# Class 8 RD Sharma Solutions- Chapter 20 Area Of Trapezium And Polygon- Exercise 20.1 | Set 1

**Question 1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m**^{2}?

^{2}?

**Solution: **

Given that,

Base of parallelogram = 24cm,

Height of parallelogram = 10cm,

Area of floor = 1080m

^{2}As we know that,

Area of parallelogram = Base Ã— Height

Area of 1 tiles = 24 Ã— 10 = 240 cm

^{2}As we know that 1m = 100cm,

So area will be 1080 m2 = 1080 Ã— 100 Ã— 100 cm

^{2}Number of tiles required = Area of floor / Area of 1 tile

Number of tiles required = (1080 Ã— 100 Ã— 100) / (24 Ã— 10) =

45000

Hence, Number of tiles required is 45000.

**Question 2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Figure. If AB = 60 m and BC = 28 m, Find the area of the plot.**

**Solution: **

From figure, we conclude that, Area of the plot = Area of the rectangle + Area of semi-circle

Radius of semi-circle = BC/2 = 28/2 = 14m

As we know that Area of the Rectangular plot = Length Ã— Breadth = 60 Ã— 28 = 1680 m

^{2}Area of the Semi-circular portion = Ï€r

^{2}/2= 1/2 Ã— 22/7 Ã— 14 Ã— 14 = 308 m

^{2}

Hence, the total area of the plot = 1680 + 308 = 1988 m^{2}

**Question 3. A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take** Ï€**= 22/7.)**

**Solution:**

From figure, we conclude that, Area of the plot = Area of the Rectangle + 2 Ã— area of one semi-circle

Radius of semi-circle = BC/2 = 24.5/2 = 12.25m

(Given)As we know that Area of the Rectangular plot = Length Ã— Breadth = 36 Ã— 24.5 = 882 m

^{2}and Area of the Semi-circular portions = 2 Ã— Ï€r

^{2}/2= 2 Ã— 1/2 Ã— 22/7 Ã— 12.25 Ã— 12.25 = 471.625 m

^{2}

Hence, the total Area of the plot = 882 + 471.625 = 1353.625 m^{2}

**Question 4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.**

**Solution: **

From figure, we conclude that, Area of the plot = Area of the rectangle â€“ 4 Ã— area of one quadrant

Radius of semi-circle = 3.5 m

(Given)Area of four quadrants = area of one circle

(from figure)Area of the plot = Length Ã— Breadth â€“ Ï€r

^{2}

Hence, Area of the plot = 20 Ã— 15 â€“ (22/7 Ã— 3.5 Ã— 3.5) = 261.5 m^{2}

**Question 5. The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.**

**Solution: **

From figure, we conclude that, Perimeter of the inner track = 2 Ã— Length of rectangle + perimeter of two semi-circular ends

Perimeter of the inner track = Length + Length + 2Ï€r

400 = 90 + 90 + (2 Ã— 22/7 Ã— r)

(2 Ã— 22/7 Ã— r) = 400 â€“ 180

(2 Ã— 22/7 Ã— r) = 220

44r = 220 Ã— 7

44r = 1540

r = 1540/44 = 35

r = 35m

Hence, the radius of inner circle = 35 m

Now we have to calculate the radius of outer track

Radius of outer track = Radius of inner track + width of the track (from figure)

Radius of outer track = 35 + 14 = 49m (given)

Length of outer track = 2Ã— Length of rectangle + perimeter of two outer semi-circular ends

Length of outer track = 2Ã— 90 + 2Ï€r

Length of outer track = 2Ã— 90 + (2 Ã— 22/7 Ã— 49)

Length of outer track = 180 + 308 = 488

Hence, the Length of outer track = 488m

Area of inner track = Area of inner rectangle + Area of two inner semi-circles

Area of inner track = Length Ã— Breadth + Ï€r

^{2}Area of inner track = 90 Ã— 70 + (22/7 Ã— 35 Ã— 35)

Area of inner track = 6300 + 3850

Hence, the Area of inner track = 10150 m

^{2}Area of outer track = Area of outer rectangle + Area of two outer semi-circles

Breadth of outer track = 35 + 35 +14 + 14 = 98 m

Area of outer track = lengthÃ— breadth + Ï€r

^{2}Area of outer track = 90 Ã— 98 + (22/7 Ã— 49 Ã— 49)

Area of outer track = 8820 + 7546

Hence, Area of outer track = 16366 m

^{2}Now, we have to calculate the Area of path

Area of path = Area of outer track â€“ Area of inner track

Area of path = 16366 â€“ 10150 = 6216

Hence, Area of path is 6216 m^{2}

**Question 6. Find the area of Fig., in square cm, correct to one place of decimal. (Take **Ï€** =22/7)**

**Solution:**

From figure we conclude that, Area of the Figure = Area of square + Area of semi-circle â€“ Area of right angled triangle

Area of the Figure = side Ã— side + Ï€r

^{2}/2 â€“ (1/2 Ã— base Ã— height)put the values in formula and we get,

Area of the Figure = 10 Ã— 10 + (1/2 Ã— 22/7 Ã— 5 Ã— 5) â€“ (1/2 Ã— 8 Ã— 6)

Area of the Figure = 100 + 39.28 â€“ 24

Area of the Figure =

115.3

Hence, Area of the Figure = 115.3 cm^{2}

**Question 7. The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take **Ï€**=22/7)**

**Solution:**

Given that,

Diameter of a wheel = 90 cm,

As we know that, Perimeter of wheel = Ï€d

Perimeter of wheel = 22/7 Ã— 90 = 282.857

Hence, the Perimeter of a wheel = 282.857 cm

Distance covered in 315 revolutions = 282.857Ã— 315 = 89099.955 cm

As we know that 1 km = 100000 cm

So, The Distance covered = 89099.955/100000 = 0.89 km

Speed in km per hour = 0.89 Ã— 60 = 53.4 km per hour.

**Question 8. The area of a rhombus is 240 cm^2 and one of the diagonal is 16 cm. Find another diagonal.**

**Solution: **

Given that,

Area of rhombus = 240 cm^2,

Diagonal = 16 cm.

As we know that,

Area of rhombus = 1/2 Ã— d1 Ã— d2

240 = 1/2 Ã— 16 Ã— d2

240 = 8 Ã— d2

d2 = 240/8 = 30

Hence the other diagonal is 30 cm.

**Question 9. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.**

**Solution: **

Given that,

Diagonal(d1) = 7.5 cm,

Diagonal(d2) = 12 cm.

As we know that,

Area of rhombus = 1/2 Ã— d1 Ã— d2

Area of rhombus = 1/2 Ã— 7.5 Ã— 12

Area of rhombus = 6 Ã— 7.5 = 45

Hence, Area of rhombus = 45 cm^{2}

**Question 10. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.**

**Solution**:

As we know that,

Area of quadrilateral = 1/2 Ã— d1 Ã— (p1 + p2)

Area of quadrilateral = 1/2 Ã— 24 Ã— (8 + 13)

Area of quadrilateral = 12 Ã— 21 = 252

Hence, Area of quadrilateral is 252 cm^{2}

**Question 11. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.**

**Solution: **

Given that,

Side of rhombus = 6 cm,

Altitude of rhombus = 4 cm.

As we know that rhombus is a parallelogram, therefore area of parallelogram = base Ã— altitude

Therefore, Area of parallelogram = 6 Ã— 4 = 24 cm

^{2}Area of parallelogram = Area of rhombus

Area of rhombus = 1/2 Ã— d1 Ã— d2

24 = 1/2 Ã— 8 Ã— d2

24 = 4 Ã— d2

d2 = 24/4 = 6

Hence, the length of other diagonal of rhombus is 6 cm.