## RS Aggarwal Class 6 Math Twenty-one Chapter Concept Of Perimeter And Area Exercise 21D Solution

## EXERCISE 21D

**(1) Find the area of a rectangle whose**

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

Solution: We know, Area = (length × breadth) sq units

So, Area = (46 × 25) = 1150 cm^{2}

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

Solution: Area = (9 × 6) = 54 m^{2}

(iii) Length = 14.5 m and breadth = 6.8 m

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

Solution: Here, Length = 2 m 5 cm = 205 cm

Area = (205 × 60) = 1230 cm^{2}

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

**(2) Find the area of square plot of side 14 m.**

Solution: We know, Area of the square = (side)^{2} = (14)^{2} = 196 m^{2}

**(3) The top of a table measures 2 m 25 cm by 1 m 20 cm. Find its area in square metres.**

Solution: Here, 2 m 25 cm = 225 cm and 1 m 20 cm = 120 cm

Area of the table = (225 × 120) = 27000 cm^{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 metres.**

Solution: Here, Length = 30 m 75 cm = 30.75 m and wide = 0.80 m

Area of the field = (30.75 × 0.80) = 24.6 m^{2}

Rate of cultivation = Rs 150 per m^{2}

∴ cost of cultivation = 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: Here, Length of the sheet = 3 m 24 cm = 324 cm and breadth of the sheet = 1 m 72 cm = 172 cm

Area of the sheet = (324 × 172) = 55728 cm^{2}

And, Length of the each envelope = 18 cm and Breadth of the each envelope = 12 cm.

**(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: Here, the area of the room = (12.5 × 8) = 100 m^{2}

And, Area of the square carpet = (8)^{2} = 64 m^{2}

Hence, the area of the floor which is not carpeted = (100 – 64) = 36 m^{2}

**(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: Here, the area of the lane = (15000 × 900) = 13500000 cm^{2}

And, the area of the bricks = (22.5 × 7.5) = 168.75 cm^{2}

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

Solution: Length of the room = 13 m = 1300 cm and its breadth = 9 m = 900 cm

Area of the floor of the room = (1300 × 900) = 1170000 cm^{2}

Area of the carpet required = 1170000 cm^{2}

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

Solution: Let, the length of the park = 5x and its breadth = 3x

So, perimeter of the park = 2(5x+3x)

.

Hence the length of the park = (5 × 8) = 40 m and its breadth = (3 × 8) = 24 m

∴ Area of the park = (40 × 24) = 960 m^{2}

**(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 70m. Find the breadth of the rectangular plot. Which plot has the greater area and by how much?**

Solution: Perimeter of the square = (4 × 64) = 256 m

Let, the breadth of the rectangle is x m.

Hence, the breadth of the rectangle plot is 58 m.

Now, the area of the square plot = (64)^{2} = 4096 m^{2}

And, the area of the rectangle plot = (70 × 58) = 4060 m^{2}

Hence, the square plot is greater than rectangle plot.

**(11) The cost of cultivating a rectangular field at Rs 35 per square meter 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 meter.**

Solution: Total cost of cultivating = Rs 71400.

Hence, the length of the field is 51 m.

∴ Perimeter of the field = 2(51+40) = 182 m

∴ Rate of fencing = Rs 50 per meter

∴ Cost of fencing = Rs (182 × 50)= Rs 9100.

**(12) The area of a rectangle is 540 cm ^{2} and its length is 36 cm. Find its width and perimeter.**

**(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: here, length of the marble tile = 12 cm

And, breadth of the marble tile = 10 cm

∴ Area of the marble tile = (12 × 10) = 120 cm^{2}

Now, length of the wall = 4 m = 400 cm

Breadth of the wall = 3 m = 300 cm

Now, rate of the tile = Rs 22.50 per tile

∴ Total cost of tiles = Rs 1000 × 22.50 = Rs 22500

**(14) Find the perimeter of a rectangle whose area is 600 cm ^{2} and breadth is 25 cm.**

**(16) Calculate the area of each one of the shaded regions given below.**

Solution: (i) Area of the given figure ABCD = AB × AC = 1 × 8 = 8 m^{2}

Area of CEFG = CG × GF = 9 × 2 = 18 m^{2}

Area of the complete figure = ABCD + CEFG = 8 + 18 = 26 m^{2}

(ii) Area of the AEDC = ED × CD = 12 × 2 = 24 m^{2}

Area of the FJIH = HI × IJ = 1 × 9 = 9 m^{2}

Here we can write {(AB = FJ – GJ) and AF = Eh – (EA + FH)}

Area of the ABGF = AB × AF = 7 × 1.5 = 10.5 m^{2}

Area of the complete figure = AEDC + FJIH + ABGF = (24+ 9 +10.5) = 43.5 m^{2}

(iii) Area of the shaded potion = Area of the figure – Area of the unshaded figure

= Area of ABCD – Area of GBFE

= (CD × AD) – (GB × BF)

= (CD × AD) – {GB × (BC – FC)}

= {(12 × 9) – (7.5 × 10)} = (108 – 75) = 33 m^{2}