RD SHARMA SOLUTION CHAPTER –20 Mensuration – I (Area of a Trapezium and a Polygon) | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 20.1

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 m² ?
Solution:
Area of floor = 1080 m²
Base of parallelogram shaped tile (b) = 24 cm
and corresponding height (h) = 10 cm
Area of one tile = b x h = 24 x 10 = 240 cm²

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

Solution:
Length of rectangular portion (l) = 60 m
and breadth (b) = 28 m
Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m²
Radius of semicircular portion (r) = b2 = 282 = 14 m
Area = 12 πr²
= 12 x 227 x 14 x 14 m²
= 308 m²
Total area of the plot = 1680 + 308 = 1988 m²

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 π = 227).
Solution:
Length of rectangular portion (l) = 36 m
and breadth (b) = 24.5 m

= 227 x 150.0625 m²
= 471.625 m²
Total area of the playground = 471.625 + 882 = 1353.625 m²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
breadth (b) = 15 m

Area of rectangular piece = l x b = 20 x 15 = 300 m²
Radius of each quadrant (r) = 3.5 m
Total area of 4 quadrants

Area of the remaining portion = 300 – 38.5 m² = 261.5 m²

Question 5.
The inside perimeter of a running track (shown in Fig.) 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:
Inner perimeter = 400 m.
Length (l) = 90 m.
Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m

Question 6.
Find the area of the Figure in square cm, correct to one place of decimal. (Take π = 227)

Solution:
Length of square (a) = 10 cm.
Area = a² = (10)² = 100 cm²
Base of the right triangle AED = 8 cm
and height = 6 cm

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 π = 227)
Solution:
Diameter of the wheel (d) = 90 cm.

Question 8.
The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.
Solution:
Area of rhombus = 240 cm²
Length of one diagonal (d₁) = 16 cm
Second diagonal (d₂)

Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
In rhombus, diagonal (d₁) = 7.5 cm
and diagonal (d₂) = 12 cm

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:
In quadrilateral shaped field ABCD,
diagonal AC = 24 m

and perpendicular BL = 13 m
and perpendicular DM on AC = 8 m
Area of the field ABED = 12 x AC x (BL + DM)
= 12 x 24 x (13 + 8) m²
= 12 x 21 = 252 m²

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:
Side of rhombus (b) = 6 cm
Altitude (h) = 4 cm

Question 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 m² is Rs 4.
Solution:
Number of rhombus shaped tiles = 300
Diagonals of each tile = 45 cm and 130 cm

Rate of polishing the tiles = Rs 4 per m²
Total cost = 202.5 x 4 = Rs 810

Question 13.
A rectangular grassy plot is 112 m long and 78 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 metre.
Solution:
Length of rectangular plot (l) = 112 m
and breadth (b) = 78 m
Width of path = 2.5 m

Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m
and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m
Area of path = outer area – inner area
= (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m²
Rate of constructing = Rs 4.50 per m²
Total cost = 925 x Rs 4.50 = Rs 4162.50

Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Side of rhombus = 20 cm.
One diagonal (d₁) = 24 cm

Diagonals of a rhombus bisect each other at right angle
AB = 20 cm
and OA = 12 AC = 12 x 24 cm = 12 cm
In right-angled ∆AOB,
AB² = AO² + BO² (Pythagoras theorem)
⇒ (20)² = (12)² + BO²
⇒ 400 = 144 + BO²
⇒ BO² = 400 – 144 = 256 = (16)²
⇒ BO = 16 cm
and diagonal BD = 2 x BO = 2 x 16 = 32 cm
Now area of rhombus ABCD

Question 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 L m ?
Solution:
Side of square = 4 m
Area of square = (a)² = 4 x 4 =16 m²

Diagonals of a rhombus bisect each other at right angles.
In right ∆AOB
AB² = QA² + BO² (Pythagoras theorem)
= (8)² + (1)² = 64 + 1 = 65
AB = √65 m.
Now, length of perpendicular AL (h)

Question 16.
Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.
Solution:
Length of each side of rhombus = 14 cm.
Length of altitude = 16 cm
Area = Base x altitude = 14 x 16 cm² = 224 cm²

Question 17.
The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.
Solution:
Cost of fencing the square field = Rs 1,200
Rate = 60 paise per m.

Question 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:
Side of a square plot = 84 m
Area = (a)² = (84)² = 84 x 84 m² = 7056 m²
Area of rectangular field = 7056 m²
Length (l) = 144 m

Question 19.
The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.
Solution:
Area of rhombus = 84 m²
Perimeter = 40 m

Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².
Solution:
Side of rhombus garden (b) = 30 m.
Altitude (h) = 16 m
Area = Base x altitude = 30 x 16 = 480 m²
Rate of levelling the garden = Rs 2 per m²
Total cost = Rs 480 x 2 = Rs 960

Question 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:
Length of side of rhombus (b) = 64 m
and altitude (h) = 16 m
Area = b x h = 64 x 16 m² = 1024 m²
Now area of square = 1024 m²
Side of the square = √Area = √1024 m = 32 m

Question 22.
The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude 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:
Base of triangle (b) = 24.8 cm
and altitude (h) = 16.5 cm
Area of triangle = 12 x base x height
= 12 x bh= 12 x 24.8 x 16.5 cm² = 204.6 cm²
Area of rhombus = 204.6 cm²
Length of one diagonal (d₁ = 22 cm
Second diagonal (d₂)

Exercise 20.2

Question 1.
Find the area, in square metres, of the trapezium whose bases and altitude are as under:
(i) bases = 12 dm and 20 dm, altitude =10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:

Question 2.
Find the area of trapezium with base 15 cm and height 8 cm. If the side parallel to the given base is 9 cm long.
Solution:
In the trapezium ABCD,

Question 3.
Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.
Solution:
Length of parallel sides of a trapezium are 16 dm and 22 dm i.e.
b₁ = 16 dm, b₂ = 22 dm
and height (h) = 12 dm

Question 4.
Find the height of a trapezium, the sum of lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm²
Solution:
Sum of parallel sides (b₁ + b₂) = 60 cm
Area of trapezium = 600 cm²

Question 5.
Find the altitude of a trapezium whose area is 65 cm² and whose bases are 13 cm and 26 cm.
Solution:
Area of a trapezium = 65 cm²
Bases are 13 cm and 26 cm
i.e. b₁ = 13 cm, b₂ = 26 cm.

Question 6.
Find the sum of the lengths of the bases of trapezium whose area is 4.2 m² and whose height is 280 cm.
Solution:
Area of trapezium = 4.2 m²
Height (h) = 280 cm = 2.8 m.

Question 7.
Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate the area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle id the sum of the areas of two triangles.
Solution:
In trapezium ABCD, parallel sides or bases are 10 cm and 15 cm and height = 6 cm
Area of trapezium

Area of trapezium = 90 – 15 = 75 cm²
= area of rectangle – areas of two triangles.

Question 8.
The area of a trapezium is 960 cm². If the parallel sides are 34 cm and 46 cm, find the distance between them:
Solution:
Area of trapezium = 960 cm²
Parallel sides are 34 cm and 46 cm
b₁ + b₂ = 34 + 46 = 80 cm

Distance between parallel sides = 24 cm

Question 9.
Find the area of the figure as the sum of the areas of two trapezium and a rectangle.

Solution:
In the figure,
One rectangle is ABCD whose sides are 50 cm and 10 cm.
Two trapezium of equal size in which parallel sides are 30 cm and 10 cm and height

Question 10.
Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Solution:
Top of a table is of trapezium in shape whose parallel sides are 1 m and 1.2 m and distance between them (h) = 0.8 m
Area of trapezium = \(\frac { 1 }{ 2 }\) (Sum of parallel sides) x height

Question 11.
The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom, and the area of the cross-section is 72 m², determine its depth.
Solution:
Area of cross-section = 72 m²
Parallel sides of the trapezium = 10 m and 6 m

Question 12.
The area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
Solution:
Area of trapezium = 91 cm²
Height (h) = 7 cm.
Sum of parallel sides

One parallel side = 9 cm
and second side = 9 + 8 = 17 cm
Hence parallel sides are 17 cm, 9 cm

Question 13.
The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.
Solution:
Area of trapezium = 384 cm²
Perpendicular distance (h) = 12 cm
Sum of parallel sides

First parallel side = 8 x 3 = 24 cm
Second side = 8 x 5 = 40 cm

Question 14.
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Solution:
Area of the trapezium shaped field = 10500 m²
and perpendicular distance between them (h) = 100 m.

Question 15.
The area of trapezium is 1586 cm² and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.
Solution:
Area of a trapezium = 1586 cm²
and distance between the parallel sides (h) = 26

Question 16.
The parallel sides of a trapezium are 25 cm and 13 cm ; Its nonparallel sides are equal each being 10 cm, find the area of the trapezium.
Solution:
Parallel sides of a trapezium ABCD are 25 cm and 13 cm
i.e. AB = 25 cm, CD = 13 cm
and each non-parallel side = 10 cm
i.e., AD = BC = 10 cm
From C, draw CE || DA and draw CL ⊥ AB
CE = DA = CB = 10 cm
and EB = AB – AE = AB – DC = 25 – 13 = 12 cm
Perpendicular CL bisects base EB of an isosceles ∆CED

Question 17.
Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm each.
Solution:
In trapezium ABCD, parallel sides are AB and DC.

AB = 25 cm, CD = 13 cm
and other sides are 15 cm each i.e. AD = CB = 15 cm
From C, draw CE || DA and CL ⊥ AB
AE = DC = 13 cm
and EB = AB – AE = 25 – 13 = 12 cm
Perpendicular CL bisects the base EB of the isosceles triangle CEB

Question 18.
If the area of a trapezium is 28 cm² and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.
Solution:
Area of trapezium = 28 cm²
Altitude (h) = 4 cm.

One of the parallel side = 6 cm
Second parallel side = 14 – 6 = 8 cm

Question 19.
In the figure, a parallelogram is drawn in a trapezium the area of the parallelogram is 80 cm², find the area of the trapezium.

Solution:
Area of parallelogram (AECD) = 80 cm²
Side AE (b) = 10 cm

Question 20.
Find the area of the field shown in the figure by dividing it into a square rectangle and a trapezium.

Solution:
Produce EF to H to meet AB at H and draw DK || EH
HF = 4 cm, KD = HE = 4 + 4 = 8 cm
HK = ED = 4 cm,
KB = 12 – (8) = 4 cm
Now, area of square AGFH = 4 x 4 = 16 cm²
area of rectangle KDEH = l x b = 8 x 4 = 32 cm²
and area of trapezium BCDK.

Total area of the figure = 16 + 32 + 22 = 70 cm²

Exercise 20.3

Question 1.
Find the area of the pentagon shown in the figure if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm and BF = 5 cm, CG = 7 cm and EH = 3

Solution:
In the figure, here are three triangles and one trapezium.

Question 2.
Find the area enclosed by each of the following figures as the sum of the areas of a rectangle and a trapezium:

Solution:
(i) In the figure ABCDEF,
Join CF, then, the figure consists one square and one trapezium ABCF is a square whose side = 18 cm

Area of the square = 18 x 18 cm² = 324 cm²
Area of trapezium FCDE = 12 (CF + ED) x 8 cm²
= 12 (18 + 7) x 8
= 12 x 25 x 8 cm²
= 100 cm²
Total area of fig. ABCDEF = 324 + 100 = 424 cm²
(ii) In the figure ABCDEF,
Join BE.
The figure consists of one rectangle BCDE and one trapezium ABEF

Area of rectangle BCDE = BC x CD = 20 x 15 = 300 cm²
Area of trapezium ABEF,
= 12 (BE + AF) x height
= 12 (15 + 6) x 8 cm²
= 12 x 21 x 8 cm²
= 84 cm²
Area of the figure ABCDEF = 300 + 84 = 384 cm²
(iii) In the figure ABCDEFGH,

HC= AB = EF = 6 cm
AH = BC = 4 cm
DE = GF = 5 cm
Join HC.
In right ∆CDE,
ED² = CD² + CE²
⇒ (5)² = (4)² + (CE)²
⇒ 25 = 16 + (CE)²
⇒ (CE)² = 25 – 16 = 9 = (3)²
CE = 3 cm
The figure consist a rectangle and a trapezium
Area of rectangle ABCH = AB x BC = 6 x 4 = 24 cm²
Area of trapezium GDEF,
= 12 (GD + EF) x CE 1
= 12 (GH + HC + CD + EF) x CE
= 12 (4 + 6 + 4 + 6) x 3 cm²
= 12 x 20 x 3 cm²
= 30 cm²
Total area of the figure ABCDEFGH = 24 + 30 = 54 cm²

Question 3.
There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways.


Find the area of this park using both ways. Can you suggest some another way of finding its area ?
Solution:
In first case, the figure ABCDE is divided into 2 trapezium of equal area.
Now area of trapezium DFBC

Total area of the pentagon ABCDE = 2 x 168.75 = 337.5 m²
In second case, the figure ABCDE is divided into two parts, namely one square and other triangle.

Total area of pentagon ABCDE = 225 + 112.5 = 337.5 m²

Question 4.
Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.

Solution:
In the figure ABCDEF,
AD = 90 cm
BL = 30 cm
AO = 60 cm
CN = 40 cm
AN = 50 cm
EO = 60 cm
AM = 20 cm
FM = 20 cm
AL = 10 cm



Area of ABCDEF = (150 + 800 + 900 + 200 + 1400 + 1600) cm² = 5050 cm²

Question 5.
Find the area of the following regular hexagon:

Solution:
In the regular hexagon MNOPQR There are two triangles and one rectangle.
Join MQ, MO and RP
NQ = 23 cm,
NA = BQ = 102 = 5 cm
MR = OP = 13 cm

In right ∆BDQ,
PQ² = BQ² + BP²
⇒ (13)² = (5)² + BP²
⇒ 169 = 25 + BP²
⇒ BP² = 169 – 25 = 144 = (12)²
BP = 12 cm
PR = MO = 2 x 12 = 24 cm
Now area of rectangle RPOM = RP x PO = 24 x 13 = 312 cm²
Area of ∆PRQ = 12 x PR x BQ
= 12 x 24 x 5 = 60 cm²
Similarly area ∆MON = 60 cm²
Area of the hexagon MNOPQR = 312 + 60 + 60 = 432 cm²