7th Maths : Term 1 Unit 2 : Measurements : Exercise 2.4 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise 2.4**

** **

__Miscellaneous Practice problems__

** **

**1. The base of the parallelogram is 16 ***cm*** and the
height is 7 ***cm*** less than its
base. Find the area of the parallelogram.**

The base of the parallelogram *b* = 16* cm*

Height *h* = 7 *cm* less than its base

= (16 – 7) *cm* = 9* cm*

Area of the parallelogram = *b
× h sq.unit*

= 16 × 9 *sq.cm*

= 144 *sq.cm*

**Area of the
parallelogram = 144 sq.cm**

** **

**2. An****
****agricultural
field is in the form of a parallelogram, whose area is 68.75 ***sq.
hm***. The distance between the parallel sides is 6.25 ***hm***. Find the
length of the base.**

The area of the parallelogram = 68.75 *sq.hm*

Height* h* = 6.25 *h.m*

base *b* = ?

Area of the parallelogram *b
× h* = 68.75

* b × *6.25 = 68.75

*b* = 68.75 / 6.25 *h.m*

* b* = 6875 / 625 *h.m* = 11 *hm*

The length of the base = 11*
hm*

** **

**3. A square and a parallelogram have the same
area. If the side of the square is 48 ***m*** and the height of the parallelogram is 18 ***m,*** find the
length of the base of the parallelogram.**

Side of the square *a* =
48 *m*

Area of the square = *a*^{2
}= 48 × 48 *sq.m*

= 2304 *sq.m*

Area of the parallelogram *b*
× *h* = 2304 *sq.m*

height *h *= 18 *m *

base = Area / height = 2304 / 18 = 128* m*

The length of the base = **128
m**

** **

**4. The height of the parallelogram is one
fourth of its base. If the area of the parallelogram is 676 ***sq.
cm***, find the height and the base.**

Let base be *b* *cm*

The height of the parallelogram = 1/4 of it base = 1/4 × *b* = *b*/4
*cm*

Area of the parallelogram = *b
× h sq. units*

= *b* × *b*/4 = 676

*b* × *b* = 676 × 4

= 2 × 2 × 13 × 13 × 2 × 2

*b *= 2 × 2 × 13 = 52* cm*

*h *= 1/4 × b = 1/~~4~~ × 52
= 13 *cm*

**The height h = 13 cm **

**The base b = 52 cm**

** **

**5. The area of the rhombus is 576 ***sq.
cm*** and the length of one of its diagonal is half of the length of the
other diagonal then find the length of the diagonals.**

Length of one diagonal d_{1} = *d cm*

Length of the other diagonal d_{2} = 1/2 of the length
of the first diagonal

= 1/2 × *d cm*

Area of the rhombus = 1/2 d_{1} × d_{2 }*sq. units*

= 576 *sq.cm*

=1/2 × d × 1/2 × d = 576

d × d = 576 × 2 × 2

= 2 × 2 × 12 × 12 × 2 × 2

d = 2 × 2 × 12 = 48 *cm*

One diagonal d_{1 }= 48 *cm*

Other diagonal d_{2} = 1/2 × 48 = 24 *cm*

**The length of the
diagonals d _{1} = 48 cm **

** d _{2} = 24 cm**

** **

**6.
A ground is in the form of isosceles trapezium with parallel sides measuring 42
***m*** and 36 ***m*** long. The distance between the parallel sides
is 30 ***m***. Find the cost of levelling it at the rate of ****₹**** 135 per ***sq.m.*

The parallel sides *a* =
42 *m*

* b* = 36 *m*

height *h* = 30 *m*

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

= 1/2 × 30 (42 + 36) *sq.m*

= 15 × 78 *sq.m*

Area of the Trapezium = 1170 *sq.
m*

The cost of levelling per *sq.m*
= ₹ 135

The cost of levelling 1170 *sq.m*
= ₹ 1170 × 135

= ₹ 1,57,950

**The cost of
levelling the ground = ₹ 1,57,950**

** **

__Challenge
Problems__

** **

**7.
In a parallelogram PQRS (see the diagram) PM and PN are the heights
corresponding to the sides QR and RS respectively. If the area of the
parallelogram is 900 ***sq. cm*** and the
length of PM and PN are 20 ***cm*** and 36 ***cm*** respectively,
find the length of the sides QR and SR.**

The area of the parallelogram *b × h* = 900 *sq. cm*

height *h*
= 20 *cm*

base b = *QR = *Area / height = 900 / 20 = 45 *cm*

QR = 45 *cm*

Height *h* = 36 *cm *

base *b = SR = *Area /
height = 900 / 36 = 25 *cm*

SR = 25 cm

**The length of the
side QR = 45 cm **

**The length of the
side SR = 25 cm**

** **

**8.
If the base and height of a parallelogram are in the ratio 7:3 and the height
is 45 ***cm ***then, find the
area of the parallelogram.**

The base and the height of a parallelogram are in the ratio = 7:3

Let the base *b* = 7*x*

The height *h* = 3*x* = 45 *cm*

*x* = 45 / 3 = 15 cm

base b = 7*x* = 7 × 15 =
105 *cm*

The area of the parallelogram = *b × h sq. units *

= 105 × 45 *sq.cm*

= 4725 *sq.cm*

**The area of the
parallelogram = 4725 sq.cm**

** **

**9.
Find the area of the parallelogram ABCD, if AC is 24 cm and BE = DF= 8 cm.**

The area of the parallelogram ABCD = Area of Δ ABC + Aea of Δ
ACD

1/2 × AC × DF + 1/2 × AC × BE *sq. units*

1/2 × AC (DF + BE) *sq.units*

1/2 × 24 (8+8) *sq.cm*

12 × 6 *sq.cm* = 192 *sq.cm*

**Area of the
parallelogram ABCD = 192 sq.cm **

** **

**10.
The area of the parallelogram ABCD is 1470 ***sq cm***. If AB = 49 ***cm*** and AD = 35 ***cm*** then, find
the heights DF and BE.**

Area of the parallelogram ABCD

= 1470 *sq.
cm*

Base AB = 49 *cm *

Base × height = 1470

49 × height =1470

height = 1470 / 49 = 30 *cm*

Height DF = 30 *cm*

Base AD = 35 *cm*

Base × height = 1470

35 × height = 1470

Height = 1470 / 35= 42 *cm*

Height BE = 42 *cm*

**Height DF = 30 cm **

**Height BE = 42 cm**

** **

**11.
One of the diagonals of a rhombus is thrice as the other. If the sum of the
length of the diagonals is 24 ***cm***, then find
the area of the rhombus.**

Let the diagonals be d_{1} and d_{2}

d_{1} = d *cm*

d_{2} = thrice of the other diagonals

d_{2} = 3 d *cm*

Sum of the length of the diagonals = d_{1 }+ d_{2}

= d + 3d = 4d *cm*

4d = 24

d = 24 / ~~4~~ = 6 *cm*

d_{1} = 6 cm

d_{2} = 3d = 3 × 6
= 18 *cm*

The area of the rhombus = 1/2 d_{1} × d_{2} *sq.cm*

= 1/2 × 6 ×
18 = 54 *sq. cm*

**Area of the rhombus
= 54 sq.cm**

** **

**12.
A man has to build a rhombus shaped swimming pool. One of the diagonal is 13 ***m*** and the other
is twice the first one. Then find the area of the swimming pool and also find
the cost of cementing the floor at the rate of ****₹**** 15 per ***sq.cm***.**

Let the diagonals be d_{1} and d_{2}

d_{1} = 13 *m*

d_{2} = 2d_{1} = 2 × l3 *m* = 26 *m*

Area of the swimming pool = 1/2 d_{1 }× d_{2} *sq.units*

= 1/2 × 13 × 26 = 169 *sq.m*

The cost of cementing the floor per *sq.m* = ₹ 15

The cost of cementing the floor 169 *sq.m* = 169 × 15 = ₹ 2535

The cost of cementing the floor = ₹ 2535

** **

**13.
Find the height of the parallelogram whose base is four times the height and
whose area is 576 ***sq. cm***.**

Let the height be *h cm*

base *b = *4 times the
height

= 4 h *cm*

Area of the parallelogram = *b
× h = 576 sq. cm*

* h* × 4* h* = 576

*h* × *h* = 576 / ~~4~~ =
144 = 12 × 12

*h* = 12 *cm*

**Height h = 12 cm**

** **

**14.
The table top is in the shape of trapezium with measurements given in the
figure. Find the cost of the glass used to cover the table at the rate of ****₹**** 6 per 10 ***sq.
cm.*

Parallel sides of the trapezium a = 150 *cm*

b = 200 *cm*

height h = 50 *cm *

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

= 1/2 × 50 (150 + 200) *sq.cm*

= 25 × 350 *sq.cm*

= 8750 *sq.cm*

The scost of the glass per 10 *sq.cm* = ₹ 6

The scost of the glass per 8750 *sq.cm* = ₹ 6/10 × 8750

= ₹5250

**The cost of the
glass used to cover the table=₹5250**

** **

**15.
Arivu has a land ABCD with the measurements given in the figure. If a portion
ABED is used for cultivation (where E is the mid-point of DC), find the
cultivated area.**

**Answer: **

Parallel sides AB = a = 24*cm*

DE = b = 12*m*

height AD = h = 18 *m *

The cultivated area ABCD = 1/2 *h*(a+b) *sq.unit s*

= 1/2
× 18 (24+12) *sq.m*

= 9 × 36
=324 *sq.m*

** The cultivated area = 324 sq.m**

**ANSWERS**

**Exercise 2.4 **

1. 144* sq.cm *

2. 11 *hm*

3. 128* m *

4. h = 13*cm* b = 52*cm*

5. *d*_{1} = 48* cm d*_{2}
= 24*cm*

6. ₹ 1,57,950

** Challenge Problems **

7. 45 *cm*; 25* cm *

8. 4725* sq.cm*

* *9. 192* sq.cm *

10. DF = 30*cm* BE = 42*cm*

11. 54* sq.cm *

12. 169 sq.*cm*; ₹ 2535

13. 12* cm *

14. ₹ 5250

15. 324* sq.m*

* *

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 2.4 | Questions with Answers, Solution | Measurements | Term 1 Chapter 2 | 7th Maths

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