6th Maths : Term 3 Unit 3 : Perimeter and Area : Exercise 3.1 : Text Book Back Exercises Questions with Answers, Solution

**Exercise 3.1**

**1.
The table given below contains some measures of the rectangle. Find the unknown
values.**

**i) **Length of rectangle (*l*) = 5 *cm*

breadth (*b*) = 8 *cm*

Perimeter P = 2 ( *l* + *b *) units

= 2 (5 + 8) *cm*

= 2 × 13 *cm* **= 26 cm**

Area A = *l* + *b* sq. units

= 5 × 8 sq. cm

**= 40 sq. cm**

**ii) **Length of rectangle (*l*) = 13 *cm*

Breadth *b* = ?

Perimeter *P* = 2 ( *l* + *b *) units = 54 *cm*

*2 *(13 + *b*) = 54

13 + *b* = 27

*b* = 27 – 13 = **14 cm**

Breadth b = 14 *cm*

Area A = *l* × *b*

= 13 × 14 sq.cm

**= 182 sq.cm**

**iii)** Length of rectangle *l* = ?

Breadth *b* = 15 *cm*

Perimeter P = 2 ( *l* + *b *) = 60 *cm*

*l* + *b * = 30

*l *= 30 – 15 = **15 cm**

Length *l *= 15 *cm*

Area A = *l* × *b* sq.units

**= 15 × 15 = 225** **sq.cm**

**iv) **Length of rectangle *l* = 10* m*

Breadth *b* = ?

Area A = *l* × *b *sq. units = 120 sq.m

10 × *b* = 120

*b *= 120/10 = 12 *m*

Breadth *b* = **12 m**

Perimeter *P* = 2 ( *l* + *b *) units

= 2 ( 10 + 12) *m*

**= 2 × 22 m = 44 m**

**v) ** Length of rectangle *l*
= ?

Breadth *b* = 4 feet

Area A = *l* × *b *sq. units = 20 sq.feet

*l* × 4 = 20

*l* = 20/4 = 5 feet

Length *l* = **5 feet**

Perimeter *P* = 2 ( *l *+ *b* ) units

= 2 ( 5 + 4 ) feets

**= 2 × 9 = 18 feets**

** **

**2.
The table given below contains some measures of the square. Find the unknown values.**

i) Side of square S = 6 *cm*

** Perimeter P = 4S = 4 × 6 = 24 cm**

** Area A = S × S = 6 × 6 =
36 sq.cm**

ii) Perimeter of square P = 4S = 100 *m*

S = **25 m**

Area A = S × S sq.unit

= 25 × 25 = **625 sq.m**

iii) Area of square A = S × S = 49 sq.feet

S × S = 7 × 7

** S = 7 feet**

Perimeter P = 4S = 4 × 7 = **28 feet**

** **

**3.
The table given below contains some measures of the right angled triangle. Find
the unknown values.**

i) In right angled triangle,

Base b = 20 cm

Height h = 40 cm

Area A = ½ bh sq.units

= 1/2 × 20 × 40 sq.cm

**Area = 400 sq.cm = 400 sq.cm **

ii) In right angled triangle,

Base *b* = 5 feet

Height *h* = ?

Area A = ½ bh sq.units = 20 sq.feet

= 1/2 × 5 × *h* = 20

h = 20/5 × 2 = 8 feet

**Height h = 8 feet**

iii) In right angle triangle,

Base b = ?

Height h = 12 *m*

Area A = ½ *bh* sq.units
= 24 sq.m.

= 1/2 × *b* × 12 = 24

*b *= 24 / 6 = 4 *m*

**Base b = 4 m **

** **

**4.
The table given below contains some measures of the triangle. Find the unknown values.**

i) Measures of sides of the triangle,

*a* = 6 *cm*

*b* = 5 *cm*

*c* = 2 *cm*

Perimeter P = *a* + *b* + *c* units

**= 6 + 5 + 2 = 13 cm**

ii) Measures of sides of the triangle,

*a* = ?

*b* = 8 *m*

*c* = 3 *m*

Perimeter P = *a* + *b* + *c* units = 17 *m*

= *a* + 8 + 3 = 17 *m*

= *a *+ 11 = 17

*a* = 17 – 11 = 6 *m*

**Measure of one side a = 6 m**

iii) Measures of sides of the triangle,

*a* =11 feet

*b* = ?

*c* = 9 feet

Perimeter P = *a* + *b* + *c* units = 28 feet

= 11+ *b* + 9 = 28

= *b* + 20 = 28

*b* = 28 − 20 = 8 feet

Measure of one side *b* = **8 feet**.

** **

**5.
Fill in the blanks.**

**i)
5 cm^{2} = **

**ii)
26 m^{2} = **

**iii)
8 km^{2} = **

** **

**6.
Find the perimeter and area of the following shapes.**

**i)**

Measure of one side S = 4 *cm *

Number of sides = 12

Perimeter P = 12 × S = 12 × 4 = 48 *cm*

Measure of one side of the square S = 4 *cm*

Area A = S × S = 4 × 4 = 16 Sq.cm

Number of squares = 5

**Area of 5 squares = 5 × 16 = 80 sq.cm**

**ii)**

Height of right angled triangle = 4* cm *

Hypoteouse = 5 *cm*

Height of one triangle + Hypotenuse = (4 + 5) *cm*

= 9 *cm*

Height and Hypotenuse of 4 right angled triangles = 4 × 9 = 36 *cm*

Perimeter P = 36 *cm*

Area of one right angled triangle A = ½ *bh* sq.units

= 1/2 × 3 × 4 = 6 sq.cm

**Area of four triangles = 4 × 6 = 24 sq.cm **

Measure of one side of the middle square

S = 3 *cm*

Area A = S × S sq.units

= 3 × 3 = 9 sq.cm

Total Area = Area of four triangle + Area of the middle square

**= (24 + 9) sq.cm = 33 sq.cm**

**iii) **

Perimeter of the shape = (50 + 12 + 13 + 40 + 10 + 10 + 15) *cm*

= 150 *cm *

Length of rectangle *l* = 50 *cm*

Breadth *b* = 5 *cm*

Area A = *l × b* = 50 × 5 = 250 sq.cm

Area of triangle A = ½ *bh*

= 1/2 × 5 × 12 = 30 sq.cm

Area of square A = S × S = 10 × 10 = 100 sq.cm

Total area of the shape =
Area of rectangle + Area of triangle + Area of square

= (250 + 30 + 100) sq.cm

**= 380 sq.cm**

** **

**7. Find the perimeter and the area of the rectangle
whose length is 6 ***m*** and breadth is 4 ***m***.**

Length of rectangle *l *= 6 *m*

Breadth *b* = 4 *m*

Perimeter P = 2 (*l* + *b*) units

= 2 (6 + 4) *m*

= 2 × 10 = 20 *m*

Area A = *l* × *b*
sq.units

= 6 × 4 = 24 sq.m

** **

**8. Find the perimeter and the area of the square
whose side is 8 ***cm***.**

Side of the square S = 8 *cm*

Perimeter P = 4 S units

= 4 × 8 = 32 *cm*

Area A = S × S sq.units

= 8 × 8 = 64 sq.cm

** **

**9. Find the perimeter and the area of a right angled
triangle whose sides are 6 feet, 8 feet and 10 feet.**

Measures of the sides of the right angled 6 feet, 8 feet and 10
feet.

Perimeter P = (6 + 8+10) feet

= 24 feet

Area A = ½ *bh* sq.
units

= 1/2 × 6 × 8 sq.cm

**= 24 sq.feet**

** **

**10. Find the perimeter of**

**i) A scalene triangle with sides 7 ***m***, 8 ***m***, 10 ***m***.**

Sides of scalene triangle *a* = 7 *m*, *b* = 8 *m*
and *c* = 10 *m*

Perimeter P = a + b + c units

= (7 + 8 + 10) *m* = **25 m**

**ii) An isosceles triangle with equal sides 10 ***cm*** each and third
side is 7 ***cm***.**

Sides of isosceles triangle, a = 10 cm, b = 10 cm and c = 7 cm

Perimeter P = *a* + *b* + *c* units

= (10 + 10 + 7) *cm* = **27 cm**

**iii) An equilateral triangle with side 6 ***cm***.**

Side of equilateral triangle, *a* = 6 *cm*

Perimeter P = 3 *a* = 3 × 6 = **18 cm**

** **

**11. The area of a rectangular shaped photo is 820
***sq***. ***cm***. and its width
is 20 ***cm***. What is its length?
Also find its perimeter.**

Area of a rectangular shaped photo = 820 sq.cm

*l* × *b* = 820 sq.cm.

width *b* = 20 *cm*

*l* × 20 = 820

*l* = 820 / 20 = 41 *cm *

length *l* = 41 *cm*

Perimeter P = 2 (*l* + *b*) units

= 2 (41+ 20) *cm*

= 2 × 61 = 122 *cm*

**Perimeter P = 122 cm**

** **

**12.
A square park has 40 m as its perimeter.
What is the length of its side? Also find its area.**

Perimeter of a square park P = 4 × S = 40 *m*

Side S = 40 / 4 = 10 *cm*

Side of the square park = 10* m*

Area of the square park = S × S sq.units

= 10 × 10 sq.m

**= 100 sq.m**

** **

**13. The scalene triangle has 40 ***cm*** as its perimeter
and whose two sides are**** ****13 ***cm*** and 15 ***cm***, find the third
side.**

Perimeter of a scalene triangle = 40 *cm*

Measure of two sides *a* = 13 *cm* and *b* = 15 *cm*

The third side C = ?

Perimeter P = *a* + *b* + C = 40 *cm*

13 + 15 + C = 40

28 + C = 40

C = 40 – 28 = 12 *cm*

**The third side = 12 cm**

** **

**14. ****A field is in the
shape of a right angled triangle whose base is 25 ***m*** and height 20
***m***. Find the cost of levelling the field at the rate
of ****₹**** 45/- per sq. ***m*^{2}**.**

Base of a right angled triangle *b* = 25 *m*

Height *h* = 20 *m*

Area A = ½ *bh* units

= 1/2 × 25 × 20 sq.m.

= 250 sq.m.

The cost of levelling the field for l sq.m = ₹ 45

∴ The cost of levelling
the field for 250 sq.m = ₹ 45 × 250

= ₹ 11250

**The cost of levelling the field = ₹ 11250**

** **

**15.
A square of side 2 ***cm*** is joined with
a rectangle of length 15 ***cm*** and breadth 10
***cm***. Find the perimeter of the combined shape.**

The Perimeter of the combined shape = 2 length + 1 breadth of rectangle + 3 sides of square + 8 *cm*

= (2 × 15 + 10 + 3 × 2 + 8) *cm*

= (30 + 10 + 6 + 8) *cm*

= 54 *cm*

**The perimeter of the combined shape = 54 cm**

** **

__Objective
Type Questions__

**16.**** ****The following figures are of equal area. Which figure
has the least perimeter?**

*Answer:* (b)

**17. If two identical rectangles of perimeter 30 cm are joined together, then the perimeter
of the new shape will be**

a) equal to 60 *cm*

b) less than 60 *cm*

c) greater than 60 *cm*

d) equal to 45 *cm*

*Answer:* b) less than 60 cm

**18.****
****If
every side of a rectangle is doubled, then its area becomes _____ times.**

a)
2

b)
3

c)
4

d)
6

*Answer:* c) 4

**19. ****The side of a square
is 10 cm. If its side is tripled, then
by how many times will its perimeter increase?**

a)
2 times

b)
4 times

c)
6 times

d)
3 times

*Answer:* d) 3 times

**20. ****The length and
breadth of a rectangular sheet of a paper are 15 ***cm*** and 12 ***cm*** respectively.
A rectangular piece is cut from one of its corners. Which of the following statement
is correct for the remaining sheet?**

a. Perimeter remains
the same but the area changes

b. Area remains the
same but the perimeter changes

c. There will be a change
in both area and perimeter.

d. Both the area and perimeter remains the same.

*Answer:* c) There will be a change in both area and perimeter.

__ANSWERS:__

__Exercise 3.1 __

**1. i) 26 cm, 40 cm^{2} ii) 14 cm,
182 cm^{2} iii) 15 cm, 225 cm^{2} iv) 12 m, 44
cm v) 5 feet, 18 feet**

**2. i) 24 cm, 36 cm^{2} ii) 25 m, 625 cm^{2}
iii) 7 feet, 28 feet **

**3. i) 400 cm^{2} ii) 8 feet iii) 4 m **

**4. i) 13 m ii) 6 m iii) 8 feet **

**5. i) 500 ii) 2,60,000
iii) 80,00,000 **

**6. i) 48 cm, 80 cm^{2} ii) 36 cm,
49 cm^{2} iii) 150 cm, 380 cm^{2} **

**7. 20 m, 24 m^{2}**

**8. 32 cm, 64 cm^{2} **

**9. 24 feet, 24 sq.
feet **

**10. i) 25 m ii) 27 cm iii) 18 cm **

**11. 41 cm, 122 cm**

**12. 10 m, 100 m^{2}
**

**13. 12 cm **

**14. 250 m^{2},
₹11250/- **

**15. 54 cm **

**16. b) **

**17. b) less than 60 cm **

**18. c) 4 times **

**19. d) 3 times **

**20. c) Both the area
& perimeter are changed**

** **

Tags : Questions with Answers, Solution | Perimeter and Area | Term 3 Chapter 3 | 6th Maths , 6th Maths : Term 3 Unit 3 : Perimeter and Area

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6th Maths : Term 3 Unit 3 : Perimeter and Area : Exercise 3.1 | Questions with Answers, Solution | Perimeter and Area | Term 3 Chapter 3 | 6th Maths

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