## RD Sharma Solutions Class 8 Chapter 20 Exercise 20.1

#### Exercise 20.1

**Q1: A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many are such tiles required to cover a floor of area 1080 m ^{2}?**

**Solution:**

Given data:

A flooring tile with the base shape of parallelogram = b = 24 cm

height = h = 10 cm.

We know that the area of a parallelogram, A = Base x Height Square units

Now, substitute the given values = 24 x 10 = 240

Therefore, the area of the parallelogram shaped tile isÂ 240 cm^{2}

It is given that the area of the floor isÂ 1080 m^{2}

Now convert metre into centimetre. We know that 1m = 100 cm

So 1080 m^{2Â }= 1800000cm

Therefore, the number of tiles required = Area of the floor / Area of the tiles

= 10800000/240

= 45000

Hence, 45000 tiles are required to cover the floor.

2. **A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. 20.23. If AB = 60 m and BC = 28 m, find the area of the plot.**

**Solution:**

From the given figure, it is observed that it has a rectangle and a semicircle.

Therefore, the total area of the plot = Area of rectangleÂ + Area of a semicircle

Total AreaÂ = Area of ABCD + Area of a semicircle with radius 14 m

Because AD = BC = 28 m

Therefore r = 28/2 = 14 m.

We know that the area of a rectangle – l x b square units

Substitute the given values from the figure.

Area of a rectangle ABCD = 60 x28 =1680 m^{2}

Area of a semicircle = (1/2) Ï€ r^{2}Â square units

A = (1/2)(22/7)(14)^{2}

A =Â 308 m^{2}

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

**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 pi = 22 / 7 ).**

**Solution:**

From the given data, it is observedÂ that the playground is in the shape of a rectangle and two smaller semicircles.

Given

length, l = 36 m

b = 24.5 m

Therefore, the total area of the playground = Area of rectangleÂ + Area of a 2 semicircles

To find the radius, r of a semicircle,

r =Â 24.5/2Â = 12.25m

We know that the area of a rectangle – l x b square units

Substitute the given values from the figure.

Area of a rectangle ABCD = 36 x24.5 =882 m^{2}

Area of a semicircle = (1/2) Ï€ r^{2}Â square units

A = (1/2)(22/7)(12.25)^{2}

A =Â 235.8 m^{2}

Therefore, the area of 2 semicircles = 2 x 235.8 =Â 471.6

Therefore, the complete area of a playground = area of the rectangular ground + 2 xÂ area of a semicircle

A =Â 882 +Â 471.6 =Â 1353.6 m^{2}

Therefore, the area of the complete playground isÂ 1353.6 m^{2}

**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:**

Given:

length of the rectangular piece = 20 m

Width = 15 m

Also, from each corner, a quadrant each of radius 3.5 m has been cut out.

A rough sketch for the given detail is given below :

So, Area of the remaining part = Area of the rectangular piece â€” (4 x Area of a quadrant with radius 3.5m)

Now, area of the rectangular piece = l x b = 20 xÂ 15 = 300 m^{2}

And, area of a quadrant with radius 3.5 m = (1/4) Ï€ r^{2}Â square units

A = (1/4)(22/7)(3.5)^{2}

A =Â 9.625 m^{2}

Hence, the area of theÂ remaining part = 300 – (4 xÂ 9.625) = 261.5 m^{2}

**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 the track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.**

**Solution:**

Given:

inside perimeter of the running track is 400m. So, the length of the inner track is 400 m.

Assume that “r” be the radius of the inner semicircles.

It is observed that : Perimeter of the inner track = Length of two straight portions of 90 m + Length of two semicircles 400

so, 400 =(2 xÂ 90) +( 2 xÂ Perimeter of a semicircle)

400 = 180 + ((2 x 22/7) xÂ r)

400 -180 = ( (44/7) xÂ r)

(44/7) x r = 220

r = (220 x 7) /44= 35 m

so, the width or breadth of the inner track = 2r = 2 xÂ 35 = 70 m

Since the track is 14 m wide at all places, so the width of the outer track : 70 + (2 x 14) = 98m

So, Radius of the outer track semicircles = 98/2 = 49 m

Therefore, the area of the outer track = (Area of the rectangular portion having the sides 90 m and 98 m) + (2 xArea of two semicircles having radius 49 m)

=(98 xÂ 90) + (2 x (Â½) x (22/7) xÂ 49^{2})

= (8820) + (7546) = 16366 m^{2}

Also, the area of the inner track = (Area of the rectangular portion having the sides 90 m and 70 m) + (2 x Area of the semicircle having radius 35 m)

= (70 xÂ 90) + (2 x (Â½)x (22/7)xÂ 35^{2})

= (6300) + (3850)

A =10150 m^{2}

Hence, the area of the running track = Area of the outer running track – Area of the inner running track

= 16366 – 10150

A= 6216 m^{2}

Also, length of the outer track = (2Â x length of the straight portion) + (2 x perimeter of the semicircles with radius 49 m)

= (2 xÂ 90) â€“(2 xÂ (22/7)Â x 49)

Therefore, the length of the outer running track = 180 +308 = 488 m

**6. Find the area of Fig. 20.25, in square cm, correct to one place of decimal. (Take it = 22/7)**

**Solution:**

From the given figure, it is observed that, when you connect the A and D, a traingle is formed. So, to find the area of the given figure,

A =Â (Area of rectangle ABCD + Area of the semicircle) – (Area of triangle AED).

Given data:

length of a rectangleÂ = 10 cm

breadth of a rectangleÂ = 10 cm

Semicircle radius = 10/2 = 5 cm

Base of a triangle = 8 cm

Height of a triangle = 6 cm

Substitute the values in the formula,

A = (10 x 10)+ ((1/2) x (22/7)Â x 5^{2}) â€“ ((1/2)Â x 6 x 8)

= 100 + 39.3 – 24 = 115.3 cm^{2}

Therefore, the area of the given figure is 115.3 cm^{2}

**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. [Use pi = 22 / 7].**

**Solution:**

Given: The diameter of the wheel is 90 cm.

Therefore, Radius, r = 90/2 = 45 cm.

So, Perimeter of the wheel = 2Â Ï€Â r

P = 2 xÂ (22/7)xÂ 45

= 282.857 cm

So, it means the wheel travels 282.857 cm in a revolution.

Therefore, to makeÂ 315 revolutions per minute.

Distance travelled by the wheel in one minute = 315 xÂ Â 282.857 = 89100 cm

Hence, Speed = 89100 cm/1 minute =Â 89100 cm per minute

So, to convert it into kilometres per hour.

Therefore, \(\frac{89100 \; cm}{1 \; minute}\) = \(\frac{89100 \times \frac{1}{100000} \; kilometer}{\frac{1}{60} \; hour}\)

= \(\frac{891000}{100000} \times \frac{60}{1} \frac{kilometer}{hour}\)

Therefore, the speed= 53.46 kilometers per hour.

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

**Solution:**

GivenÂ Data:

Area of the rhombus is 240 cm

One of its diagonals length = 16 cm

We know that the area of the rhombus if the diagonals of a rhombus are d_{1} and d_{2}Â are given

Area of a rhombus= \(\frac{1}{2} \left ( d_{1} \times d_{2} \right )\) square units

Substitute the given values :

240 = \(\frac{1}{2} \left ( 16 \times d_{2} \right )\)

240 \(\times\) 2 = 16 \(\times\) d_{2}

It is written as

16 xÂ d_{2} = 480

d_{2} = 480/16

Therefore, the length of the other diagonal of a rhombus = 30 cm

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

**Solution:**

Given :

The diagonal length of a rhombus is 7.5 cm and 12 cm respectively.

We know that,

Area of a rhombus = \(\frac{1}{2} \left ( d_{1} \times d_{2} \right )\) square units

= \(\frac{1}{2} \left ( 7.5 \times 12 \right )\)

Therefore, the area of rhombus, A= 45 cm^{2}

**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:**

Given Data:

Quadrilateral shaped field Diagonal = 24 m

When the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m.

Now, we know : Area = \(\frac{1}{2} \times d \times \left ( h_{1} + h_{2} \right )\)

Therefore, Area of the field =

(1/2) x 24 x (8+13)

\(\frac{1}{2}Â

= 12 x 21

= 252 m^{2}

Therefore, the area of afield = 252m^{2}

**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 :

Side of the rhombus = 6 cm

Altitude = 4 cm

One of the diagonals = 8 cm

Area of the rhombus = Side \(\times\) Altitude = 6 \(\times\) 4 = 24 cm^{2}

We know :

Area of rhombus = \(\frac{1}{2} \left ( d_{1} \times d_{2} \right )\)

Using (i) :

24 = \(\frac{1}{2} \left ( d_{1} \times d_{2} \right )\)

24= \(\frac{1}{2} \left ( 8 \times d_{2} \right )\)

D_{2} = 6 cm .

**12. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m ^{2} is Rs 4.**

**Solution:**

Given :

The floor consists of 3000 rhombus-shaped tiles.

The lengths of the diagonals of each tile are 45 cm and 30 cm.

Area of a rhombus shaped the = \(\frac{1}{2} \left ( 45 \times 30 \right )\) = 675 cm^{2}

Therefore, Area of the complete floor = 3000 \(\times\) 675 = 2025000 cm^{2}

Now, we need to convert this area into m^{2} because the rate of polishing is given as per m^{2}.

Therefore, 2025000 cm^{2} = 2025000 \(\times\) cm \(\times\) cm

= 2025000 \(\times\) 1/100 m \(\times\) 1/100 m

= 202.5 m^{2}

Now, the cost of polishing 1 m^{2} is Rs 4.

Therefore, Total cost of polishing the complete floor = 202.5 \(\times\) 4 = 810

Thus, the total cost of polishing the floor is Rs 810.

**13. A rectangular grassy plot is 112 m long and 78 m broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square meter.**

**Solution:**

Given: The length of a rectangular grassy plot is 112 m and its width is 78 m.

Also, it has a gravel path of width 2.5 m around it on the sides

Its rough diagram is given below :

Length of the inner rectangular field = 112 – (2 \(\times\) 2.5) = 107 m

The width of the inner rectangular field = 78 – (2 \(\times\) 2.5) = 73 m

Area of the path = (Area of the rectangle with sides 112 m and 78 m) – (Area of the rectangle with sides 107 m and 73 m)

= (112 \(\times\) 78) – (107 \(\times\) 73)

= 8736 – 7811 = 925 m^{2}

Now, the cost of constructing the path is Rs 4.50 per square meter.

Cost of constructing the complete path = 925 \(\times\) 4.50 = Rs 4162.5

Thus, the total cost of constructing the path is Rs 4162.5.

**14. Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.**

**Solution:**

Given that,

Rhombus side= 20 cm

diagonal length = 24 cm

We know that : If d_{1} and d_{2} are the lengths of the diagonals of the rhombus, then side of the rhombus =\(\frac{1}{2} \sqrt{d_{1}^{2} + d_{2}^{2}}\)

So, substitute the given data to find the length of the other diagonal of the rhombus

20 = \(\frac{1}{2} \sqrt{24^{2} + d_{2}^{2}}\)

40 = \(\sqrt{24^{2} + d_{2}^{2}}\)

To remove the square root, take sqaures on both the sides

So, it becomes

40^{2} = 24^{2 }â€“ d_{2}^{2}

d_{2}^{2 }= 1600 – 576 = 1024

d_{2} = \(\sqrt{1024}\) = 32 cm

The other diagonal of the rhombus is 32 cm

Therefore,

The area of the rhombus = (1/2) (24 x32)Â = 384 c m^{2}

**15. The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?**

**Solution:**

Given:

The length of the square field is 4 m

So, the area of the square field = 4 xÂ 4 = 16 m^{2}

Given thar: Area of the rhombus = Area of the square field

one of the diagonal length of the rhombus = 2 m

To find the Side of the rhombus = \(\frac{1}{2} \sqrt{d_{1}^{2} + d_{2}^{2}}\)

We know that, the area of the rhombus

A = \(\frac{1}{2} \times \left ( d_{1} \times d_{2} \right )\) square units

Therefore, Area :

16 = \(\frac{1}{2} \times \left ( 2 \times d_{2} \right )\)

First we need to find the rhombus side length

So, Side of the rhombus = \(\frac{1}{2} \sqrt{2^{2} + 16^{2}}\)

= \(\frac{1}{2} \sqrt{260}\)

= \(\frac{1}{2} \sqrt{4 \times 65}\)

= \(\frac{1}{2}\times 2 \sqrt{ 65}\)

= \(\sqrt{ 65}\)

Also, we know: Area of the rhombus = Side xÂ Altitude

Substitite the obtained values

16 = \(\sqrt{ 65}\) \(\times\) Altitude

Therefore, the altitude of a rhombus= \(\frac{16}{\sqrt{ 65}}\) m .

**16. Find the area of the field in the form of a rhombus, if the length of each side be 14 on and the altitude is 16 cm.**

**Solution:**

Given: Rhombus length of each side = 14 cm

The altitude of a rhombus = 16 cm

We know that,

Area of the rhombus = Side x Altitude

Therefore,

A = 14 xÂ Â 16

Area of the field = 224 cm^{2}

**17. The cost of fencing a square field at 60 paise per meter is Rs 1200. Find the cost of reaping the field at a rate of 50 paise per 100 sq. metres.**

**Solution:**

It is given that the cost of fencing 1 metre of a square field is 60 paise

Therefore, the total cost of fencing the whole field = Rs 1200 = 1,20,000 paise

So, the perimeter of the square field =120000/60Â = 2000 metres

We know that, the perimeter of a square = 4 xÂ side units

To find the side for the given square field :

4 x Side = 2000 m

Side = 2000/4= 500 metres

Therefore, Area of the square field = (side)^{2Â }Â = 500 x 500 = 250000 m^{2}

Also given that, the cost of reaping per 100 m^{2} = 50 paise

Therefore, Cost of reaping/1 m^{2} = 50/100Â paise

Thus, Cost of reaping 250000 m^{2} =(50/100)xÂ 250000 = 125000 paise .

Therefore, the total cost of reaping the whole square field is 125000 paise, i.e. RS. 1250.

**18. In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.**

**Solution:**

Given that, the side of the square plot is 84 m

The man wants to exchange the square plot with a rectangular plot of the same area having a length of 144m.

We know that, the area of the square plot = 84 x 84 = 7056 m^{2}

Also w.k.t, Area of the rectangular plot = Length xÂ Width

Therefore, it becomes

7056 = 144 xÂ Width

Width = 7056/144 = 49 m

Thus, the width of the rectangular plot is 49 m.

**19. The area of a rhombus is 84 m ^{2}. If its perimeter is 40 m, then find its altitude.**

**Solution:**

Given that,

Area of the rhombus is 84 m^{2}

The rhombus perimeter= 40 m

We know that, the perimeter of the rhombus = 4 xÂ Side

40 = 4 x Side

Side of rhombus= 40/4 = 10 m

Now, substitute the values in the rhombus formula,

Area of the rhombus = Side xÂ Altitude square units

84 = 10 xÂ Â Altitude

Altitude = 84/10 = 8.4 m

Therefore, the altitude of the rhombus is 8.4 m.

**20. A garden is in the form of a rhombus whose side is 30 meters and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m ^{2}**

**Solution:**

Given that:

Side of the rhombus-shaped garden = 30 m

The altitude of a rhombus = 16 m

We know that, the area of a rhombus = side xÂ Altitude square units

Area of the garden = 30 x16 = 480 m^{2}

Also given that, the rate of levelling the garden is Rs 2 / 1m^{2}.

So, the total cost of levelling the complete garden of area 480 m^{2 }= 480 x 2

Therefore, the cost of levelling the garden is Rs 960.

**21. A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus?**

**Solution:**

Given data

Side of rhombus-shaped fields = 64 m

The altitude of a rhombus =16 m

We know that the area of rhombus = Side x Altitude square units

Therefore, Area of the field = 64 x 16 = 1024 m^{2}

It is given that, Area of the square field = Area of the rhombus

Area of a square = (Side)^{2}

1024 = (Side)^{2}

Side = \(\sqrt{1024}\) = 32 m

Therefore, the side of the square field is 32 m.

**22. The area a rhombus is equal to the area of a triangle whose base and the corresponding altitudes are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.**

**Solution:**

Given that:

Area of the given rhombus is equal to the area of the triangle with base 24.8 cm and altitude 16.5 cm

Area of the triangle = (1/2) xÂ base x altitude

=(1/2)xÂ 24.8 x 16.5

A = 204.6 cm^{2}

Thus, the area of the rhombus = 204.6 cm2

Also it is given that the length of one of the diagonals of the rhombus is 22 cm

It is known that : Area of rhombus = \(\frac{1}{2} \times \left ( d_{1} \times d_{2} \right )\) square units

204.6 = \(\frac{1}{2} \times \left ( 22 \times d_{2} \right )\)

22 x d_{2} = 409.2

d_{2} = 409.2/22 = 18.6 cm

Thus, the length of the other diagonal of the rhombus is 18.6 cm.