Laxmi

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.3

Miscellaneous Practice Problems

Question 1.
Choose the correct relationship between x and y for the given table.

(i) y = x + 4
(ii) y = x + 5
(iii) y = x + 6
(iv) y = x + 7
Answer:
(iii) y = x + 6

Question 2.
Find the triangular numbers from the Pascal’s Triangle and colour them.

Solution:
Triangular numbers are numbers the objects of which can be arranged in the form of equilateral triangle.
Example : 1, 3, 6, 10, 15,…
From Pascal’s Triangle, the triangular numbers are

Question 3.
Write the first five numbers in the third slanting row of the Pascal’s Triangle and find their squares. What do you infer?
Solution:

Numbers in the 3rd slanding row are 1, 3, 6, 10, 15, 21,….
The squares are 1², 3², 6², 10². 15², 21²,…. = 1, 9, 36, 100, 225, 441,…

From the above table we can conclude that the squares of the triangular numbers are the sum of cubes of natural numbers.

Challenge Problems

Question 4.
Tabulate and find the relationship between the variables (x and y) for the following patterns.

Solution:
(i) Let the number of steps be x and the number of shapes be y.
Tabulating the values of x and y

From the table
x = 1 ⇒ y = 1 = 12
x = 2 ⇒ y = 4 = 22
x = 3 ⇒ y = 9 = 32
x = 4 ⇒ y = 16 = 42
Hence the relationship between x and y is y = x2.

(ii) Let the number of steps be x and the number of shapes be y.
Tabulating the values of x and y

From the table x = 1 ⇒ y = 1 = 1
x = 2 ⇒ y = 2 + 1 = 3
x = 3 ⇒ y = 3 + 2 = 5
x = 4 ⇒ y = 4 + 3 = 7
x = 5 ⇒ y = 5 + 4 = 9
Hence the relationship between x and y is y = 2x-1.

Question 5.
Verify whether the following hexogonal shapes form a part of the Pascal’s Triangle.

Solution:
In Pascal’s Triangle product of the 3 alternate numbers given around the hexagon is equal to the product of remaining three numbers.

1 × 13 × 66 = 11 × 1 × 78 = 858
∴ It form a part of Pascal’s Triangle.

5 × 21 × 20 = 10 × 6 × 35 = 2100
∴ It form a part of Pascal’s Triangle

8 × 45 × 84 = 28 × 9 × 120 = 30240
∴ It form a part of Pascal’s Triangle

56 × 210 × 126 = 70 × 84 × 252 = 1481760
∴ It form a part of Pascal’s Triangle

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.2

Question 1.
Complete the Pascal’s Triangle.

Solution:

Question 2.
The following hexagonal shapes are taken from Pascal’s Triangle. Fill in the missing numbers.

Solution:

Question 3.
Complete the Pascal’s Triangle by taking the numbers 1,2,6,20 as line of symmetry.

Solution:
Corresponding numbers are equal about the line of symmetry.

Objective Type Questions

Question 1.
The elements along the sixth row of the Pascal’s Triangle is
(i) 1,5,10,5,1
(ii) 1,5,5,1
(iii) 1,5,5,10,5,5,1
(iv) 1,5,10,10,5,1
Answer:
(iv) 1,5,10,10,5,1

Question 2.
The difference between the consecutive terms of the fifth slanting row containing four elements of a Pascal’s Triangle is
(i) 3,6,10,…
(ii) 4,10,20,…
(iii) 1,4,10,…
(iv) 1,3,6,…
Answer:
(ii) 4,10,20,…

Question 3.
What is the sum of the elements of ninth row in the Pascal’s Triangle?
(i) 128
(ii) 254
(iii) 256
(iv) 126
Answer:
(iii) 256

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.1

Question 1.
Match the given patterns of shapes with the appropriate number pattern and its generalization.


Solution:
(i) (d)
(ii) (a)
(iii) (c)
(iv) (c)
(v) (b)

Objective Type Questions

Question 2.
Identify the correct relationship between x andy from the given table.

(i) y = 4x
(ii) y = x + 4
(iii) y = 4
(iv) y = 4 × 4
Answer:
(i) y = 4x

Question 3.
Identify the correct relationship between x and y from the given table.

(i) y = -2x
(ii) y = +2x
(iii) y = +3x
(iv) y = -3x
Answer:
(iv) y = -3x

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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:

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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