Prasanna

Students can Download Maths Chapter 2 Measurements Intext 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 1 Chapter 2 Measurements Intext Questions

Try this Page No. 35

Question 1.
\(\frac{22}{7}\) and 3.14 are rational numbers. Is ‘π’ a rational number? Why?
Solution:
\(\frac{22}{7}\) and 3.14 are rational numbers n has non-terminating and non -repeating decimal expansion. So it is not a rational number. It is an irrational number.

Try this Page No. 38

Question 1.
The given circular figure is divided into six equal parts. Can we call the parts as sectors? Why?

Solution:
No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle.
Here the boundaries are not radii.

Try these Page No. 38

Question 1.
Fill the central angle of the shaded sector (each circle is divided into equal sectors)

Try this Page No. 44

Question 1.
If the radius of a circle is doubled, what will the area of the new circle so formed?
Solution:
If r = 2r₁ ⇒ Area of the circle = πr² = π(2r₁)² = π4r₁² = 4πr₁²
Area = 4 × old area.

Try this Page No. 49

Question 1.
All the sides of a rhombus are equal. Is it a regular polygon?
Solution:
For a regular polygon all sides and all the angles must be equal. But in a rhombus all the
sides are equal. But all the angles are not equal
∴ It is not a regular polygon.

Try this Page No. 53

Question 1.
In the above example split the given mat as into two trapeziums and verify your answer.
Solution:
Area of the mat = Area of I trapezium + Area of II trapezium

∴ Cost per sq.feet = ₹ 20
Cost for 28 sq. feet = ₹ 20 × 28 = ₹ 560
∴ Total cost for the entire mat = ₹ 560
Both the answers are the same.

Try these Page No. 54

Question 1.
Show that the area of the unshaded regions in each of the squares of side ‘a’ units are the same in all the cases given below.

Solution:

Question 2.
If π = \(\frac{22}{7}\), show that the area of the unshaded part of a square of side ‘a’ units is approximately \(\frac{3}{7}\) a² sq. units and that of the shaded part is approximately \(\frac{4}{7}\) a² sq. units for the given figure.

Solution:

Try this Page No. 57

Question 1.
List out atleast three objects in each category which are in the shape of cube, cuboid,
cylinder, cone and sphere.
Solution:
(i) Cube – dice, building blocks, jewel box.
(ii) Cuboid – books, bricks, containers.
(iii) Cylinder – candles, electric tube, water pipe.
(iv) Cone – Funnel, cap, ice cream cone
(v) Sphere – ball, beads, lemon.

Try this Page No. 58

Question 1.
Tabulate the number of faces(F), vertices(V) and edges(E) for the following polyhedron. Also find F + V – E
Solution:

From the table F + V – E = 2 for all the solid shapes.

Try this Page No. 58

Question 1.
Find the area of the given nets.

Solution:
Area = 6 × Area of a square of side 6 cm
= 6 × (6 × 6) cm²
= 216 cm²
(ii) Area = Area of 2 rectangles of side (8 × 6) cm² + Area of 2 rectangles of side (8 × 4) cm² + Area of 2 rectangles of side (6 × 4) cm²
= (8 × 6) + (8 × 4) + (6 × 4)cm²
= 48 + 32 + 24 cm²
= 104 cm²

Students can Download Maths Chapter 3 Algebra Intext 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 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Recap Page No. 66 and 67

Question 1.
Write the numbers of terms in the following expressions.
(i) x + y + z – xyz
Solution:
4 terms

(ii) m²n²c
Solution:
1 term

(iii) a²b²c – ab²c² + a²bc² + 3abc
Solution:
4 terms

(iv) 8x² – 4xy + 7xy²
Solution:
3 terms

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
Question 1.
2x² – 5xy + 6y² + 7x – 10y + 9
Solution:
Numerical co efficient in 2x² is 2
Numerical co efficient in -5xy is -5
Numerical co efficient in 6y² is 6
Numerical co efficient in 7x is 7
Numerical co efficient in -10y is – 10
Numerical co-efficient in 9 is 9

Question 2.
\(\frac{x}{3}+\frac{2 y}{5}-x y+7\)
Solution:

Numerical co efficient in -xy is -1
Numerical co efficient in 7 is 7

Question 3.
Pick out the like terms from the following.

Solution:

Question 4.
Add : 2x, 6y, 9x – 2y
Solution:
2x + 6y + 9x – 2y = 2x + 9x + 6y – 2y = (2 + 9)x + (6 – 2)y = 11x + 4y

Question 5.
Simplify : (5x³ y³ – 3x² y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
Solution:
(5x³y³ – 3x²y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
= 5x³y³ + x³y³ – 3x²y² + 2x²y² + xy + 2xy + 7 – 5
= (5 + 1)x³y³ + (-3 + 2)x²y² +(1 +2)xy + 2
= 6x³y³ – x²y² + 3xy + 2

Question 6.
The sides of a triangle are 2x – 5y + 9, 3y + 6x – 7 and -4x + y +10 . Find perimeter of the triangle.
Solution:
Perimeter of the triangle = Sum of three sides
= (2x – 5y + 9) + (3y + 6x – 7) + (-4x + y + 10)
= 2x – 5y + 9 + 3y + 6x – 7 – 4x + y + 10
= 2x + 6x – 4x – 5y + 3y + y + 9 – 7 + 10
= (2 + 6 – 4)x + (-5 + 3 + 1)y + (9 – 7 + 10)
= 4x – y + 12
∴ Perimeter of the triangle = 4x – y + 12 units.

Question 7.
Subtract -2mn from 6mn.
Solution:
6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn

Question 8.
Subtract 6a² – 5ab + 3b² from 4a² – 3ab + b².
Solution:
(4a² – 3ab+ b²) – (6a²– 5ab + 3b²)
= (4a² – 6a²) + (- 3ab -(-5 ab)] + (b²– 3b²)
= (4 – 6) a² + [-3ab + (+ 5ab)] + (1 – 3) b²
= [4 + (- 6)] a² + (-3 + 5) ab + [1+ (-3)]b²
= -2a² + 2ab – 2b²

Question 9.
The length of a log is 3a + 4b – 2 and a piece (2a – b) is remove from it. What is the length of the remaining log?
Solution:
Length of the log = 3a + 4b – 2

Length of the piece removed = 2a – b
Remaining length of the log = (3a + 4b – 2) – (2a – b)
= (3a – 2a) + [4b – (-b)] – 2
= (3 – 2)a + (4 + 1)b – 2
= a + 5b – 2

Question 10.
A tin had ‘x’ litre oil. Another tin had (3x² + 6x – 5) litre of oil. The shopkeeper added (x + 7) litre more to the second tin. Later he sold (x² + 6) litres of oil from the second tin. How much oil was left In the second tin?
Solution:
Quantity of oil in the second tin = 3x² + 6x – 5 litres.
Quantity of oil added = x + 7 litres
∴ Total quantity of oil in the second tin
= (3x² + 6x – 5) + (x + 7) litres
= 3x² + (6x + x) + (-5 + 7)
= 3x² + (6 + 1)x + 2
= 3x² + 7x + 2 litres
Quantity of oil sold = x + 6 litres
∴ Quantity of oil left in the second tin = (3x² + 7x + 2) – (x² + 6)(3x² – x² ) + 7x + (2 – 6)
= (3 – 1)x² + 7x + (-4) = 2x² + 7x – 4
Quantity of oil left = 2x² + 7x – 4 litres

Try this Page No. 70

Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Solution:
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
Eg. 2y² + 5y^-1 – 3 is a an algebraic expression. But not a polynomial.

Try this Page No. 71

Question 2.
-(5y² + 2y – 6) Is this correct? If not, correct the mistake.
Solution:
Taking -(5y² + 2y – 6) = 5y² + [(-)(+) 2y] + [(-) × (-)6]
= -5y² – 2y + 6
≠ -5y² – 2y + 6
∴ Correct answer is -5y² + 2y – 6 = -(5y² + 2y + 6)

Try this Page No 71

(i) 3ab², -2a²b³
(ii) 4xy, 5y²x, (-x²)
(iii) 2m, -5n, -3p
Solution:
(i) (3ab²) × (-2a²b²) = (+) × (-) × (3 × 2) × (a × a²) × (b² × b³) = -6a³ b⁵

(ii) (4xy) × (5y²x) × (-x²)
= (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x²) × (y × y²)
= -20x⁴y³

(iii) (2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30mnp = 30 mnp

Try this Page No. 71

Question 1.
Why 3 + (4x – 7y) ≠ 12x – 21y?
Solution:
Addition and multiplication are different
3 + (4x – 7y) = 3 + 4x – 7y
We can add only like terms.

Try this Page No. 72

Question 1.
Which is corrcet? (3a)² is equal to
(i) 3a²
(ii) 3²a
(iii) 6a²
(iv) 9a²
Solution:
(3a) =3²a² = 9a²
(iv) 9a² is the correct answer

Try These Page No.72

Question 1.
Multiply
(i) (5x² + 7x – 3) by 4x²
Solution:
(5x² + 7x – 3) × 4x²
= 4x²(5x² + 7x – 3) Multiplication is commutative
= 4x² (5x² + 4x² (7x) + 4x² (-3)
= (4 × 5)(x² × x²) + (4 × 7)(x² × x) + (4 × -3)(x²)
= 20x⁴ + 28x³ – 12x²

(ii) (10x – 7y + 5z) by 6xyz
Solution:
(10x – 7y + 5z) by 6xyz
(10x – 7y + 5z) × 6xyz = 6xyz (10x – 7y + 5z) [∵ Multiplication is commutative]
= 6xy (10x) + 6xyz (-7y) + 6xyz (5z)
= (6 × 10)(x × x × y × z) + (6 × -7) + (x × y × y × z) + (6 × 5)(x × y × z × z)
= 60x²yz + (-42xy²z) + 30xyz²
= 60x²yz – 42x²z + 30xyz²

(iii) (ab + 3bc – 5ca) by – 3abc
Solution:
(ab + 3bc – 5ca) × (- 3abc) = (-3abc) (ab + 3bc – 5ca)
[∵ Multiplication is commutativel
= (-3abc) (ab) + (-3abc) (3bc) + (-3abc) (5ca)
= (-3)(a × a × b × b × c) + (- 3 × 3) + (a × b × b × c × c)
= -3a²b²c – 9ab²c² – 30a²bc²

Try these Page No. 74

Question 1.
Multiply
(i) (a – 5) and (a + 4)
Solution:
(a – 5) (a + 4) = a(a + 4) – 5 (a + 4)
= (a × a) + (a × 4) + (-5 × a) + (-5 × 4)
= a² + 4a – 5a – 20 = a² – a – 20

(ii) (a + b) and (a – b)
Solution:
(a + b) (a – b) = a(a – b) + b (a – b)
= (a × a) + (a × -b)+(b × a) + b(-b)
= a² – ab + ab – b² = a² – b²

(iii) (m⁴ + n⁴) and (m – n)
Solution:
(m⁴ + n⁴)(m – n) = m⁴(m – n) + n⁴(m – n)
= (m⁴ × m) + (m⁴ × (-n)) + (n⁴ × m (n⁴ × (-n))
= m⁵ – m⁴n + mn⁴ – n⁵

(iv) (2x + 3)(x – 4)
Solution:
(2x + 3)(x – 4) = 2x(x – 4) + 3(x – 4)
= (2x² × x) – (2x × 4) + (3 × x) – (3 × 4)
= 2x² – 8x + 3x – 12 = 2x² – 5x – 12

(v) (x – 5)(3x + 7)
Solution:
(x – 5)(3x + 7) = x(3x + 7) – 5(3x + 7)
= (x × 3x) + (x × 7) + (-5 × 3x) + (-5 × 7)
= 3x² + 7x – 15x – 35
= 3x² – 8x – 35

(vi) (x – 2)(6x – 3)
Solution:
(x – 2)(6x – 3) × (6x – 3) – 2(6x – 3)
= (x × 6x)+(x × (-3) × (2 × 6x) – (2 × 3)
= 6x² – 3x – 12x + 6
= 6x² – 15x + 6

Try this Page No. 74

Question 2.
3x² (x⁴ – 7x³ + 2), what is the highest power in the expression.
Solution:
3x²(x⁴ – 7x³ + 2) = (3x²) (x⁴) + 3x² (-7x³)+ (3x²)²
= 3x⁶ – 21x⁵ + 6x²
Highest power is 6 in x⁶.

Exercise 3.2

Try this Page No. 77

Question 1.
Are the following correct?
(i) \(\frac{x^{3}}{x^{8}}=x^{8-3}=x^{5}\)
(ii) \(\frac{10 m^{4}}{10 m^{4}}=0\)
(iii) When a monomial is divided by itself, we will get I?
Solution:

Try this Page No. 77

Question 1.
Divide

Solution:

Try this Page No. 78

Question 1.
Are the following divisions correct ?

Solution:

Try this Page No. 78

Question 1.

Solution:

Exercise 3.3

Try these Page No. 81

Question 1.
1. (p + 2)² = …….
2. (3 – a)² = …….
3. (6² – x²) = ………
4. (a + b)² – (a – b)² = …….
= a² + 2ab + b² – a² – 2ab – b²
= (1 – 1)a² + (2 + 2)ab + (+1 – 1 )b² = 4ab
5. (a + b)² = (a + b) × (a + b)
6. (m + n)( m – n) = m² – n²
7. (m + 7)² = m² + 14m + 49
8. (k² – 36) ≡ k² – 6² = (k + 6)(k – 6)
9. m² – 6m + 9 = (m – 3)²
10. (m – 10)(m + 5) = m² + (-10 + 5)m + (-10)(5) = m² – 5m – 50
Solution:

Try these page No. 83

Question 1.
Expand using appropriate identities
Question 1.
(3p + 2q)²
Solution:
(3p + 2q)²
Comparing (3p + 2q)² with (a + b)², we get a = 3p and b = 2q.
(a + b)² = a² + 2ab + b²
(3p + 2q)² = (3p)²+ 2(3p) (2q) + (2q)²
= 9p² + 12pq + 4q²

Question 2.
(105)²
Solution:
(105)² = (100 + 5)²
Comparing (100 + 5)² with (a + b)², we get a = 100 and b = 5.
(a + b)² = a² + 2ab + b²
(100 + 5)² = (100)² + 2(100)(5) + 5² = 1oooo + 1000 + 25
105² = 11,025

Question 3.
( 2x – 5d)²
Solution:
(2x – 5d)²
Comparing with (a – b)², we get a = 2x b = 5d.
(a – b)² = a² – 2ab + b²
(2x – 5d)² = (2x)² – 2(2x)(5d) + (5d)²
= 2x² – 20 xd + 5²d² = 4x² – 20 xd + 25d²

Question 4.
(98)²
Solution:
(98)² = (100 – 2)²
Comparing (100 – 2)² with (a – b)² we get
a = 100, b = 2
(a – b)² = a² – 2ab + b²
(100 – 2)² = 100² – 2(100)(2) + 2²
= 10000 – 400 + 4 = 9600 + 4 = 9604

Question 5.
(y – 5)(y + 5)
Solution:
(y – 5)(y + 5)
Comparing (y – 5) (y + 5) with (a – b) (a + b) we get
a = y; b = 5
(a – b)(a + b) = a² – b²
(y – 5)(y + 5) = y² – 5² = y² – 25

Question 6.
(3x)² – 5²
Solution:
(3x)² – 5²
Comparing (3x)² – 5² with a² – b² we have
a = 3x; b = 5
(a² – b²) = (a + b)(a – b)
(3x)² – 5² = (3x + 5)(3x – 5) = 3x(3x – 5) + 5(3x – 5)
= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)
= 9x² – 15x + 15x – 25 = 9x² – 25

Question 7.
(2m + n)(2m +p)
Solution:
(2m + n) (2m + p)
Comparing (2m + n) (2m + p) with (x + a) (x + b) we have
x = 2n; a = n ;b = p
(x – a)(x + b) = x² + (a + b)x + ab
(2m +n) (2m +p) = (2m²) + (n +p)(2m) + (n) (p)
= 2²m² + n(2m) + p(2m) + np
= 4m² + 2mn + 2mp + np

Question 8.
203 × 197
Solution:
203 × 197 = (200 + 3)(200 – 3)
Comparing (a + b) (a – b) we have
a = 200, b = 3
(a + b)(a – b) = a² – b²
(200 + 3)(200 – 3) = 200² – 3²
203 × 197 = 40000 – 9
203 × 197 = 39991

Question 9.
Find the area of the square whose side is (x – 2)
Solution:
Side of a square = x – 2
∴ Area = Side × Side
= (x – 2) (x – 2) = x(x – 2) – 2(x – 2)
= x(x) + (x)(-2) + (-2)(x) + (-2)(-2)
= x – 2x – 2x + 4x² – 4x + 4

Question 10.
Find the area of the rectangle whose length and breadth are (y + 4) and (y – 3).
Solution:
Length of the rectangle = y+ 4
breadth of the rectangle = y – 3
Area of the rectangle = length × breadth
= (y + 4)(y – 3) = y² + (4 +(-3))y + (4)(-3)
= y² + y – 12

Try these Page No. 88

Question 1.
Expand :
Question 1.
(x + 4)³
Solution:
Comparing (x + 4)³ with (a + b)³, we have a = x and b = 4.
(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 4)³ = x³ + 3x²(4) + 3(x)(4)² + 4³
= x³ + 12x² + 48x + 64

Question 2.
( y – 2)²
Solution:
Comparing (y – 2) with (a – b)³ we have a = y b = z
(a – b)³ = a³ – 3a²b + 3ab² – b³
(y – 2)² = y³ – 3y(2) + 3y(2)² + 2³
= y³ – 6y² + 12y + 8

Question 3.
(x + 1)(x + 3)(x + 5)
Solution:
Comparing (x + 1) (x + 3) (x + 5) with (x + a) (x + b) (x + c) we have
a = 1
b = 3
and c = 5

Exercise 3.4

Try These Page No.92

Question 1.
Factorize the following:
Question 1.
3y + 6
Solution:
3y + 6
3y + 6 = 3 × y + 2 × 3
Taking out the common factor 3 from each term we get 3 (y + 2)
∴ 3y + 6 = 3(y + 2)

Question 2.
10x² + 15y
Solution:
10x² + 15y²
10x² + 15y² = (2 × 5 × x × x) + (3 × 5 × y × y)
Taking out the common factor 5 we have
10x² + 15y² = 5(2x² + 3y²)

Question 3.
7m(m – 5) + 1(5 – m)
Solution:
7m(m – 5) + 1(5 – m)
7m(m – 5) + 1(5 – m) = 7m(m – 5) + (-1)(-5 + m)
= 7m(m – 5) – 1 (m – 5)
Taking out the common binomial factor (m – 5) = (m – 5)(7m – 1)

Question 4.
64 – x²
Solution:
64 – x²
64 – x² = 8² – x²
This is of the form a² – b²
Comparing with a² – b² we have a = 8, b = x
a² – b² = (a + b)(a – b)
64 – x² = (8 + x)(8 – x)

Students can Download Maths Chapter 2 Measurements 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 1 Chapter 2 Measurements Additional Questions

Additional Questions And Answers

Exercise 2.1

Very Short Answers [2 Marks]

Question 1.
Find the length of arc if the perimeter of sector is 45 cm and radius is 10 cm.
Solution:
Given Radius of the sector = 10 cm
Perimeter of the sector P = 45 cm
l + 2r = 45
l + 2(10) = 45
l + 20 = 45
l = 45 – 20
l = 25 cm
Length of the arc l = 25 cm

Question 2.
Find the radius of sector whose perimeter and length of arc are 30 cm and 16 cm respectively.
Solution:
Given length of the arc = 16 cm
Perimeter of the arc = 30 cm
l + 2r = 30
16 + 2 r = 30
2 r = 30 – 16
2 r = 14
r = \(\frac{14}{2}\)
r = 7 cm
Radius of the sector = 7 cm

Question 3.
Find the length of arc whose radius is 7 cm and central angle 90°.
Solution:
Here θ = 90°; radius r = 7cm
Length of the arc = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
= \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × \(\frac{22}{7}\) × 7 = 11 cm
∴ Length of the arc = 11 cm

Short Answers [3 Marks]

Question 1.
Find the length arc whose radius is 42 cm and central angle is 60°.
Solution:
Length of the arc = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
Given central angle 0 = 60°
Radius of the sector r = 42 cm
l = \(\frac{60^{\circ}}{360^{\circ}}\) × 2 × \(\frac{22}{7}\) × 42 = 44 cm
∴ Length of the arc = 44 cm

Question 2.
Find the length of the arc whose radius is 10.5 cm and central angle is 36°.
Solution:

∴ Length of the arc = 6.6 cm

Long Answers [5 Marks]

Question 1.
A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of its arc and area of the sector.
Solution:
Radius of the sector = 21 cm
Length of the arc

∴ Length of the arc = 55 cm
Area of the sector = 577.5 cm²

Question 2.
Find the perimeter of sector whose area is 324 sq. cm and radius is 27 cm.
Solution:
Radius of the sector = 27 cm
Area of the sector =324 cm²

Perimeter of the sector P = (l + 2r) units = 24 + 2(27) cm = (24 + 54) cm = 78 cm

Exercise 2.2

Question 1.
PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal semi-circles drawn on PQ and Question as diameters. Find the p perimeter and area of the shaded region.

Solution:
PS = Diameter of a circle of radius 6 cm = 12 cm
PQ = QR = RS = \(\frac{12}{3}\) = 4 cm Question = QR + RS = 4 + 4 = 8 cm
∴ Perimeter of the shaded part = Arc length of semi-circle of radius 6 cm + Arc length of semicircle of radius 4 cm + Arc length of semi-circle of radius 2 cm.
= (π × 6) + (π × 4) + (π × 2) cm
P = 12 π cm
Area required = Area of semicircle with PS as diameter + Area of semi circle with PQ as diameter – Area of semi-circle with Question as diameter.

Question 2.
In the figure AOBCA represents a quadrant of a circle of radius 3.5cm with center ‘O’ calculate the area of the shaded portion (π = \(\frac{22}{7}\))

Solution:

∴ Area of shaded region = Area of the quadrant – Area of triangle
= 9.625 – 3.5 cm² = 6.125 cm²

Question 3.
Find the area of the shaded region in the figure

Solution:
Radius of the big semicircle = 14 cm

∴ Required area = 308 + 154 cm² = 462 cm²

Exercise 2.3

Very Short Answers [2 Marks]

Question 1.
What is the least number of planes that can enclose a solid? What is the name of the solid?
Solution:
Least number of planes = 4, the solid is tetrahedron.

Question 2.
Can a polyhedron have for its faces = 12 edges = 16 and vertices = 6.
Solution:
Verifying Euler’s formula
F + V – E = 12 + 6 – 16 = 18 – 16 = 2
Yes, the polyhedron can have F = 12, E = 16 and V = 6

Short Answers [3 Marks]

Question 1.
Verify Euler’s formula for a pyramid.
Solution:
A pyramid has faces = 5, Vertices = 5, Edges = 8
By Euler’s formula F + V – E = 5 + 5 – 8 = 10 – 8 = 2

Question 2.
Verify Eulers formula for a triangular prism.
Solution:
For a triangular prism
Faces = 5, Edges = 9, Vertices = 6
By Euler’s formula F + V – E = 5 + 6 – 9 = 11 – 9 = 2

Long Answers [5 Marks]

Question 1.
(a) Dice are cubes where the numbers on the opposite faces must total 7. Is the following a die.

(b) The following shows a net with areas of faces. What can be the shape?

Solution:
(a) 2 + 5 = 6 + 1 = 3 + 4 = 7
∴ It can be a die.
(b) It is a cuboid

Students can Download Maths Chapter 2 Measurements Ex 2.4 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 1 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet istance from the wall ito which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (π = 3.14).

Solution:
Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.
Now the arc length BC gives the distance moved by the wheel.
Length of the arc
= \(\frac{\theta}{360^{\circ}}\) × 2πr units
= \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 6 feets
= 3.14 × 3 feets
= 9.42 feets
∴ Distance moved by the wheel = 9.42 feets.

Question 2.
With his usual speed, if a person covers a circular track of radius 150 ra in 9 minutes, find the distance that he covers in 3 minutes (π = 3.14).
Solution:
Radius of the circular track = 150m
Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units
Distance covered in 9 min = 2 × 3.14 × 150 m

Distance he covers in 3 min = 314 m

Question 3.
Find the area of the house drawing given in the figure.

Solution:
Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, h = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm
= (side × side) + (l × b) + (\(\frac{1}{2}\) × b × h) + 6h cm²
= (6 × 6) + (8 × 6) + (\(\frac{1}{2}\) × 6 × 4) + (8 × 4) cm²
= 36 + 48 + 12+ 32 cm²
Required Area = 128 cm²

Question 4.
Draw the top, front and side view of the following solid shapes.

Solution:

Question 5.
Draw the net for the cube of side 4 cm in a graph sheet.
Solution:

Challenging Problems

Question 6.
Guna has fixed a single door of 3 feet wide in his room whereas Nathan has fixed a double door, each 1 \(\frac{1}{2}\) feet wide in his room. From the closed state, if each of the single and double doors can open up to 120°, whose door requires a minimum area?

Solution:
(a) Width of the door that Guna fixed = 3 feet.
When the door is open the radius of the sector = 3 feet
Angle covered = 120°
∴ Area required to open the door = \(\frac{120^{\circ}}{360^{\circ}}\) × πr² = \(\frac{120^{\circ}}{360^{\circ}}\) × π × 3 × 3 = 37π feet²

(b) Width of the double doors that Nathan fixed = 1\(\frac{1}{2}\) feet.
Angle described to open = 120°
Area required to open = 2 × Area of the sector
= 2 × \(\frac{120^{\circ}}{360^{\circ}} \times \pi \times \frac{3}{2} \times \frac{3}{2} \text { feets }^{2}=\frac{3 \pi}{2}\) feet²
= \(\frac{1}{2}\) (3π) feet²
∴ The double door requires the minimum area.

Question 7.
In a rectangular field which measures 15 m × 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (π = 3.14).
Solution:
Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circle

Area of the rectangle = l × b units²
= 15 × 8 m² = 120 m²
Area of 4 quadrant circles = 4 × \(\frac{1}{4}\) πr² units
Radius of the circle = 3 m
Area of 4 quadrant circles = 4 × \(\frac{1}{4}\) × 3.14 × 3 × 3 = 28.26m²
Area of the circle at the middle = πr² units
= 3.14 × 3 × 3m² = 28.26m²
∴ Area where none of the cows can graze
= [120 – 28.26 – 28.26]m² = 120 – 56.52 m² = 63.48m²

Question 8.
Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins, (π = 3.14) ( \(\sqrt{3}\) = 1.732)

Solution:
Given diameter of the coins = 6 cm
∴ Radius of the coins = \(\frac{6}{2}\) = 3 cm
Area of the shaded region = Area of equilateral triangle – Area of 3 sectors of angle 60°
Area of the equilateral triangle

∴ Area of the shaded region = 15.588 – 14.13 cm² = 1.458 cm²
Required area 1.458 cm² (approximately)

Question 9.
Using graph sheet, draw the net for the cuboid whose length is 5cm, breadth is 4cm and height is 3cm and also find its area.
Solution:
Net for the cuboid is:

One of the possible nets for a cuboid of length = 5 cm, breadth = 4 cm, height = 3 cm is given above
Area of the cuboid
= 20 cm² + 15 cm² + 20 cm² + 15 cm² + 12 cm² + 12 cm² = 94 cm²
Using formula,
Surface area of a cuboid
= 2 (lb + bh + lh) unit²
= 2(5 × 4 + 4 × 3 + 5 × 3) cm²
= 2(20 + 12 + 15) cm²
= 94 cm²

Question 10.
Using Euler’s formula, find the unknowns.

Solution:
Euler’s formula is given by F + V- E = 2
(i) V = 6, E = 14
By Euler’s formula
= F + 6 – 14 = 2
F = 2 + 14 – 6
F = 10

(ii) F = 8, E = 10
By Euler’s formula
= 8 + V – 10 = 2
V = 2 – 8 + 10
V = 4

(iii) F = 20, V = 10
By Euler’s formula
= 20 + 10 – E = 2
30 – E = 2
E = 30 – 2
E = 28
Tabulating the required unknowns

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3

Question 1.
Fill in the blanks:
(i) The three dimensions of a cuboid are _____.
(ii) The meeting point of more than two edges- is called as ______.
(iii) A cube has _____ faces.
(iv) The cross section of a solid cylinder is ______.
(v) If a net of a 3-D shape has six plane squares, then it is called ______.
Solution:
(i) length, breadth and height
(ii) vertex
(iii) six
(iv) circle
(v) cube

Question 2.
Match the following:

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

Question 3.
Which 3-D shapes do the following nets represent? Draw them.

Solution:
(i) The net represents cube, because it has 6 squares

(ii) The net represents cuboid

(iii) The net represents Triangular prism

(iv) The net represents square pyramid

(v) The net represents cylinder

Question 4.
For each solid, three views are given. Identify for each solid, the corresponding top, front and side (T, F & S) views.

Solution:

Question 5.
Verify Euler’s formula for the table given below

Solution:
Euler’s formula is given by F + V – E
(i) F = 4 ; V = 4; E = 6
F + V – E = 4 + 4 – 6 = 8 – 6
F + V – E = 2
∴ Euler’s formula is satisfied.

(ii) F = 10; V = 6; E = 12
F + V – E = 10 + 6 – 12
= 16 – 12 = 4 ≠ 2
∴ Euler’s formula is not satisfied.

(iii) F = 12 ; V = 20 ; E = 30
F + V – E = 12 + 20 – 30
= 32 – 30 = 2
∴ Euler’s formula is satisfied.

(iv) F = 20 ; V = 13 ; E = 30
F + V – E = 20 + 13 – 30
= 33 – 30 = 3 ≠ 2
∴ Euler’s formula is not satisfied.

(v) F = 32 ; V = 60 ; E = 90
F + V – E = 32 + 60 – 90
= 92 – 90 = 2
∴ Euler’s formula is satisfied.

Question 6.
Find the area of the given nets.

Solution:
(i) Area = Area of 6 squares of side 4 cm
= 6 × a² sq. units
= 6 × 4 × 4 cm²
= 96 cm²
(ii) Area = Area of 2 rectangles of
l = 10, b = 6 + Area of 2 rectangles of l = 6, b = 4 + Area of 2 rectangles of l= 10,b = 4
= (10 × 6) + (6 × 4)+ (10 × 4) cm²
= 60 + 24 + 40 cm²
= 124 cm²

Question 7.
Can a polyhedron have 12 faces, 22 edges and 17 vertices?
Solution:
By Euler’s formula F + V- E = 2 fora polyhedron.
Here F = 12, V = 17, E = 22
F + V – E = 12 + 17 – 22
= 29 – 22
= 7 ≠ 2
∴ The polyhedron cannot have 12 faces 22 edges and 17 vertices.

You can Download Samacheer Kalvi 9th Maths 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 Maths Solutions Chapter 4 Geometry Ex 4.4

Question 1.
Find the value of x in the given figure.

Solution:
In the cyclic quadrilateral ABCD
∠ABC = 180° – 120° = 60°
∠BCA = 90°
∴ x = ∠BAC = 180°- (90° + 60°) = 30°

Question 2.
In the given figure, AC is the diameter of the circle with centre O. If ∠ADE = 30°; ∠DAC = 35° and ∠CAB = 40°.
Find
(i) ∠ACD
(ii) ∠ACB
(iii) ∠DAE

Solution:
(i) ∠ACD = 180°- (90° + 35°) = 180°- 125° = 55°
(ii) ∠ACB = 180°- (90°+ 40°)= 180° – 130° = 50°
(iii) ∠ADC = 90°
∠CAE = 180° – 120° = 60°
∴ ∠DAE = 60°- 35° = 25°

Question 3.
Find all the angles of the given cyclic quadrilateral ABCD in the


Solution:

Question 4.
In the given figure, ABCD is a cyclic quadrilateral where diagonals intersects at P such that ∠DBC = 40° and ∠BAC = 60° find
(i) ∠CAD
(ii) ∠BCD

Solution:

Question 5.
In the given figure, AB and CD are the parallel chords of a circle with centre O. Such that AB = 8 cm and CD = 6 cm. If OM ⊥ AB and OL ⊥ CD distance between LM is 7 cm. Find the radius of the circle?

Solution:
In the figure LM = 7 cm

Question 6.
The arch of a bridge has dimensions as shown, where the arch measure 2 m at its highest point and its width is 6 m. What is the radius of the circle that contains the arch?

Solution:

Question 7.
In figure ∠ABC = 120°, where A,B and C are points on the circle with centre O. Find ∠OAC ?

Solution:

Question 8.
A school wants to conduct tree plantation programme. For this a teacher allotted a circle of radius 6 m ground to nineth standard students for planting sapplings. Four students plant trees at the points A, B, C and D as shown in figure. Here AB = 8 m, CD = 10 m and AB ⊥ CD. If another student places a flower pot at the point P, the intersection of AB and CD, then find the distance from the centre to P.

Solution:

ONPM is a rectangle with all the angles 90° and with length \(\sqrt{20}\) cm, breadth \(\sqrt{11}\) cm.

We need to find OP which is the diagonal of the rectangle ONPM.

Question 9.
In the given figure, ∠POQ = 100° and ∠PQR = 30°, then find ∠RPO.
Solution:

You can Download Samacheer Kalvi 9th Maths 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 Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Solution:
The distance of the chord from the centre O

Question 2.
The chord of length 30 cm is drawn at the distance of 8cm from the centre of the circle. Find the radius of the circle.
Solution:

Question 3.
Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4 \(\sqrt{2}\) cm and also find ∠OAC and ∠OCA.
Solution:
∆OAC is an isoceles triangle with one angle 90°
∴ ∠OAC + ∠OCA = 180° – 90°

2∠OAC = 90°
∠OAC = 45°
∴ ∠OCA = 45°

Question 4.
A chord is 12cm away from the centre of the circle of radius 15 cm. Find the length of the chord.
Solution:

Question 5.
In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?

Solution:
The distance between the two chord FE = OE + OF

∴ Distance between the chords is 14 cm

Question 6.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:

The length of the common chord AB = AD + BD = (3 + 3) cm = 6 cm

Question 7.
Find the value of x° in the following

Solution:

Question 8.
In the given figure, ∠CAB = 25°, find ∠BDC, ∠DBA and ∠COB

Solution:

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions

Question 1.
From the given figure, name the parallel lines

Solution:
(i) Parallel Lines:
\(\overrightarrow{\mathrm{CD}}\) and \(\overrightarrow{\mathrm{EF}}\) ; \(\overrightarrow{\mathrm{CD}}\) and \(\overrightarrow{\mathrm{IJ}}\) ; \(\overrightarrow{\mathrm{EF}}\) and \(\overrightarrow{\mathrm{IJ}}\) are parallel lines.
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}}\)
(b) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(e) \(\overrightarrow{\mathrm{GH}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(iii) Points of Intersection:
P, Q and R are the points of intersection.

Question 2.
(a) Name the line segments in the figure.
(b) Is Q, the endpoint of each line segment?

Solution:
(a) \(\overline{\mathrm{QP}} \text { and } \overline{\mathrm{QR}}\) are the line segments
(b) Yes, Q is the end point of each line segment

Question 3.
How many lines can pass through
(a) one given point
(b) two given points.
Solution:

(a) An infinite number of lines can pass through one given point.
(b) Exactly one and only one line can pass through two given points.

Question 4.
A line contains how many points?
(a) minimum?
(b) maximum?
Solution:
(a) A line contains a minimum of two points.
(b) A line contain a maximum of infinitely many points.

Question 5.
Write the (a) maximum and (b) the minimum number of point of intersection of three lines.
Solution:

Maximum – 3 points of intersection
Minimum – No point of intersection

Fill in the blanks.

Question 6.
Complementary angle of 20° is _____
Solution:
70°

Question 7.
The supplementary angle of 90° is _____
Solution:
90°

Question 8.
78°, 12°, ______
Solution:
Complementary angle

Answer the following question.

Question 9.
∠ABD =?

Solution:
On Sum of complementary angles = 90°
∠ABC = 90°
∠CBD = 30°
∠ABD = ∠ABC – ∠DBC = 90° – 30° = 60°
∠ABD = 60°
Complementary angle of 30° = 60°

Question 10.
In the following figure, name the angles.

Solution:
∠AOB, ∠BOZ, ∠AOZ

Question 11.
Write the alternate name of the angle ∠XYZ in the given figure.

Solution:
∠Y or ∠ZYX

Question 12.
Draw the diagram of two angles having only one common point.

Solution:
∠COD and ∠AOB have the point ‘O’ in common

Question 13.
What are the supplementary and complementary angles of 60°?
Solution:
Supplementary angle is 120°
Complementary angle is 30°

Question 14.
How many lines can you draw passing through three collinear points? Draw the figure also.
Solution:
Only one.

Question 15.
Write the maximum number of lines that can pass through a single point.
Solution:
Infinite.

Question 16.
Use a protractor to draw an angle 45°.
Solution:

Construction:
1. Drawn the base ray PQ.
2. Placed the centre of the protractor at the vertex P. Lined up the ray \(\overrightarrow{\mathrm{PQ}}\) with the 0° line. Then drawn and labelled a pointed (R) at the 45° mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)
3. Removed the protractor and drawn at \(\overrightarrow{\mathrm{PR}}\) to complete the angle
Now ∠P = ∠QPR = ∠RPQ = 45°.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.2

Question 1.
The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find all the angles.
Solution:
In a quadrilateral the angles add upto 360°.
Let’s call the angles 2x, 4x, 5x, 7x
2x + 4x + 5x + 7x = 360°
18x = 360°

A = 2x = 2 × 20° = 40°
B = 4x = 4 × 20° = 80°
C = 5x = 5 × 20° = 100°
D = 7x = 7 × 20° = 140°

Question 2.
In a quadrilateral ABCD, ∠A = 12° and ∠C is the supplementary of ∠A. The other two angles are 2x – 10 and x + 4. Find the value of x and the measure of all the angles.
Solution:
∠A = 72°
∠C = 180° – 72° (∵ Supplementary at ∠A) = 108°
The other two angles are 2x – 10 and x + 4.
2x – 10 + x + 4 + 108° + 12° = 360°

3x + 174° = 360°
3x = 360° – 174°

∴ ∠A = 72°
∠B = 2x – 10 = 2(62)- 10 = 124 – 10 = 114°
∠C = 108°
∠D = x + 4 = 62 + 4 = 66°

Question 3.
ABCD is a rectangle whose diagonals AC and BD intersect at O. If ∠OAB = 46°, find ∠OBC.
Solution:
∠ABC = 90°

∠OAB + ∠OBC = 90°
46° + ∠OAB = 90°
∠OBC = 90° – 46° = 44°

Question 4.
The lengths of the diagonals of a Rhombus are 12 cm and 16 cm. Find the side of the rhombus.
Solution:
Let ABCD be a rhombus with AC and BD as its diagonals.
We know that the diagonals of a rhombus bisect each other at right angles.
Let O be the intersecting point of both the diagonals
Let AC = 16 cm and BD = 12 cm


use Pythagoras theorem, we have
AB² = OA² + OB²
AB² = 100
∴ AB = 10 cm

Question 5.
Show that the bisectors of angles of a parallelogram form a rectangle.
Solution:
Given ABCD is a parallelogram. Draw the angular bisectors AP, BP, CR and DR of the angles ∠A, ∠B, ∠C and ∠D respectively.
Now to prove : PQRS is a rectangle.
Proof: A rectangle is a parallelogram with one angle 90°.
First we will prove PQRS is a parallelogram.
Now AB || CD and AD is transversal. [ ∴ Interior angles on the same side of transversal are supplementary]
[Opposite sides of a parallelogram are parallel]

Also lines AP and DR intersects
So ∠PSR = ∠DS A
∴ ∠PSR = 90° [∵ Vertically opposite angles]
Similarly we can prove that ∠SPQ = 90°, ∠PQR = 90° and ∠SRQ = 90°
∴ ∠PSR = ∠PQR and ∠SPQ = ∠SRQ
∴ Both pair of opposite angles of PQRS is a parallelogram.
Also ∠PSR = ∠PQR = ∠SPQ = ∠SRQ = 90°
∴ PQRS is a parallelogram with one angle 90°.
∴ PQRS is a rectangle. Hence proved.

Question 6.
If a triangle and a parallelogram lie on the same base and between the same parallels, then prove that the area of the triangle is equal to half of the area of parallelogram.
Solution:
Given: ∆ABE and parallelogram ABCD have the same base and are between the same parallel lines (i.e) l₁ || l₂.

Perpendicular distance between l₁ and l₂ = P (say).
Prove that: area of (∆ABE) = \(\frac{1}{2}\) × area of the parallelogram ABCD
Proof: Area of ∆ABE = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × AB × (Perpendicular distance between l₁
= \(\frac{1}{2}\) × AB × P ….(1)
Area of parallelogram ABCD = base × height.
∴ Area of parallelogram ABCD = AB × P …. (2)
From (1) and (2),
Area of ∆ABE = \(\frac{1}{2}\) × Area of parallelogram ABCD.
Hence proved.

Question 7.
Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure. If a || b, c || d, e || f, find the marked angles between
(i) b and c
(ii) d and e
(iii) d and f
(iv) c and f.

Solution:
Since l, m are two parallel lines and PQ, RS, TU, VW are transversal.
Then ∠1 = ∠QOR [vertically opposite angles]
∠1 = 30° [∴ ∠QOR = 30°]
Also, PQ and TU are parallel and m and l are transversal.

Also ∠3 + ∠4 = 180°
⇒ 75° + ∠4 = 180°
∠4 = 180° – 75° = 105°
Hence,
(i) 30°
(ii) 105°
(iii) 75°
(iv) 105°

Question 8.
In the given figure ∠A = 64° , ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ∆ABC, find x° and y°.
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180°
⇒ 64° + 58° + ∠C = 180°
⇒ 122°+ ∠C = 180°
⇒ ∠C = 180°- 122° = 58°

Also BO and CO are the bisectors of ∠ABC and ∠ACB respectively

Question 9.
In the given Fig. if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ACDF.

Solution:
Given: AB = 2,
BC = 6,
AE = 6,
BF = 8,
CE = 7 and
CF = 7
Consider ∆AEC and ∆BCF,
In ∆AEC,
AC = 8,
AE = 6,
CE = 7
In ∆BCF,
BF = 8,
BC = 6,
CF = 7
∴ ∆AEC ≅ ∆BCF
∴ Area of ∆AEC = Area of ∆BCF
Subtract.area of ∆BDC both sides, we get
Area of ∆AEC – Area of ∆BDC = Area of ∆BCF – Area of ∆BDC
⇒ Area of quadrilateral ABDE = Area of ∆CDF
∴ The required ratio is 1 : 1

Question 10.
In the figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to \(\overline{\mathbf{H E}}\) and \(\overline{\mathbf{F G}}\) ?

Solution:
Area of ABCD = length × breadth.
= DC × BC = 10 × 8 = 80.
Area of ∆AEH = Area of ∆CGF [since they are congruent by RHS rule]
Similarly, Area of ∆BEF = Area of ∆DGH
∴ Area of parallelogram = EFGH = Area of rectangle ABCD – 2(area of ∆AEH) – 2(area of ∆BEF)

Question 11.
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is \(\frac{9}{8}\) of the area of the parallelogram ABCD.

Solution:

Area of ∆QDP = Area of ∆QMA + Area of ∆MNO + Area of MNBS + Area of ∆MAB
= Area of ∆DCM + Area of ∆MNO + Area of MNBA + Area of ∆NDC
= 2Area of ∆OMN + Area of ∆MNO + 4 Area of ∆OMN + 2 Area of ∆OMN

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
In the figure, AB is parallel to CD, find x

Solution:
(i) From the figure
∠1 = 140° (∴ corresponding angles are equal)
∠2 = 40° (∴ ∠1 + ∠2= 180°)
∠3 = 30° (∵ ∠3 + 150= 180°)
∠4 = 110° (∵ ∠2 + ∠3 + ∠4 = 180°)
∴ ∠x = 70° (∵ ∠4 + ∠x = 180°)

(ii) From the figure
∠1 = 48°
∠3 = 108° (∠1 +24° + ∠3 = 180°)
∠4 = 108° (If two lines are intersect, then the vertically the opposite angles are equal)
∠5 = 72° (∵ ∠3 + ∠5 = 180°)
∴ ∠3 + ∠4 + ∠5 = 108° + 108° + 72°
x = 288°

(iii) From the figure
∠D = 53° ( ∵ ∠B and ∠D are alternate interior angles)
Sum of the three angles of a triangle is 180°
∠x° = 180°- (38°+ 53°)
= 180°- 91° = 89°

Question 2.
The angles of a triangle are in the ratio 1 : 2 : 3, find the measure of each angle of the triangle.
Solution:
Let the angles be x, 2x and 3x respectively.
Sum of the three angles of a triangle = 180°

The 3 angles of the triangle are 30°, 60°, 90°.

Question 3.
Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

Solution:
(i) Consider ∆PQR and ∆ABC
Given, RQ = BC
PQ = AB
∆ABC is not congruent to ∆PQR
If PR = AC, then ∆ABC ≅ ∆PQR

(ii) Consider ∆ABD and ∆BCD for the triangles to be congruent.
Given, AB = DC
AD = BC and AB is common side.
∴ By SSS rule ∆ABD ≅ ∆BCD.

(iii) Consider ∆PXY and ∆PXz,
Given, XY = XZ
PY = PZ and PX is common
∴ By SSS rule ∆PXY ≅ ∆PXZ.

(iv) Consider ∆OAB and ∆ODC,
Given, OA = OC

∠ABO = ∠ODC and ∠AOB = ∠DOC (vertically opposite angles)
∴ By AAS rule, AOAB = AODC.

(v) Consider ∆AOB and ∆DOC,
Given, AO = OC
OB = OD


and ∠AOB = ∠DOC [vertically opposite angles]
∴ By SAS rule, ∆AOB = ∆DOC.

(vi) Consider ∆AMB and ∆AMC,
Given, AB = AC

∠AMB = ∠AMC = 90°
∴ AM is common.
∴ By RHS rule
∆AMB ≅ ∆AMC.

Question 4.
∆ABC and ∆DEF are two triangles in which AB = DF, ∠ACB = 70°, ∠ABC = 60°; ∠DEF = 70° and ∠EDF = 60°. Prove that the triangles are congruent.
Solution:

∴ By ASA rule ∆ABC ≅ ∆FDE

Question 5.
Find all the three angles of the ∆ABC

Solution:
Exterior angle = Sum of the two opposite interior angles.