7th Maths : Term 1 Unit 2 : Measurements: Exercise 2.3 (Area of the Trapezium) : Text Book Back Exercises Questions with Answers, Solution

**Exercise 2.3**

**1. Find the missing values.**

(i) Height *h* = 10 *m*

Parallel sides a = 12*m*,
*b* = 20 *m*

Area of the trapezium = 1/2 *h
(a + b)* *sq. units*

= 1/2 × 10 (12 + 20) *sq.m*

= 5 × 32 *sq.m* = 160 *sq.m*

(ii) Height *h *= ?

Parallel sides *a* = 13 *cm*, *b*
= 28 *cm*

Area of the trapezium = 1/2 *h*(*a + b*) = 492 *sq.cm*

= 1/2 *h *(13 + 28) = 492

= 1/2 *h*
× 41 = 492

* h* = 492 / 41 × 2 = 24 *cm*

**Height h = 24 cm**

(iii) Height *h* = 19 *m*

Parallel sides *a* = ?, *b *= 16 *m*

Area = 323 *sq.m*

1/2 *h×*(*a + b*) = 323 *sq.m*

1/2 × 19 (*a* + 16) = 323

* a* + 16 = [ 323 × 2 ]
/ 19 = 34

* a* = 34 – 16 = 18 *m*

* a ***= 18 m**

(iii) Height *h* = 16 c*m*

Parallel sides *a* = 15c*m*, *b *= ?

Area = 360 *sq.cm*

1/2 *h×*(*a + b*) = 360 *sq.cm*

1/2 × 16 (15 + *b*) = 360

15 + *b* = [ 360 × 2 ] / 16 = 45

* b* = 34 – 15 = 30 c*m*

* a *= 30 c*m*

** **

**2. Find the area of a trapezium whose parallel
sides are 24 ***cm*** and 20 ***cm*** and the
distance between them is 15 ***cm***.**

Parallel sides *a* = 24 *cm*, b = 20 *cm*

Height *h* = 15 *cm*

Area of the trapezium = 1/2 *h
(a + b) sq.units*

Area of the trapezium = 1/2 × 15 (24 + 20) *sq. units*

= 1/2 × 15 × ~~44~~
*sq.cm*

= 330 *sq.cm*

**Area of the
trapezium = 330 sq.cm**

** **

**3. The area of a trapezium is 1586 ***sq.
cm.*** The distance between its parallel sides is 26 ***cm***. If one of
the parallel sides is 84 ***cm*** then, find
the other side.**

Area of the trapezium = 1586 *sq.
cm*

Height *h* = 26 *cm*

One of the parallel sides *a*
= 84 *cm*

The other side *b *= ?

1/2 *h* (*a + b*) = 1586

1/2 × 26 (84 + b) = 1586

84 + b = 1586 / 13 = 122

*b* = 122 – 84 = 38 *cm*

**The other side = 38
cm**

** **

**4. The area of a trapezium is 1080 ***sq.
cm***. If the lengths of its parallel sides are 55.6 ***cm*** and 34.4 ***cm***, find the
distance between them.**

The area of the trapezium = 1080 *sq.cm*

Parallel sides a = 55.6cm

* b* = 34.4 cm

Height *h *= ?

1/2 *h* (*a + b*) = 1080

1/2 *h* (55.6+ 3.4) =
1080

½ *h* × 90.0 = 1080

*h *=1080 / 45 = 24 cm.

**The distance
between them = 24** *cm*

** **

**5. The area of a trapezium is 180 ***sq.
cm*** and its height is 9 ***cm***. If one of
the parallel sides is longer than the other by 6 ***cm***, find the
length of the parallel sides.**

The area of the trapezium = 180 *sq. cm*

Height *h* = 9 *cm*

Parallel sides a = * x cm*

* b* = *x* + 6 *cm*

Area of the trapezium = 1/2 *h*
(*a + b*) = 180

= 1/2 × 9(*x + x* + 6) = 180

= 1/2 × 9 (2*x* + 6) = 180

= 2*x* + 6 = ( 180 / 9) × 2 = 40

2*x* = 40 – 6 = 34

* x* = 34 / 2 = 17

* x* = 17

* a* = 17 *cm*

* b* = (17 + 6) *cm*

= 23 *cm*

**The length of the
parallel sides = 17 cm, 23 cm.**

** **

**6. The sunshade of a window is in the form of
isosceles trapezium whose parallel sides are 81 ***cm*** and 64 ***cm*** and the
distance between them is 6 ***cm***. Find the
cost of painting the surface at the rate of ****₹**** 2 per ***sq. cm.*

The parallel sides *a* =
81*cm*, b = 64* cm*

Height *h* = 6 *cm*

Area of isosceles trapezium = 1/2 *h *(*a + b*) *sq. units*

= 1/2 × 6 (81+ 64) *sq.cm*

= 3 × 145 *sq.cm*

= 435 *sq.cm*

Cost of painting per *sq.
cm* = ₹ 2

Cost of painting 435 *sq.
cm* = ₹ 435 × 2 = 870

**Cost of painting
the surface = ₹ 870 **

** **

**7.
A window is in the form of trapezium whose parallel sides are 105 ***cm*** and 50 ***cm*** respectively
and the distance between the parallel sides is 60 ***cm***. Find the
cost of the glass used to cover the window at the rate of ****₹**** 15 per 100 ***sq.
cm.*

The parallel sides *a* =
105 *cm*, *b* = 50 *cm*

Height *h*
= 60 *cm*

Area of the trapezium = 1/2 *h*
(*a + b*) *sq.units*

= 1/2 × 60 (105+ 50) *sq.cm*

= 30 × 155* sq.cm*

= 4650 *sq.cm *

Cost of the glass used to cover 100 sq.cm = ₹ 15

Cost of the glass used to cover 4650 sq.cm = [ ₹ 15 × 4650 ] /100

= ₹ 6975 / 10 = ₹ 697.50

The cost of the glass used to cover the window = **₹ 697.50**

** **

__Objective
type questions__

**8. The area of the trapezium, if the parallel
sides are measuring 8 ***cm*** and 10 ***cm*** and the
height 5 ***cm*** is**

(i)
45 *sq. cm*

(ii)
40 *sq. cm*

(iii)
18 *sq. cm*

(iv)
50 *sq. cm*

*Answer :***(i) 45 sq cm**

**9. In a trapezium if the sum of the parallel
sides is 10 ***m*** and the area is 140 ***sq.
m,*** then the height is**

(i)
7*cm*

(ii)
40 *cm*

(iii)
14 *cm*

(iv)
28 *cm*

*Answer :***(iv) 28 cm**

**10.
when the non-parallel sides of a trapezium are equal then it is known as**

(i) a
square

(ii) a
rectangle

(iii) an
isosceles trapezium

(iv) a
parallelogram

*Answer :***(iii) an
isosceles trapezium**

**ANSWERS**

**Exercise 2.3 **

1. (i) 160* sq.cm *(ii) 24* cm *(iii)
18* m *(iv) 30 *cm*

2. 330* sq.cm *

3. 38* cm *

4. 24* cm *

5. 23* cm *and 17 *cm*

6. ₹ 870

7. ₹ 697.50

** ****Objective type questions **

8. (i) 45* sq.cm *

9. (iv) 28* cm *

10. (iii) isosceles trapezium

Tags : Questions with Answers, Solution | Measurements | Term 1 Chapter 2 | 7th Maths , 7th Maths : Term 1 Unit 2 : Measurements

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7th Maths : Term 1 Unit 2 : Measurements : Exercise 2.3 (Area of the Trapezium) | Questions with Answers, Solution | Measurements | Term 1 Chapter 2 | 7th Maths

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