RS Aggarwal Solutions Class 6 Ex 21B

1. Find the area of a rectangle whose

(i) Length = 46 cm and Breadth = 25 cm

Solution:

Length = 46 cm

Breadth = 25 cm

Area of rectangle = (length × breadth) sq units

= (46 × 25) cm2 = 1150 cm2

(ii) Length = 9 m and Breadth = 6 m

Solution:

Length = 9 m

Breadth = 6 m

Area of rectangle = (length × breadth) sq units

= (9 × 6) m2 = 54 m2

= (46 × 25) cm2 = 1150 cm2

(iii) Length = 9 m and Breadth = 6 m

Solution:

Length = 14.5 m

Breadth = 6.8 m

Area of rectangle = (length × breadth) sq units

\( \left ( \frac {145} {10} \times \frac {68} {10} \right ) m^{2} = \frac {9860} {100} m^{2} = 98.60 m^{2} \)

(iv) Length = 2 m 5 cm and Breadth = 60 cm

Solution:

Length = 2 m 5 cm

= (200 + 5) cm (1 m = 100 cm)

= 205 cm

Breadth = 60 cm

Area of rectangle = (length × breadth) sq units

= (205 × 60) cm2 = 12300 cm2

(v) Length = 3.5 km and Breadth = 2 km

Solution:

Length = 3.5 km

Breadth = 2 km

Area of rectangle = (length × breadth) sq units

= (3.5 × 2) km2

= \( \left ( \frac {35} {10} \times 2\right ) km^{2} = 7 km^{2} \)

2. Find the area of the square plot of side 14 m.

Solution:

Side of the square plot = 14 m

Area of the plot = (side) 2sq units

= (14)2 m2 = 196 m2

3. The top of the table measures 2 m 25 cm by 1 m 20 cm. find its area in square meters.

Solution:

Length of the table = 2 m 25 cm

= (2 + 0.25) m (100 cm = 1 m)

= 2.25 m

Breadth of the table = 1 m 20 cm

= (1 + 0.20) m (100 cm = 1 m)

= 1.20 m

Area of the table = (length × breadth) sq units

= (2.25 × 1.20) m2

= \( \left ( \frac {225} {100} \times \frac {120} {100}\right ) m^{2} = 2.7 m^{2} \)

4. A carpet is 30 m 75 cm long and 80 cm wide. Find its cost at Rs 150 per square metre.

Solution:

Length of the carpet = 30 m 75 cm

= (30 + 0.75) cm (100 cm = 1 m)

= 30.75 m

Breadth of the table = 80 cm

(100 cm = 1 m)

= 0.80 m

Area of the carpet = (length × breadth) sq units

= (30.75 × 0.80) m2

= \( \left ( \frac {3075} {100} \times \frac {80} {100}\right ) m^{2} = 24.6 m^{2} \)

Cost of 1 m2 carpet = Rs 150

Cost of 24.6 m2 carpet = Rs (24.6 × 150) = Rs 3690

5. How many envelopes can be made out of a sheet of paper 3 m 24 cm by 1 m 72 cm? If each envelope requires a piece of paper of size 18 cm by 12 cm?

Solution:

Length of the sheet of paper = 3 m 24 cm = 324 cm

Breadth of the sheet of paper = 1 m 72 cm = 172 cm

Area of the sheet = (length × breadth)

= (324 × 172) cm2

= 55728 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 × 12) cm2

= 216 cm2

\( No. \; of \; envelopes \; that \; can \; be \; made = \frac {Area \; of \; the \; sheet} {Area \; of \; the \; piece \; required \; to \; make \; 1 \; envelope} \)

\( \Rightarrow No. \; of \; envelopes \; that \; can \; be \; made = \frac {55728} {216} = 258 \; envelopes \)

6. A room is 12.5 m long and 8 m wide. A square carpet of side 8 m is laid on its floor. Find the area of the floor which is not carpeted.

Solution:

Length of the room = 12.5 m

Breadth of the room = 8 m

Area of the room = (length × breadth)

= (12.5 × 8) m2

= 100 m2

Side of the square carpet = 8 m

Area of the carpet = (side) 2

= 82 m2

= 64 m2

Area of the floor which is not carpeted = Area of the room – Area of the carpet

= (100 – 64) m2

= 36 m2

7. A lane, 150 m long and 9 m wide, is to be paved with bricks, each measuring 22.5 cm by 7.5 cm. Find the number of bricks required.

Solution:

Length of the road = 150 m = 15000 cm

Breadth of the road = 9 m = 900 cm

Area of the road = (length × breadth)

= (15000 × 900) cm2

= 13500000 cm2

Length of the brick = 22.5 cm

Breadth of the brick = 7.5 cm

Area of the brick = (length × breadth)

= (22.5 × 7.5) cm2

= 168.75 cm2

\( No. \; of \; bricks = \frac {Area \; of \; the \; road} {Area \; of \; one \; brick} = \frac {13500000} {168.75} = 80000 \; bricks \)

8. A room is 13 m long and 9 m broad. Find the cost of carpeting the room with carpet 75 cm broad at the rate of Rs 65 per metre.

Solution:

Length of the room = 13 m

Breadth of the room = 9 m

Area of the room = (length × breadth)

= (13 × 9) m2

= 117 m2

Let the length of the required carpet be x m.

Breadth of the carpet = 75 cm

= 0.75 cm (100 cm = 1 m)

Area of the carpet = (0.75 × x) m2

= 0.75x m2

For carpeting the room:

Area covered by the carpet = Area of the room

\( \Rightarrow 0.75x = 117 \)

\( \Rightarrow x = \frac {117} {0.75} \)

\( \Rightarrow x = 117 \times \frac {4} {3} \)

\( \Rightarrow x = 156 m \)

So, the length of the carpet is 156 m.

Cost of 1 m carpet = Rs 65

Cost 156 m carpet = Rs (156 × 65)

= Rs 10140

9. The length and the breadth of the rectangular park are in a ratio 5 : 3 and its perimeter is 128 m. Find the area of the park.

Solution:

Let the length of the rectangular park be 5x.

Therefore, breadth of the rectangular park = 3x

Perimeter of the rectangular field = 2 (Length + Breadth)

= 2 (5x + 3x)

= 16x

It is given that the perimeter of the rectangular park is 128 m.

\( \Rightarrow 16 x = 128 \)

\( \Rightarrow = \frac {128} {16} \)

\( \Rightarrow x = 8 \)

Length of the park = (5 × 8) m

= 40 m

Breadth of the park = (3 × 8)

= 24 m

Area of the park = (length × Breadth) sq units

= (40 × 24) m2

= 960 m2

10. Two plots of land have the same perimeter. One is a square of side 64 m and the other is a rectangle of length 70 m. Find the breadth of the rectangle plot. Which plot has the greater area and by how much?

Solution:

Side of the square plot = 64 m

Perimeter of the square plot = (4 × side) m = (4 × 64) m = 256 m

Area of the square plot = (side) 2

= (64)2 m2

= 4096 m2

Let the breadth of the rectangular plot be x m.

Perimeter of the rectangular plot = 2(Length + Breadth) m

= 2(70 + x) m

Perimeter of the rectangular plot = Perimeter of the square plot (given)

\( \Rightarrow 2 \left ( 70 + x \right ) = 256 \)

\( \Rightarrow 140 + 2x = 256 \)

\( \Rightarrow 2x = 256 – 140 \)

\( \Rightarrow 2x = 116 \)

\( \Rightarrow x = \frac {116} {2} = 58 \)

So, the breadth of the rectangular plot is 58 m.

Area of the rectangular plot = (length × breadth)

= (70 × 58) m2

= 4060 m2

Area of the square plot – Area of the rectangular plot

= (4096 – 4060)

= 36 m2

Area of the square plot is 36 m2 greater than rectangular plot.

11. The cost of the cultivating a rectangular field at Rs 35 per sq m is Rs 71400. If the width of the field is 40 m, find its length. Also, find the cost of fencing the field at Rs 50 per metre.

Solution:

Total cost of cultivating the field = Rs 71400

Rate of cultivating the field = Rs 35/m2

\( Area \; of \; the \; field = \frac {Total \; cost \; of \; cultivating \; the \; field} {Rate \; of \; cultivating} = \frac {Rs \; 71400} {Rs \; 35/m^{2}} = 2040 m^{2} \)

Let the length of the field be x m.

\( Area \; of \; the \; field = \left ( Length \times Width \right ) m^{2} = \left ( x \times 40 \right ) m^{2} = 40x m^{2} \)

It is given that the area of the field is 2040 m2.

\( \Rightarrow 40x = 2040 \)

\( \Rightarrow x = \frac {2040} {40} = 51 \)

Therefore, \(Length \; of \; the \; field = 51 m \)

Perimeter of the field = 2(L + B)

= 2(51 + 40) m

= 182 m

Cost of fencing 1 m of the field = Rs 50

Cost of fencing 182 m of the field = Rs (182 × 50)

= Rs 9100

12. The area of the rectangle is 540 cm2 and its length is 36 cm. Find its width and perimeter

Solution:

Let the width of the rectangle be x cm.

Length of the rectangle = 36 cm

Area of the rectangle = (Length × Width) = (36 × x) cm2

It is given that the area of the rectangle is 540 cm2

\( \Rightarrow 36 \times x = 540 \)

\( \Rightarrow x = \frac {540} {36} \)

\( \Rightarrow x = 15 \)

Therefore, Width of the rectangle = 15 cm

Perimeter of the rectangle = 2(Length + Breadth) cm

= 2(36 + 15) cm

= 102 cm

13. A marble tile measures 12 cm × 10 cm. How many tiles will be required to cover a wall of size 4 m by 3 m? Also, find the total cost of the tiles at Rs 22.50 per tile.

Solution:

Length of the wall = 4 m = 400 cm

Breadth of the wall = 3 m = 300 cm

Area of the room = (length × breadth)

= (400 × 300) cm2

= 120000 cm2

Length of the tile = 12 cm

Breadth of the tile = 10 m

Area of the tile = (length × breadth)

= (12 × 10) cm2

= 120 cm2

\( Number \; of \; tiles \; required \; to \; cover \; the \; wall = \frac {Area \; of \; the \; wall} {Area \; of \; one \; tile} = \frac {120000} {120} = 1000 \; tiles \)

Cost of 1 tile = Rs 22.50

Cost of 1000 tiles = (1000 × 22.50) = Rs 22500

Thus, the total cost of the tiles is Rs 22500.

14. Find the perimeter of the rectangle whose area is 600 cm2 and breadth is 25 cm.

Solution:

Let the length of the rectangle be x cm.

Breadth of the rectangle is 25 cm

Area of the rectangle = (length × breadth) cm2

= (x × 25) cm2

= 25x cm2

It is given that the area of the rectangle is 600 cm2

\( \Rightarrow 25 \times x = 600 \)

\( \Rightarrow x = \frac {600} {25} \)

\( \Rightarrow x = 24 \)

Therefore, length of the rectangle is 24 cm

Perimeter of the rectangle = 2(Length + Breadth) units

= 2(25 + 24) cm

= 98 cm

15. Find the area of a square whose diagonal is \( \sqrt [5] {2} cm. \)

Hint. \( Area = \left [ \frac {1} {2} \times \left ( diagonal \right )^{2} \right ] sq \; units. \)

Solutions:

\( Area \; of \; the \; square = \left [ \frac {1} {2} \times \left ( diagonal \right )^{2} \right ] sq \; units. \)

= \( \left [ \frac {1} {2} \times \left ( \sqrt [5] {2} \right )^{2} \right ] cm^{2}. \)

= \(\left [ \frac {1} {2} \times \left ( 5 \right )^{2} \times \left ( \sqrt {2} \right )^{2}\right ] cm^{2}. \)

= \( \left [ \frac {1} {2} \times 25 \times 2\right ] cm^{2}. \)

= \( \left ( \frac {1} {2} \times 50 \right ) cm^{2} = 25 cm^{2} \)

16. Calculate the area of each one of the shaded region given below:

(i)

Solution:

Area of the rectangle ABCD = Length × Breadth

= AB × AC (AC = AE – CE)

= (1 × 8) m2

= 8 m2

Area of the rectangle CEFG = Length × Breadth

= CG × GF (CG = GD – CD)

= (9 × 2) m2

= 18 m2

Area of the complete figure = Area of the rectangle ABCD + Area of the rectangle CEFG

= (8 + 18) m2

= 26 m2

(ii)

Solution:

Area of the rectangle AEDC = Length × Breadth

= ED × CD

= (12 × 2) m2

= 24 m2

Area of the rectangle FJIH = Length × Breadth

= HI × IJ

= (1 × 9) m2

= 9 m2

Area of the rectangle ABGF = Length × Breadth

= AB × AF {(AB = FJ – GJ) and AF = EH – (EA + FH)}

= (7 × 1.5) m2

= 10.5 m2

Area of the complete figure = Area of the rectangle AEDC + Area of the rectangle FJIH + Area of the rectangle ABGF

= (24 + 9 + 10.5) m2

= 43.5 m2

(iii)

Solution:

Area of the shaded portion = Area of the complete figure – Area of the unshaded figure

= Area of the rectangle ABCD – Area of the rectangle GBFE

= (CD × AD) – (GB × BF)

= {(12 × 9) – (7.5 × 10)} m2 (BF = BC – FC)

= (108 – 75) m2

= 33 m2

17. Calculate the area of each one of the shaded region given below (all measures are in cm):

(i)

Solution:

Area of the square BCDE = (side) 2

= (CD) 2

= (3)2 cm2

= 9 cm2

Area of the rectangle ABFK = Length × Breadth

= AB × AK {(AB = AC – BC) and (AK = AL + LK)}

= (5 × 1) cm2

= 5 cm2

Area of the rectangle MLKG = Length × Breadth

= ML × MG

= (2 × 3) cm2

= 6 cm2

Area of the rectangle JHGF = Length × Breadth

= JH × HG

= (2 × 4) cm2

= 8 cm2

Area of the complete figure = Area of the rectangle ABFK + Area of the rectangle MLKG + Area of the rectangle JHGF + Area of the square BCDE

= (9 + 5 + 6 + 8) cm2

= 28 cm2

(ii)

Solution:

Area of the rectangle CEFG = Length × Breadth

= EF × CE

= (1 × 5) cm2 (CE = EA-AC)

= 5 cm2

Area of the rectangle ABDC = Length × Breadth

= HI × IJ

= (1 × 2) cm2

= 2 cm2

Area of the rectangle HIJG = Length × Breadth

= HI × IJ

= (1 × 2) cm2

= 2 cm2

= 8 cm2

Area of the complete figure = Area of the rectangle CEFG + Area of the rectangle HIJG + Area of the rectangle ABDC

= ( 5 + 2 + 2) cm2

= 9 cm2

(iii)

Solution:

In the figure, there are 5 squares, each of whose sides are 6 cm in length.

Area of the figure = 5 × Area of square

= 5 × (side) 2

= 5 × (6)2cm2

= 180 cm2


Practise This Question

You got Homework from your tutor that, how can you draw a triangle that will have exactly two equal length sides or what if you need to find the centre of a circle that passes through a given set of points?
It turns out that a single line called the perpendicular bisector can be useful in both these problems.
The given statements show how perpendicular bisector of a line segment AB behaves:

Statement 1: Cut the line segment AB into two equal pieces.
Statement 2       : Makes 180° angle with the line segment AB.