Mensuration - 8th math

8th-math-home

8th-math hindi-home

NCERT Exercise 11.2


Mensuration Class Eighth Math NCERT Exercise 11.2 Question (1) The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1m and 1.2m and perpendicular distance between them is 0.8m.

mensuration class eighth ncert exercise 11.2 q1

Solution

Given, parallel sides of trapezium are 1 m and 1.2 m

Perpendicular distance between parallel sides (h) = 0.8m

Thus, Area of trapezium = ?

We know that, Area of a trapezium = 1/2 × (a + b) × h

Where a and b are parallel sides and h is the height of trapezium.

Thus, Area of given trapezium = 1/2 × (1 + 1.2) × 0.8 m2

`=1/2xx2.2xx0.8m^2`

= 1.1 × 0.8 m2

= 0.88 m2

Thus Area of given trapezium shaped table top = 0.88m2 Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (2) The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Solution

Given, Area of a trapezium = 34 cm2

Height of trapezium = 4 cm

And, one parallel side of trapezium (a) = 10 cm

Thus, second parallel side of trapezium (b) = ?

We know that, Area of a trapezium = 1/2 × (a + b) × h

Where a and b are parallel sides and h is the height of trapezium.

Thus, area of given trapezium = 1/2 × (10 + b) × 4

⇒ 34 = (10 + b) × 2

⇒ (10 + b) × 2 = 34

⇒ 10 + b = 34/2

⇒ 10 + b = 17

⇒ b = 17 – 10

⇒ b = 7 cm

Thus, other parallel side of given trapezium = 7 cm Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (3) Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17m and AD = 40m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

mensuration class eighth ncert exercise 11.2 q2

Solution

mensuration class eighth ncert exercise 11.2 answer of q3

Given, ABCD is a trapezium.

And length of fence of this trapezium = 120 m

And, AD (a) = 40 m

And BC (b) = 48 m

CD = 17 m

And side AB is perpendicular to the parallel sides AD and BC.

Thus, Area of this trapezium = ?

We know that, length of fence = perimeter of given trapezium

And, now, perimeter of the given trapezium = BC + CD + AD + AB

⇒ 120 m = 48m + 17m + 40m + AB

⇒ 120m = 105 m + AB

⇒ 105m + AB = 120m

⇒ AB = 120 m – 105m

⇒ AB = 15m

Thus, height of the trapezium (AB) (h) = 15 m

Now, we know that, Area of a trapezium = 1/2 × (a + b) × h

Thus, area of given trapezium = 1/2 × (40 + 48) × 15 m2

= 1/2 × 88 × 15 m2

= 44 × 15 m2

= 660 m2

Thus, area of given trapezium = 660 m2 Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (4) The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

mensuration class eighth ncert exercise 11.2 q4

Solution

We know that area of a general quadrilateral which is split into two triangles = 1/2 (diagonal × sum of altitudes drawn on the diagonal from the other two vertices)

That means, area of such general quadrilateral = 1/2 × d(h1 + h2) square unit

Here, diagonal (d) = 24 m

And one altitude (h1) = 13 m

And other altitude (h2 = 8 m

Thus, area of given general quadrilateral

= 1/2 × 24 × (13 + 8)

= 12 × 21

= 252 m2

Thus area of given quadrilateral = 252 m2 Answer

Alternate Method

mensuration class eighth ncert exercise 11.2 answer of q4

This quadrilateral is made of two triangles, i.e. Δ ACD and Δ ABC

For triangle ACD

Base of the triangle (b) = 24 m

And height of the triangle (h) = 13m

Now, we know that, Area of a triangle = 1/2 × Base × Height

Thus, area of given triangle = 1/2 × 24 m × 13m

= 12 m × 13m

= 156 m2

Thus, area of Δ ACD = 156 m2

For triangle ABC

Base of the triangle (b) = 24 m

And height of the triangle (h) = 8m

Now, we know that, Area of a triangle = 1/2 × Base × Height

Thus, area of given triangle = 1/2 × 24 m × 8m

= 12 m × 8m

= 96 m2

Thus, area of Δ ACD = 96 m2

Thus, area of given quadrilateral = Area of Δ ACD + Area of Δ ABC

= 156 m2 + 96 m2

= 252 m2

Thus area of given quadrilateral = 252 m2 Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (5) The diagonals of a rhombus are 7.5 cm and 12cm. Find its area.

Solution

Given, Diagonals of a rhombus = 7.5 cm and 12 cm

Thus, area of given rhombus = ?

We know that, Area of a rhombus = 1/2 × product of diagonals

Thus, Area of given rhombus = 1/2 × 7.5 cm× 12 cm

= 7.5 cm × 6 cm

= 45.0 cm2

Thus, Area of given rhombus = 45 cm2 Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (6) Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Solution

Given, side of rhombus = 5 cm

And Altitude of rhombus = 4.8 cm

One diagonal of the rhombus = 8 cm

Then area and other diagonal = ?

We know that, Area of a rhombus whose sides and altitude are given = side × altitude

Thus area of given rhombus = 5cm × 4.8cm

= 24 cm2

Thus, area of given rhombus = 24 cm2

Now, again we know that, Area of a rhombus whose diagonals are given = 1/2 × product of diagonals

Here, as given one diagonal (d1) = 8 cm

And area of rhombus = 24 cm2 (as calculated above)

Thus, second diagonal (d2) = ?

Thus, Area of given rhombus = 1/2 × 8 cm × d2

⇒ 24 = 4 cm × d2

`=>d_2=24/4`

⇒ d2 = 6 cm

Thus, Area of the rhombus = 24 cm2 and length of other diagonal = 6 cm Answer

8-science-home


Reference: