RS Aggarwal Class 6 Solutions Chapter 21 - Concept Of Perimeter And Area Ex 21B(21.3)

RS Aggarwal Class 6 Chapter 21 - Concept Of Perimeter And Area Ex 21B(21.3) Solutions Free PDF

Mark correct against the correct answer in each of the following:

1. The sides of a rectangle are in a ratio 7 : 5 and its perimeter is 96 cm. the length of the rectangle is

(a) 21 cm (b) 28 cm (c) 35 cm (d) 14 cm

(b) 28 cm

Let the length and the breadth of the rectangle be 7x cm and 5x cm, respectively.

It is given that the perimeter of the rectangle is 96 cm.

Perimeter of the rectangle = 2(7x + 5x) cm

$\Rightarrow 2\left ( 7x + 5x \right )= 96$

= 2(12x) = 96

= 24x = 96

$\Rightarrow x = \frac {96} {24} = 4$

Therefore, $Length = \left ( 7\times 4 \right )cm = 28 cm$

2. The area of the rectangle is 650 cm2 and the breadth is 13 cm. the perimeter of the rectangle is

(i) 63 cm (ii) 130 cm (iii) 100 cm (iv) 126 cm

(iv) 126 cm

Let the length of the rectangle be L cm

Area of the rectangle = 650 cm2

Area of the rectangle = (L Ã— 13) cm2

$\Rightarrow \left ( L\times 13 \right )= 650$ $\Rightarrow L = \frac {650} {13} = 50$

Length of the rectangle is 50 cm

Perimeter of the rectangle = 2(length + breadth) cm

= 2(50 + 13) cm

= 126 cm

3. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 22.50 per meter is

(a) Rs 2430 (b) Rs 2340 (c) 2400 (d) 3340

(b) Rs 2340

Perimeter of the rectangle = 2(length + breadth) cm

= 2(34 + 18) m

= 104 m

Cost of fencing 1 m = Rs 22.50

Cost of fencing 104 m = Rs (22.50 Ã— 104) = Rs 2340

4. The Cost of fencing a rectangular field at Rs 30 per meter is Rs 2400. If the length of the field is 24 m, then its breadth is

(a) 8 m (b) 16 m (c) 18 m (d) 24 m

(b) 16 m

Total cost of fencing = Rs 2400

Rate of fencing = Rs 30/m

$Perimeter \; of \; the\; rectangular \; field = \frac {Total \; cost} {Rate} = \frac {Rs \; 2400} {Rs 30/m} = 80 m$

Let the breadth of the rectangular field be x m.

Perimeter of the rectangle = 2(24 + x) m

$\Rightarrow 2\left ( 24 + x \right )= 80$ $\Rightarrow 48 + 2x = 80$ $\Rightarrow 2x = \left ( 80 – 48 \right )$ $\Rightarrow 2x = 32$ $\Rightarrow x = 16$

So, the breadth of the rectangular field is 16 m.

5. The area of the rectangular carpet is 120 m2 and its perimeter is 46 m. The length of its diagonal is

(i) 15 m (ii) 16 m (iii) 17 m (iv) 20 m

Hint. L + B = 23 and lb = 120

(l2 + b2) = (l + b)2 â€“ 2lb = (23)2 â€“ 2 Ã— 120 = 289

$Diagonal = \sqrt{l^{2} + b^{2}} = \sqrt{289} = \sqrt{17\times 17} = 17$

(iii) 17 m

Let the length and the breadth of the rectangle be L m and B m, respectively.

Area of the rectangular carpet = (L Ã— B) m2

= LB = 120â€¦â€¦â€¦â€¦.(i)

Perimeter of the rectangular carpet = 2(L + B)

= 2(L + B) = 46

= (L + B) = 46/2

= (L + B) = 23â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

Diagonal of the rectangle = $\sqrt{l^{2} + b^{2}} m$ $\sqrt{l^{2} + b^{2} – 2LB} m$ $\sqrt{\left ( 23 \right )^{2} – 240} m\; \left ( from \; equations \left ( i \right )and \left ( ii \right )\right )$

= $\sqrt{ 529 – 240} m$

= $\sqrt{289} m$

6. The length of a rectangle is three times its width and the length of its diagonal is $\sqrt [6] {10}$ cm. the perimeter of the rectangle is

(a) 48 cm (b) 36 cm (c) 24 cm (d) $\sqrt [24] {10}$

(a) 48 cm

Let the width and the length of the rectangle be x cm and 3x cm, respectively.

Applying the Pythagoras theorem:

(Diagonal) 2 = (Length) 2 + (width) 2

$\Rightarrow \left ( \sqrt [6] {10} \right )^{2} = \left ( 3x \right )^{2} + \left ( x \right )^{2}$ $\Rightarrow 360 = 9x^{2} + x^{2}$ $\Rightarrow 360 = 10x^{2}$ $\Rightarrow x^{2} = \frac {360} {10}$ $\Rightarrow x^{2} = 36$ $\Rightarrow x = pm 6$

Since, the width cannot be negative, we will neglect -6.

So, width of the rectangle is 6 cm

Length of the rectangle = (3 Ã— 6) = 18 cm

Perimeter of the rectangle = 2(length + breadth)

= 2(18 + 6) = 48 cm

7. if the ratio between the length and the perimeter of a rectangular plot is 1 : 3, then the ratio between the length and breadth of the plot is

(a) 1 : 2 (b) 2 : 1 (c) 3 : 2 (d) 2 : 3

Hint. Let the length be x cm, then its perimeter is 3x cm

Therefore, $2\left ( x + b \right )= 3x \Rightarrow 2b = \left ( 3x – 2x \right )\Rightarrow b = \frac {x} {2}$

Therefore, $l : b = x : \frac {x} {2} = 2x : x = 2 : l$

(b) 2 : 1

Let the breadth of the plot be b cm.

Let the length of the plot be x cm.

Perimeter of the plot = 3x cm

Perimeter of the plot = 2(length + breadth) = 2(x + b) cm

= 2(x + b) = 3x

2x + 2b = 3x

2b = 3x â€“ 2x

2b = x

b = x/2

Therefore, Ratio ofÂ  the length and the breadth ofÂ  the plot = $\frac {x} {\frac {x} {2}} = \frac {x} {x}\times 2 = \frac {2} {1}$

Therefore, Ratio ofÂ  the length and the breadth ofÂ  the plot = 2 : 1

8. The length of the diagonal of a square is 20 cm. Its area is

(a) 400 cm2 (b) 200 cm2 (c) 300 cm2 (d) $\sqrt [100] {2} cm^{2}$

(b) 200 cm2

$Area \; of \; the \; square = \left [ \frac {1} {2}\times \left ( Diagonal \right )^{2} \right ] sq \; units$ $\left [\frac {1} {2}\times 20^{2} \right ] cm^{2}$ $\left [\frac {1} {2}\times 20 \timesÂ 20 \right ] cm^{2}$

(20 Ã— 10) cm2

= 200 cm2

9. The cost of putting the fence around a square field at Rs 25 per meter is Rs 2000. The length of each side of the field is

(a) 80 m (b) 40 m (c) 20 m (d) none of these

(c) 20 m

Let one side of the square field be x m.

Total cost of fencing a square field = Rs 2000

Rate of fencing the field = Rs 25/m

$Perimeter : of \; the \; square \; field = \frac {Total \; cost \; of \; fencing \; the \; field} {Rate \; of \; fencing \; the \; field} = \frac {Rs 2000} {Rs 25/m} = \frac {2000} {25} m = 80 m$

Perimeter of the square field = (4 Ã— side) = 4x m

= 4x = 80

= x = 20

Each side of the field is 20 m

10. The diameter of a circle is 7 cm. its circumference is

(a) 44 cm (b) 22 cm (c) 28 cm (d) 14 cm

(b) 22 cm

$Radius = \frac {Diameter} {2} = \frac {7} {2} cm$ $Circumference \; of \; the \; circle = 2\pi r = \left ( 2\times \frac {22} {7}\times \frac {7} {2} \right )cm$

= 22 cm

11. The circumference of a circle is 88 cm. its diameter is

(a) 28 cm (b) 42 cm (c) 56 cm (d) none of these

(a) 28 cm

Circumference of the circle is 88 cm

Let the radius be r cm.

It is given that the circumference of the circle is (2Ï€r) cm

= 2Ï€r = 88

$\Rightarrow 2\times \frac {22} {7}\times r = 88$ $\Rightarrow r = \frac {1} {2}\times \frac {7} {22}\times 88$

r = 14

Diameter = (2 Ã— radius) = (2 Ã— 14) cm = 28 cm

12. The diameter of a wheel of a car is 70 cm. how much diameter will it cover in making 50 revolutions?

(a) 350 m (b) 110 m (c) 165 m (d) 220 m

(b) 110 m

$Radius \; of \; the \; wheel = \frac {Diameter} {2} = \frac {70} {2} = 35 cm$ $Circumference \; of \; the \; wheel = 2\pi r = \left ( 2\times \frac {22} {7}\times 35 \right )cm = 220 cm$

The distance covered by the wheel in one revolution is equal to its circumference.

Distance covered by the wheel in 1 revolution = 220 cm

Therefore, Distance covered by the wheel in 50 revolution = (50 Ã— 220) cm = 11000 cm = 110 m

13. A lane 150 m long and 9 m wide is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. how many bricks are required?

(a) 65000 (b) 70000 (c) 75000 (d) 80000

(d) 80000

Length of the road = 150 m = 15000 cm

= (15000 Ã— 900) cm2

= 13500000 cm2

Length of the brick = 22.5 cm

Breadth of the brick = 7.5 cm

Area of one brick = (length Ã— breadth)

= (22.5 Ã— 7.5) cm2

= 168.75 cm2

$Number \; of \; bricks = \frac {Area \; of \; the \; road} {Area \; of \; one \; brick} = \frac {13500000} {168.75} = 80000 bricks$

14. A room is 5 m 40 cm long and 4 m 50 cm broad. Its area is

(a) 23.4 m2 (b) 24.3 m2 (c) 25 m2 (d) 98.01 m2

(b) 24.3 m2

Length of the room is = 5 m 40 cm = 5.40 m

Breadth of the room is = 4 m 50 cm = 4.50 m

Area of the room = (length Ã— breadth) = (5.40 Ã— 4.50) m2

= $\left ( \frac {540} {100}\times \frac {450} {100} \right )m^{2}$

= $\left ( \frac {27} {5}\times \frac {9} {2} \right )m^{2}$ $\left ( \frac {243} {10} \right )m^{2} = 24.3 m^{2}$

15. How many envelopes can be made out of a sheet of paper 72 cm by 48 cm. if each envelope requires a paper of size 18 cm by 12 cm?

(a) 4 (b) 8 (c) 12 (d) 16

(d) 16

Length of the sheet paper is = 72 cm

Breadth of the sheet of paper is = 48 cm

Area of the sheet = (length Ã— breadth) = (72 Ã— 48) cm2

= 3456 cm2

Length of the piece of the paper required to make 1 envelope = 18 cm

Breadth of the piece of the paper required to make 1 envelope = 12 cm

Area of the piece of the paper required to make 1 envelope = (18 Ã— 12) cm2 = 216 cm2

$No. \; of \; envelope \; that \; can \; be \; made = \frac {Area \; of \; the \; sheet} {Area \; of \; the\; piece\; of \; paper \; required\; to \; make \; 1 \; envelope}$ $\Rightarrow No. \; of \; envelope \; that \; can \; be \; made = \frac {3456} {216} = 16 \; envelopes$’

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