voice and manner hurt me much.

16. They were welcomed back by the city on their return.
Answer:
The city welcomed back them on their return.

17. He was punished by the master for speaking in the class. ‘
Answer:
The master punished him for speaking in class.

18. Many objections were raised by us to the plan that was proposed by him.
Answer:
He proposed many plans that we raised many objections.

19. Will he not be persuaded to work harder by a sense of duty?
Answer:
Will a sense of duty not persuade him to work?

20. I was called upon by the meeting to give my reasons.
Answer:
The meeting called upon me to give my reasons.

21. He was known by me by his voice when I was spoken to by him in the dark.
Answer:
I knew him by his voice when he spoke to me in the dark.

22. His return was not expected by us.
Answer:
He did not expect his return.

(iv) Complete these sentences with the verbs in brackets. Use the present simple passive.

I. E-mails _________ and received by most internet users. (send)
Answer:
are sent

2. Information about goods and services _________ (find)
Answer:
is found

3. Go&ls and services _________ in e-shops. (buy)
Answer:
are bought

4. Online newspapers and magazines _________ mostly by adult users. (read)
Answer:
are read

5. The internet _________ for social networking, especially by young people. (use)
Answer:
is used

6. relephone and video calls _________ (make)
Answer:
are made

7. Videos and films _________ (watch)
Answer:
are watched

8. Listening and music streaming activities _________ (cany out)
Answer:
are carried out

9. Hotel accommodation _________ by travellers. (search for)
Answer:
is searched for

10. Financial transactions through internet banking _________ (do)
Answer:
are done

Students can Download Maths Chapter 3 Geometry Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.2

Miscellaneous Practice Problems

Question 1.
The sides of a triangle are 1.2 cm, 3.5 cm and 3.7cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
The sides of a triangle are a = 1.2 cm; b = 3.5 cm; c = 3.7 cm
By Pythagoras theorem,
c² = a² + b²
a² + b² = 1.2² + 3
5² – 1.44 +12.25 = 13.69
c² = 3.7² = 13.69
(1) = (2)
Yes, it is a right angled triangle. The hypotenuse c = 3.7 cm.

Question 2.
Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Why?
Solution:

Take the sides of a right angled triangle ∆ ABC as
a = 7 inches
b = 25 inches
c = ?
By Pythagoras theorem,
b² = a² + c²
25 = 7² + c²
c² = 25² – 7² = 625 – 49 = 576
∴ c² = 24²
⇒ c = 24 inches
∴ Width of TV cabinet is 20 inches which is lesser than the width of the screen ie.24 inches. The TV will not fit into the cabinet.

Question 3.
Find the length of the support cable required to support the tower with the floor.

Solution:
From the figure, by Pythagoras theorem,
x² = 20² + 15² = 400 + 225 = 625
x² = 25² ⇒ x = 25ft.
∴ The length of the support cable required to support the tower with the floor is 25 ft.

Question 4.
A ramp is constructed in a hospital as shown. Find the length of the ramp.

Solution:
Take a = 7 ft ; b = 24 ft.
By Pythagoras theorem,
l² = c² + b²
= 7² + 24² = 49 + 576
l = 25ft

Question 5.
In the figure, find MT and AH

Solution:
By Pythagoras theorem,
MT² = MH² + HT²
= 60² + 80² = 3600 + 6400
= 10000 = 100²
∴ MT = 100
By the altitude – on – hypotenuse theorem,
MH² = MA x MT
60² = MA x 100
MA = \(\frac{3600}{100}\) = 36
∴ In ∆ MAH, by Pythagoras theorem,
AH² = MH² – MA²
= 60² – 36² = 3600 – 1296 = 2304 = 48²
∴ AH = 48
∴ Ans MT = 100
AH = 48

Challenging Problems

Question 6.
Mayan travelled 28 km due north and then 21 km due east. What is the least distance that he could have travelled from his starting point?
Solution:
From the figure AC is to be found.

By using Pythagoras theorem,
AC² = AB² + BC²
= 28² + 21² = 784 + 441 = 1225 = 352
∴ AC = 35 km
This is the least distance that he could have travelled from his starting point.

Question 7.
If ∆ APK is an isosceles right angled triangle, right angled at K. Prove that AP² = 2AK².
Solution:
From the figure, ∆ APK is a right triangle.

By using Pythagoras theorem,
AP² = AK² + PK²
= AK² + AK² (since it is an isosceles A)
= 2AK²
Hence it is proved.

Question 8.
The diagonals of the rhombus is 12 cm and 16 cm. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).
Solution:

Here AO = CO = 8 cm
BO = DO = 6 cm
(∴ the diagonals of rhombus bisect each other at right angles)
∴ In ∆ AOB,
AB² = AO² + OB² = 8² + 6² = 64 + 36
= 100 = 10²
∴ AB = 10
Since it is a rhombus, all the four sides are equal.
AB = BC = CD = DA
∴ Its Perimeter = 10 + 10 + 10 + 10 = 40 cm

Question 9.
In the figure, find AR.

Solution:
∆ AFI, ∆ FRI are right triangles.
By Pythagoras theorem,
AF² = AT² – FT² = 25² – 15²
= 625 – 225 = 400 = 20²
∴ AF = 20 ft ….(1)
FR² = RI² – FI² = 17² – 15² = 289 – 225 = 64 = 8²
FR = 8 ft.
∴ AR = AF + FR = 20 + 8 = 28 ft.

Question 10.
∆ ABC is a right angled triangle in which ∠A = 90° and AM ⊥ BC. Prove that AM = \(\frac{ABxAC}{BC}\). Also if AB = 30 cm and AC = 40 cm, find AM.

Solution:
Given: ∆ ABC is a right angled triangle.
∠A = 90, AM ⊥ BC
To Prove: AM = \(\frac{ABxAC}{BC}\)
Proof:
In the given figure, Area of the triangle A ABC [taking AB as base and AC as height]
= \(\frac{1}{2}\) x AB x AC ….(1)
And also, Area of the triangle ∆ ABC [taking BC as base and AM as height]
= \(\frac{1}{2}\) x BC x AM ……(2)
Since (1) & (2) are the same, we can equate (2) and (1)
i.e. \(\frac{1}{2}\) x BC x AM = \(\frac{1}{2}\) x AB x AC
AM = \(\frac{ABxAC}{BC}\)
Hence proved. If AB = 30 cm, AC = 40 cm
By Pythagoras theorem, BC² = AB² + AC² = 30² + 40² = 900 + 1600 = 2500 = 50²
∴ BC = 50
Using the altitude – on – hypotenuse theorem

Here AC² = MC x BC
40² = MC x 50
∴ BM = BC – MC = 50 – 32 = 18 cm
AM² = BM x MC = 18 x 32 = 576 = 24²
∴ AM = 24 cm

You can Download Samacheer Kalvi 9th English Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th English Grammar Determiners

Observe the nouns in the following sentences and words before them.

• An apple is a healthy fruit.
• Two cats have drunk a bowl of milk,
• My father has many cars.

Determiners are the words that introduce a noun and provide some information about it (but do not describe it). .
Determiners are followed by a noun.

• The ball
• Five cats
• His son
• Some students

Types of Determiners

• The quantifiers all, any, enough, less, a lot of, more, most, no, none of, some etc., are used with both countable and uncountable nouns.
• The quantifiers both, each, either, fewer, neither etc., are used only with countable nouns.

ARTICLES

Articles are bifurcated into
(i) Definite Article (the) and
(ii) indefinite articles (a, an).

(i) DEFINITE ARTICLE : When to use “THE”

❖ General Rules
Use the to refer to something which has already been mentioned.

Examples :
On Monday, an unarmed man stole 10,000 from the bank. The thief hasn’t been caught yet.
I was walking past Benny’s Bakery when I decided to go into the bakery to get some bread.

Use the when you assume there is just one of something in that place, even if it has not been mentioned before.

Examples :
We went on a walk in the forest yesterday.
Where is the bathroom?
Turn left and go to number 45. Our house is near the restaurant.
My father enjoyed the book you gave him.

(ii) INDEFINITE ARTICLES : When to use “a” and “an”.

In English, the two indefinite articles are a and an.
Like other articles, indefinite articles are invariable.

You use one or the other, depending on the first letter of the word following the article, for pronunciation reasons. Use a when the next word starts with a consonant, or before words starting in u and eu when they sound like you. Use an when the next word starts with a vowel (a, e, i, o, u) or with a mute h.

Examples :
a boy, an apple, a car, a helicopter, an elephant, a big elephant, an itchy sweater, an ugly duck, a European, a university, a unit, an hour, an honour.

The indefinite article is used to refer to something for the first time or to refer to a particular member of a group or class. Some use cases and examples are given below.

1. Would you like a drink?
2. I’ve finally got a good job.
3. An elephant and a mouse fell into the well.

Example :

1. Mary is training to be an engineer.
2. John is an Englishman
3. I’d like an orange and two lemons please.

Demonstratives
Demonstratives show where an object, event, or person is in relation to the speaker. They can refer to a physical or a psychological closeness or distance. When talking about events, the near demonstratives are often used to refer to the present while the far demonstratives often refer to the past.

Examples:

Possessive Adjectives (Determiners)

Possessive adjectives are not pronouns, but rather determiners. It is useful to learn them at the same time as pronouns, however, because they are similar in form to the possessive pronouns. Possessive adjectives function as adjectives, so they appear before the noun they modify. They do not replace a noun as pronouns do.

Examples :
Did mother find my shoes?
The teacher wants to see your homework.
Samantha will fix her bike tomorrow.
The cat broke its leg.
This is our house.
Where is their school?

Possessive Pronouns

Possessive pronouns replace possessive nouns as either the subject or the object of a clause. Because the noun being replaced doesn’t appear in the sentence, it must be clear from the context.

Examples :
This bag is mine.
Yours is not blue.
That bag looks like his.
These shoes are not hers.
That car is ours.
Theirs is parked in the garage.

Reflexive & Intensive Pronouns

Reflexive and intensive pronouns are the same set of words but they have different functions in a sentence.

Reflexive pronouns refer back to the subject of the clause because the subject of the action is also the direct or indirect object. Only certain types of verbs can be reflexive. You cannot remove a reflexive pronoun from a sentence because the remaining sentence would be grammatically incorrect.

Examples :
I told myself to calm down.
You cut yourself on this nail?
He hurt himself on the stairs.
She found herself in a dangerous part of town.
The cat threw itself under my car!
We blame ourselves for the fire.
The children can take care of themselves.

Quantifiers
Quantifiers are adjectives and adjectival phrases that give approximate or specific answers to the questions “How much?” and “How many?” The pages in this section will teach you more about the different quantifiers in English and how they are used.

• Numbers in English: ordinal, cardinal, and percentages
• To answer the questions How much? and How many? certain quantifiers can be used with countable nouns (friends, cups, people), others with uncountable nouns (sugar, tea, money) and still others with all types of nouns.

Examples :
Would you like some tea and a few cookies?
I always put a little milk and some carrots in my soup.
He has several apples. I don’t have any fruit at all.
She has plenty of clothes for the winter.
I received a large amount of feedback from my survey.

Cardinal Numbers
A Cardinal Number says how many of something there are, such as one, two, three, four, five, etc.
A Cardinal Number answers the question “How Many?”

Examples :
There are twenty children in the class.
He purchased two books.
There are five plates on the table.
Sumitha is fourteen years old.

Ordinal Numbers

An Ordinal Number tells us the position of something in a list, such as first, second, third, fourth, fifth and so on.
Examples :
Let us begin with the first chapter.
We live in the fourth house on the right.
I was bom on the fifteenth of January.
He got the second prize.

Exercises

1. Fill in the blanks with appropriate determiners,
1. Our garden looks awful this summer. There are too ____________ weeds
Answers:
many

2. There aren’t ____________ flowering plants in our garden.
Answers:
many

3. How ____________ pages did you read?
Answers:
many

4. They say ____________ knowledge is a dangerous thing.
Answers:
a little

5. I am having ____________ trouble passing my driving exam.
Answers:
a lot of

6. ____________ people can afford a home these days.
Answers:
few

7. You have ____________ patience
Answers:
little

8. She earns ____________ money than i do
Answers:
less

9. ____________ of the information proved to be outdated.
Answers:
some

10. I didn’t use ____________ fertilizer last spring.
Answers:
much

Choose the correct determiner.
1. There are chairs in this room than in the other room, (more / much)
Answer:
more

2. The assistant didn’t give information, (much / many)
Answer:
much

3. After the negotiations, they made changes in their proposal, (little/few)
Answer:
few

4. mosquitoes appeared after the rain. (A large amount of / A great number of)
Answer:
A great number of

5. Toned Milk has calories than Full Cream Milk, (less / fewer)
Answer:
fewer

6. students taking TOEFL is increasing. (The amount of / The number of)
Answer:
The number of

7. The case had to be reconsidered with new evidence, (these / this)
Answer:
this

8. I like to eat food, (many / a lot of)
Answer:
A lot of

9. She ate French–fries than usual, (fewer / less)
Answer:
fewer

10. He wants to make as money as possible, (much / many)
Answer:
much

11. Vini invited a large of people to the party, (amount/number)
Answer:
number

12. Raji will drink an endless of milk if you let her. (amount / number)
Answer:
Amount

Fill in the blanks with a, an, the, or leave the blank.
1. Han is earning Rupees 1000 _____ hour at the Food court.
Answer:
an

2. Janani makes it ______ habit to buy clothes on sale.
Answer:
a

3. To tell ______ truth, a bank savings account may not be the best place for your money.
Answer:
the

4. Hemá showed initiative when she decided to start a business of her own.
Answer:
an

5. Losing as little as ______ quart of blood can result in shock and unconsciousness.
Answer:
a

6. Over last 20 years, more than 3 million people have visited ______ theme park.
Answer:
the, the

7. Major changes have taken place in ______ Educational services.
Answer:
the

8. Dr. Richards predicts ______ extinction of the whooping crane.
Answer:
th

9. Taking a hot bath is ______ good way to relax.
Answer:
a

You can Download Samacheer Kalvi 9th English Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th English Grammar Reported Speech (Direct to Indirect Speech)

Quoting the exact words of the speaker is called “The Direct Speech”.

David said, “I am writing a letter now”.

Reporting of what a speaker said without quoting his exact words is called ‘Indirect Speech’, (also called Reported Speech).

David said that he was writing a letter then.

The following table suggests the words that change during a transformation.

If the reporting verb is in the Present or Future tense (e.g., say, will say) there is no change in the tense of the verb in the Indirect speech.

Anto says, “I eat a mango”. (Direct Speech)
Anto says that he eats a mango”. (Indirect Speech)

If Reporting Verb is in the Past Tense, the tense of the verbs in the reported speech or Indirect Speech must be generally changed.
1. Present Tense in the Direct becomes past tense.
Janis said, “I write a letter”. (D.S)
Janis said that she wrote a letter. (I.S)

2. Past Tense in the Direct becomes past perfect or remains unchanged.
Arun said, “I bought a pen yesterday”. (D.S)
Arun said that he had bought a pen the previous day. (I.S)

3. Present Continuous in the Direct becomes past continuous.
Juber said, “I am going to mosque”. (D.S)
Juber said that he was going to mosque. (I.S)

4. Past Continuous in the Direct becomes past perfect continuous. *
Nelson said, “I was playing cricket”. (D.S) > M
Nelson said that he had been playing cricket. (I.S)

5. Present Perfect in the Direct becomes past perfect.
Kamal said, “I have done my home work”. (D.S)
Kamal said that he had done his home work. (I.S)

6. Present Perfect Continuous in the Direct becomes past perfect continuous.
He said, “I have been reading a novel”. (D.S)
He said that he had been reading a novel. (I.S)

7. ‘Will’ and ‘Shall’ are changed to ‘would’.
He said, “I will go to Trichy tomorrow”. (D.S)
He said that he would go to Trichy the next day. (I.S)

8. may – might
can – could
must – had to (or) must
Nisha said, “I must go now”. (D.S)

Nisha said that she must go then, (or) Nisha said that she had to go then. (I.S)

Hint: Past Perfect and Past Perfect Continuous in the Direct Speech do not take any change *, in their tenses.

Exception to the above rule:
If the direct speech contains the Universal Truth, the tense of the direct speech remains unchanged even if the reporting verb is in the past.

The teacher said, “The sun rises in the East”. (D.S)
The teacher said that the sun rises in the East. (I.S)

Statement (Or) Assertive Sentence

Rules:
Remove the quotation marks in the statement
Use the conjunction ‘that’
Change the reporting verb ‘say to’ into ‘tell’
Change the reporting verb ‘said to’ into ‘told’ ‘

Note: He said that (correct)
He told me that (correct)
He told that (Incorrect)

1. “I will work hard to get first-class” said Nasser (D.S.)
Nasser said that he would work hard to get first class. (I.S.)

2. “You can do this work” said Sundar to Kumar (D.S.)
Sundar told Kumar that he could do that work. (I.S.)

3. He says, “I am glad to be here this evening”(D.S.)
He says that he is glad to be there that evening. (I.S.)

4. “I‘m going to the library now” said David (D.S.)
David said that he was going to the library then. (I.S.)

Imperative Sentence (Order Or Request)

Rules:
Remove the quotation marks in an imperative sentence.
Use ‘to’ if it is an affirmative sentence, (without don’t)
Use ‘not to’ if the sentence is negative, (with don’t)
Don’t use ‘that’
Omit the word ‘please’. Use the word ‘request’ instead of ‘say’.

If the direct speech contains a request or a command, the reporting verb (say, said) change to tell, request, order, command etc. in its correct tense.

1. “Don’t talk in the class” said the teacher to the boys. (D.S.)
The teacher advised the boys not to talk in the class. (I.S.)

2. “Please give me something to eat. I am hungry” the old man said to them. (D.S.)
The old man requested them to give him something to eat and said that he was hungry (I.S.)

3. “Be careful” said he to her. (D.S.)
He warned her to be careful. (I.S.)

4., “Bring me a cup of tea” said Danush to Andrea. (D.S.)
Danush asked Andrea to bring him a cup of tea. (I.S.)

Interrogative Sentence (Questions)

Rules:
Remove the quotation marks and the question mark in the interrogative sentence.
Use ‘if’ or ‘whether’ if the sentence inside the quotation marks begins with a helping verb (Auxiliary verb).
Use the given interrogative word (what, when, where, why, who, whom, whose, which, now etc.) if it does not begin with the helping verb.
Don’t use ‘that’
Changing the reporting verb (say, said) into ‘ask’ or ‘enquire’ in its correct tense.
Omit helping verb like ‘do, does, did’. But don’t omit them when they are with ‘not’.

1. “Won’t you help me to carry this box?” said I to my friend. (D.S.)
I asked my friend if he would not help me to carry that box. (I.S.)

2. Mohan said to Stalin, “Why did not you attend the meeting yesterday”? (D.S.)
Mohan asked Stalin why he had not attended the meeting the day before. (I.S.)

3. “How often do you go to the theatre?” said David to John. (D.S.)
David asked John how often he went to the theatre. (I.S.)

4. Mohamed said to Sultan, “Do you like mangoes?” (D.S.)
Mohamed asked Sultan if he liked mangoes. (I.S.)

Exclamatory Sentence

Rules:
Remove the quotation marks and exclamatory mark.
Change the exclamatory sentence into Statement or Assertive.
Use the conjunction ‘that’.
Omit the interjections such as Oh, O, Alas, how, what, hurrah.
Add the word ‘very’ to the adjective or adverb if necessary.
If the verb is not given, use ‘Be’ form verb (is, was, are, were, am) in its correct tense according to the subject.
Change the reporting verb (say, said) to ‘exclaim joyfully’
Use ‘exclaim’ sorrowfully for sorrowful incidents.

1. “O, what a beautiful flower that is!” said she. (D.S.)
She exclaimed joyfully that that was a very beautiful flower. (I.S.)

2. “What a horrible sight!” we all exclaimed. (D.S.)
We all exclaimed that it was a very horrible sight. (I.S.)

3. “Alas! I have broken my brother’s watch” said he.
He exclaimed sorrowfully that he had broken his brother’s watch. (I.S.)

4. “How beautiful she is!” said Banu. (D.S.)
Banu exclaimed joyfully that she was very beautiful. (I.S.)

Direct Speech Into Indirect Speech

I. Look at the following examples of Direct and Indirect Speech:

1. He said, “Danny will be in London on Tuesday.”
He said that Danny would be in London on Tuesday.

2. “I never eat meat”, he explained.
He explained that he never ate meat.

3. He said, “I wish I knew.”
He said that he wished he knew.

4. She says, “I shall be there.”
She says that she would be there.

5. He said, “She is coming this week.”
He said that she was coming that week.

6. He said, “I bought this pearl for my mother.”
He said that he had bought that pearl for his mother.

7. He said, “Where is she going?”.
He asked where she was going.

8. He said, “Jaya, when is the next bus.”
He asked Jaya when the next bus was.

9. “Is anyone there?” she asked.
She asked if anyone was there.

10. The mother said, “Lie down, David.”
The mother asked David to lie down.

11. He said, “Don’t move, boys.”
He asked the boys not to move.

12. He said, “Please–say nothing about this.”
He asked her to say nothing about that.

Reported Speech–Statement–Rules

Whatever may be the tense of the Reporting Sentence, if the Reported Sentence tells a universal fact, no change is made in the tense of the Reported Sentence.

Example No. 1:
The mother is saying to the child, “The third day of the week is Tuesday. ”.
Step 1 : The Reported Sentence is: The third day of the week is Tuesday.
Step 2 : It is a Statement and a universal fact.
Step 3 : So, the conjunction word is — “that”.
Step 4 : ‘is saying to’ changes into ‘is telling’.
Step 5 : No change of pronoun.
Step 6 : It is a universal fact. So, no change of tense is necessary.
Step 7 : No change of extension.
‘Now, the Indirect Speech is:
The mother is telling the child that the third day of the week is Tuesday.

Example No. 2:
The History teacher says, “Megellan was the first navigator to come around the world. ”.

Step 1 : The Reported Sentence is: “Megellan was the first navigator to come around the world.”
Step 2 : It is a statement.
Step 3 : The conjunction word is — “that”.
Step 4 : ‘Says’ does not change. Use it as it is.
Step 5 : There are no pronouns to get changed.
Step 6 : No change of tense is made.
Step 7 : No extensive word to get changed.

Now, the Indirect Speech is:
The History teacher says that Megellan was the first navigator to come around the world.

Example No. 3:
You know the two types of Interrogative Sentences:
Inverted questions requiring ‘Yes’ or ‘No’ answers — 1st type.
Questions that begin with interrogative words — 2nd type
At first we shall deal with the First type:

Direct Speech: The boy said to the fruit–seller, “Are all these mangoes sweet?”
Step 1 : Identify the Reported Sentence.
Step 2 Know what kind of sentence the Reported Sentence is.
Step 3 Look for the correct Conjunction, (The Conjunction of the First type is “If or Whether”)
Step 4 Change of ‘said to’ — Since it is an interrogative sentence ‘said to ’ changes into ‘asked’.
Step 5 Step Look for the change of pronouns.
Step 6 Look for the change of tenses.

The Reporting Sentence is in past tense. The Reported Sentences is in present tense. So, the Reported Sentence should be changed into past tense, corresponding to the tense, of the Reporting Sentence.

The verb is ‘are’ — Its past tense is ‘were’.

Step 7 : Look for the change of extension words.
‘These’ changes into ‘those’.

The Indirect Speech is : The boy asked the fruit–seller if all those mangoes were sweet.

Example No 4:
Direct Speech: The grandfather said fo his grandsons, “Did you not like my story yesterday?”
SIeR 1 : The Reported Sentence is: “Did you not like my story yesterday?”
Step 2 : ‘It is an interrogative sentence. It is of the First type.
Step 3 : So its conjunction word is: ¡for Whether.
Step 4 : So ‘said to’ changes into ‘asked’.
Step 5 : Look for the pronouns.

(i) The first one is: ‘You’ (subject)
‘You’ — refers to grandsons. They are in the third person plural number.
So the third person of ‘You’ (subject, plural)
It is — ‘they’. ‘You’ changes into ‘they’.
‘You’ — grandsons — ‘They’
‘You’–they.

(ii) The next one is ‘My’.
‘My’ — refers to ‘the grandfather’ — in the third person.
So, take the third person of ‘My’
‘My’ changes into ‘His’.

Step 6 : Look for the change of tenses.
Step 7 : Extensive word ‘Yesterday’ changes into ‘the day before’. Now, the Indirect Speech is-

The grandfather asked his grandsons if they had not liked his story the day before.

Exercises
Transform the following questions into indirect form :

1. Radha asked, “Where are you going?”
Answer:
Radha asked (me) where I was going.

2. Mani asked, “Where are you going to spend the holiday?”
Answer:
Mani asked (me) where I was going to spend the holiday.

3. Jenifer said, “What will you do when you leave school?”
Answer:
Jennifer asked (me) what I would do when I left school.

4. The nurse asked the doctor, “How did you know my name?”
Answer:
The nurse wanted to know how the doctor had known her name.

5. The clerk said, “Do you have an appointment?”
Answer:
The clerk asked (me) whether/if I had an appointment.

6. Bharath said to his wife, “Have you seen my car keys?”
Answer:
Bharath asked his wife whether she had seen his car keys.

7. Rekha’s brother asked her, “Why didn’t you call me?”
Answer:
Rekha’s brother wanted to know why she hadn’t called him.

8. Rayan said to James, “Will you carry my briefcase for me please?”
Answer:
Rayan asked James to carry his briefcase.

9. Charles asked the receptionist, “When can I see the doctor?”
Answer:
Charles asked the receptionist when he could see the doctor.

10. Devi asked, “Where do you live?” ‘
Answer:
Devi asked (me) where I lived.

Change the following sentences into Indirect Speech:

1. He said, “I will do it now.”
Answer:
He said that he would do it then.

2. He says, “Honesty is the best policy.”
Answer:
He says that honesty is the best policy.

3. Ramesh says, “I have written a letter.”
Answer:
Ramesh says that he has written a letter.

4. She said, “Mahesh will be reading a book.”
Answer:
She said that Mahesh would be reading a book.

5. She said, “Where is your father?”
Answer:
She inquired where his father was.

6. He said to me, “Please take your book.”
Answer:
He requested me to take my book.

7. The Principal said to the peon, “Let this boy go out.”
Answer:
The Principal ordered the peon to let that boy go out.

8. He said to me, “May you live long!”
Answer:
He prayed that I might live long.

9. She said, “Goodbye friends!”
Answer:
She bade goodbye to her friends.

10. The student said, “Alas! I wasted my time last year.”
Answer:
The student regretted that he had wasted his time the previous year.

(iii) Change the following sentences into direct speech.

1. She asked me who was the best player..
Answer:
“Who is the best player here?”, she asked.

2. He asked me if I was going home with him.
Answer:
“Are you coming home with me?”, he asked.

3. She asked me what I wanted.
Answer:
She said to me, “What do you want?”.

4. He asked me what the matter was.
Answer:
He said to me, “What is the matter?”.

5. She enquired which her seat was.
Answer:
She said, “Which is my seat?”.

6. I asked whether he did not know the way home.
Answer:
I said, “Don’t you know the way home?”.

7. Aladdin asked the magician what he had done to deserve so severe a blow.
Answer:
Aladdin said to the magician, “What have I done to deserve so severe a blow?”

8. Ulysses asked the little bird whether it had anything to tell him.
Answer:
“Have you anything to tell me, little bird?”, asked Ulysses.

9. He asked me what my name was.
Answer:
He asked me, “What is your name?”.

10. The kind man asked the boy what he could do for him.
Answer:
“What can I do for you?”, the kind man asked the boy.

Students can Download Maths Chapter 3 Geometry Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.1

Question 1.
Fill in the blanks:

Question (i)
If in a ∆ PQR, PR² = PQ² + QR², then the right angle of ∆ PQR is at the vertex ………
Answer:
Q
Hint:

We have also given Lines and Angles concepts and problems for free of cost on our website.

Question:
(ii) If ‘l’ and ‘m’ are the legs and is the hypotenuse of a right angled triangle then, l² = ……….
Answer:
n² – m²
Hint:

Question (iii)
If the sides of a triangle are in the ratio 5:12:13 then, it is ………
Answer:
a right angled triangle.
13² = 169
5² = 25
12² = 144
169 = 25 + 144
132 = 5² + 12²
By Pythagoras theorem, In a right triangle, square of the hypotenuse is equal to the sum of the squares of other two sides.

Question (iv)
If a perpendicular is drawn to the hypotenuse of a right angled triangle, then each of the three pairs of triangles formed are …………
Answer:
Similar.

Question (v)
In the figure BE² = TE x ………

Answer:
EN

Question 2.
Say True or False.

Question (i)
8, 15, 17 is a Pythagorean triplet.
Answer:
True
Hint:
17² = 289
15² = 225
8² = 64
64 + 225 = 289 ⇒ 17² = 15² + 8²

Question (ii)
In a right angled triangle, the hypotenuse is the greatest side.
Answer:
False
Hint:

Question (iii)
One of the legs of a right angled triangle PQR having ∠R = 90° is PQ.
Answer:
False
Hint:

In ∆ PQR, QR and RP are legs and PQ is the hypotenuse

Question (iv)
The hypotenuse of a right angled triangle whose sides are 9 and 40 is 49.
Answer:
False
Hint:

49² = 2401
9² = 81
40² = 1600
40² + 9² = 1600 + 81 + 1681
49² = 2401
2401 ≠ 1681

Question (v)
Pythagoras theorem is true for all types of triangles.
Answer:
False
Hint:
Pyhtagoras theorem is true for only right angled triangles.

Question 3.
Check whether given sides are the sides of right – angled triangles, using Pythagoras theorem,

1. 8, 15, 17
2. 12, 13, 15
3. 30, 40, 50
4. 9, 40, 41
5. 24, 45, 51

Solution:
1. 8, 15, 17
Take a = 8,
b = 15 and
c = 17
Now a² + b² = 8² + 15² = 64 + 225 = 289
17² = 289 = c²
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
Answer:
yes.

2. 12, 13, 15
Take a = 12
b = 13 and
c = 15
Now a² + b² = 12² + 13² = 144 + 169 = 313
15² = 225 ≠ 313
By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.
Answer:
No.

3. 30, 40, 50
Take a = 30
b = 40 and
c = 50
Now a² + b² = 30² + 40² = 900 + 1600 = 2500
c² = 50² = 2500
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
Answer:
yes.

4. 9, 40, 41
Take a = 9
b = 40 and
c = 41
Now a² + b² = 9² + 40² = 81 + 1600= 1681
c² = 41² = 1681
∴ a² + b² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
Answer:
Yes.

5. 24, 45, 51
Take a = 24
b = 45 and
c = 51 Now
a² + b² = 24² + 45² = 576 + 2025 = 2601
c² = 51² = 2601
a² + b² = c²
By the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.
Answer:
Yes.

Question 4.
Find the unknown side in the following triangles.

Solution:
(i) From ∆ ABC, by Pythagoras theorem

BC² = AB² + AC²
Take AB² + AC² = 9² + 40² = 81 + 1600 = 1681
BC² = AB² + AC² = 1681 = 41²
BC² = 41² ⇒ BC = 41
∴ x = 41

(ii) From ∆ PQR, by Pythagoras theorem,

PR² = PQ² + QR²
34² = y² + 30²
⇒ y² = 34² – 30²
= 1156 – 900
= 256 = 16²
y² = 16² ⇒ y = 16

(iii) From ∆ XYZ, by Pythagoras theorem,

YZ² = XY² + XZ²
⇒ XY² = YZ² – XZ²
Z² = 39² – 36²
= 1521 – 1296 = 225 = 15²
Z² = 15²
⇒ Z = 15

Question 5.
An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height.
Solution:
In an isosceles triangle the altitude dives its base into two equal parts. Now in the figure, ∆ ABC is an isosceles triangle with AD as its height.

In the figure, AD is the altitude and Δ ABD is a right triangle.
By Pythagoras theorem,
AB² = AD² + BD²
⇒ AD² = AB² – BD²
= 13² – 12² = 169 – 144 = 25
AD² = 25 = ²
Height: AD = 5 cm

Question 6.
In the figure, find PR and QR.

Solution:
In the figure ∆ PQR is a right triangle.
By Pythagoras theorem,
PR² = PQ² + QR²
(x + 1)² = 7² + x²
x² +2 × x × 1 + 1² = 49+ x²
2x + 1 = 49
2x = 49 – 1 = 48
x = \(\frac{48}{2}\) = 24
∴ PR = x + 1 = 24 + 1 = 25
QR = x = 24

Question 7.
The length and breadth of the screen of an LED – TV are 24 inches and 18 inches. Find the length of its diagonal.

Solution:
The length and breadth of a LED TV form a right angled triangle with its diagonal.
Therefore by Pythagoras theorem,
AC² = AB² + BC²
= 24² + 18² = 576 + 324 = 900 = 30²
∴ AC = 30 ⇒ The length of the diagonal is 30 inches.

Question 8.
Find the distance between the helicopter and the ship.

Solution:
From the figure AS is the distance between the helicopter and the ship.
∆ APS is a right angled triangle, by Pythagoras theorem,
AS² = AP² + PS²
= 80² + 150²
= 6400 + 22500 = 28900 = 170²
∴ The distance between the helicopter and the ship is 170 m

Question 9.
From the figure, 1. If TA = 3 cm and OT = 6 cm, find TG.

Solution:
1. From the figure, if, TA = 3 cm, OT = 6 cm

By Pythagoras theorem,
OA² = OT² – TA² = 6² – 3²
i.e. h² = 36 – 9 = 27 cm.
Now, by altitude – on – hypotenuse theorem
h² = xy
27 = x × 3
x = \(\frac{27}{3}\) = 9 cm
TG = x + 3 = 9 + 3 = 12 cm

Question 10.
If RQ = 15 cm and RP = 20 cm, find PQ, PS and SQ.

Solution:

RQ = 15cm
RP = 20 cm and ∆ PQR is a right angled triangle
By Pythagoras theorem, p 20 cm
PQ² = PR² + RQ² = 20² + 15²
= 400 + 225 = 625 = 25²
∴ PQ = 25 cm
Now by altitude – on – hypotenuse theorem,
RQ² = q x r
15² = q x 25
q = \(\frac{225}{25}\) = 9 cm ⇒ SQ = 9 cm
PR² = P x r
20² = P x 25
P = \(\frac{400}{25}\) = 16 cm ⇒ PS = 16 cm
Answer:

1. PQ = 25 cm
2. PS = 16 cm
3. SQ = 9 cm

Objective Type Questions

Question 11.
If ∆ GUT is isosceles and right angled, then ∠TUG is …………

(a) 30°
(b) 40°
(c) 45°
(d) 55°
Answer:
(c) 45°
Hint:
∠U ∠T = 45° (∆ GUT is an isosceles given)
∴ ∠TUG = 45°

Question 12.
The hypotenuse of a right angled triangle of sides 12 cm and 16 cm is ……….
(a) 28 cm
(b)20cm
(c) 24 cm
(d)21cm
Answer:
(b) 20 cm
Hint:
Side take a = 12 cm
b = 16 cm
The hypotenuse c² = a² + b² = 12² + 16^2
2 = 400 ⇒ c = 20 cm

Question 13.
The area of a rectangle of length 21 cm and diagonal 29 cm is ………. cm²
(a) 609
(b) 580
(c) 420
(d) 210
Answer:
(c) 420
Hint:

Length = 21 cm
Diagonal = 29 cm
By the converse of Pythagoras theorem,
AB² + BC² = AC²
21² + x² = 29²
x² = 841 – 441 = 400 = 20²
x = 20 cm
Now area of the rectangle = length x breadth.
i.e. AB x BC = 21 cm x 20 cm = 420 cm²

Question 14.
if the square of the hypotenuse of an isosceles right triangle is 50 cm2, the length of each side is ………..
(c) 10 cm
(a) 25 cm
(b) 5 cm
(c) 10 cm
(d) 20 cm
Answer:
(b) 5 cm
Hint:
By Pythagoras theorem
c² = a² + b²
In an isosceles triangle, a = b
c² = a² + a² = 2a²
⇒ 2a² = 50
a² = 25 ⇒ a = 5cm
∴ The length of each sides a = 5cm, b = 5 cm.

Question 15.
The sides of a right angled triangle are in the ratio 5 : 12 : 13 and its perimeter is 120 units then, the sides are .
(a) 25, 36, 59
(b) 10, 24, 26
(c) 36, 39, 45
(d) 20, 48, 52
Answer:
(d) 20, 48, 52
Hint:
The sides of a right angled triangle are in the ratio 5 : 12 : 13
Take the three sides as 5a, 12a, 13a
Its perimeter is 5a + 12a + 13a = 30a
It is given that 30a = 120 units
a = 4 units
∴ The sides 5a = 5 x 4 = 20 units
12a = 12 x 4 = 48 units
13a = 13 x 4 = 52 units