Home ≫ Arithmetic Aptitute ≫ Elementary Mensuration-I ≫ Page 10
91. The sides of a rectangular field of 726sq m are in the ratio of 3:2. Find the sides.
  A.  30 m; 24 m
  B.  33 m; 22 m
  C.  35 m; 27 m
  D.  31 m; 25 m
     
   
View Answer

Shortcut:
If the sides of a rectangular field of 'A sq m area are in the ratio of m : n, then the sides are given by
√A x
m / n
or √(Area x Ratio) and √A x
n / m
or √(Area x Inverse Ratio)
Here, A = 726, m = 3, n = 2
Using these values in the shortcut, we get:
First Side = √(Area x Ratio) = √(726 x
3 / 2
) = √(1089) = 33 m
Second side = √(Area x Inverse Ratio) = √(726 x
2 / 3
) = √(484) = 22 m

92. The base of a triangular field is three times its height. If the cost of cultivation the field at Rs 36.72 per hectare is Rs 495.72, find its base and height.
  A.  Base:900 m; Height:300 m
  B.  Base:850 m; Height:350 m
  C.  Base:930 m; Height:320 m
  D.  Base:910 m; Height:280 m
     
   
View Answer

Shortcut:
If the base and the height of a triangle are in the ratio m:n and the area of the triangle is 'A' sq metre, then the base is given by √2 x A x
m / n
metre or √(2 x Area x Ratio) and height is given by √2 x A x
n / m
metre or √(2 x Area x Inverse Ratio)
Here, Area (A) =
495.72 / 36.72
=
27 / 2
hectares.
The ratio between base and height is 3 : 1
∴ Base = √(2 x
27 / 2
x
3 / 1
) = 900 m
Height = √(2 x
27 / 2
x
1 / 3
) = 300 m A = 726, m = 3, n = 2
Using these values in the shortcut, we get:
First Side = √(Area x Ratio) = √(726 x
3 / 2
) = √(1089) = 33 m
Second side = √(Area x Inverse Ratio) = √(726 x
2 / 3
) = √(484) = 22 m

93. Find the area of a rhombus one side of which measures 20 cm and one diagonal 24 cm.
  A.  574 sq cm
  B.  400 sq cm
  C.  375 sq cm
  D.  384 sq cm
     
   
View Answer

Shortcut:
To find the area of a rhombus if one side and one diagonal are given. Area of a rhombus = diagonal x √[(side)2 −
(diagoanl)2 / 22

Here, side = 20 , diagonal = 24
Using these values in the shortcut, we get:
Area = 24 x √[(20)2 −
(24)2 / 22

= 24 x √[20 x 20 −
24 x 24 / 2 x 2

= 24 x √[400 − 6 x 24]
= 24 x √[400 − 144]
= 24 x √(256)
= 24 x 16 = 384
Hence, the area of rhombus is 384 sq cm.

94. The perimeter of a rhombus is 146 cm and one of its diagonal is 55 cm. Find the other diagonal and the area of the rhombus.
  A.  Other diagonal: 52 cm; Area: 1440 sq cm
  B.  Other diagonal: 38 cm; Area: 1420 sq cm
  C.  Other diagonal: 48 cm; Area: 1320 sq cm
  D.  Other diagonal: 44 cm; Area: 1360 sq cm
     
   
View Answer

Shortcut:
To find the other diagonal of a rhombus, if perimeter of rhombus and one of its diagonals are given. Other diagonal = 2 x √[(side)2 −
(diagoanl)2 / 22

where side =
perimeter / 4

Here, one side of diagonal =
144 / 4
= 36.5, diagonal = 55
Using these values in the shortcut, we get:
Other diagonal = 2 x √[(36.5)2 −
(55)2 / 22
]
= 2 x √[36.5 x 36.5 −
55 x 55 / 2 x 2
]
= 2 x √[1332.25 − 756.25]
= 2 x √(576) = 2 x 24 = 48 cm
Now, area =
1 / 2
x product of diagonals
=
1 / 2
x 48 x 55 = 1320 sq cm.

95. Find the area of a regular hexagon whose side measures 9 cm.
  A.  215.6 sq cm
  B.  210.4 sq cm
  C.  225.6 sq cm
  D.  208.4 sq cm
     
   
View Answer

Shortcut:
To find the area of a regular polygon if length of its each side is given. Area of a regular polygon = [
n / 4
cot(
π / n
)] x (side)2
When 'n' = Number of sides
Now consider the following regular polygons. Hexagon: A hexagon has 6 sides. (i) Area of a regular Hexagon = [
6 / 4
cot(
180 / 6
)] x (side)2
=
3 / 2
cot30° x (side)2
=
3√3 / 2
x (side)2 [∵ cot30° = √3]
Here, side = 9
Using this value in the shortcut, we get:
Area =
3√3 / 2
x (9)2
=
3√3 x 9 x 9 / 2
= 210.4
Hence, the area is 210.4 sq cm.

96. Find the side of a regular hexagon whose area is 24√3 sq m.
  A.  8 m
  B.  2 m
  C.  4 m
  D.  5 m
     
   
View Answer

Shortcut:
To find the area of a regular polygon if length of its each side is given. Area of a regular polygon = [
n / 4
cot(
π / n
)] x (side)2
When 'n' = Number of sides
Now consider the following regular polygons. Hexagon: A hexagon has 6 sides. (i) Area of a regular Hexagon = [
6 / 4
cot(
180 / 6
)] x (side)2
=
3 / 2
cot30° x (side)2
=
3√3 / 2
x (side)2 [∵ cot30° = √3]
Here, area = 24√3, side = ?
Using this value in the shortcut, we get:
Area =
3√3 / 2
x (side)2
24√3 =
3√3 / 2
x (side)2
8 =
1 / 2
x (side)2
(side)2 = 8 x 2
side = 4
Hence, the area is 210.4 sq cm.

97. Find the area of an octagon whose side measures 6 m.
  A.  174.25 sq m
  B.  170.12 sq m
  C.  176.44 sq m
  D.  173.82 sq m
     
   
View Answer

Shortcut:
To find the area of a regular polygon if length of its each side is given. Area of a regular polygon = [
n / 4
cot(
π / n
)] x (side)2
When 'n' = Number of sides
Now consider the following regular polygons. Octagon: An octagon has 8 sides. (i) Area of a regular Octagon = [
8 / 4
cot(
180 / 8
)] x (side)2
= 2(√2 + 1) x (side)2
Here, side = 6
Using this value in the shortcut, we get:
Area = 2(√2 + 1) x (6)2
= 2(√2 + 1) x 36
= (1.414 + 1) x 72 = 2.414 x 72 = 173.82
Hence, the area is 173.82 sq cm.

98. Find to the nearest metre the side of a regular octagonal enclosure whose area is 1 hectare.
  A.  50 m
  B.  46 m
  C.  42 m
  D.  54 m
     
   
View Answer

Shortcut:
To find the area of a regular polygon if length of its each side is given. Area of a regular polygon = [
n / 4
cot(
π / n
)] x (side)2
When 'n' = Number of sides
Now consider the following regular polygons. Octagon: An octagon has 8 sides. (i) Area of a regular Octagon = [
8 / 4
cot(
180 / 8
)] x (side)2
= 2(√2 + 1) x (side)2
Here, area = 10000
Using this value in the shortcut, we get:
Area = 2(√2 + 1) x (side)2
1 = 2(√2 + 1) x (side)2
(side)2 =
10000 / 2(√2 + 1)

(side)2 = 2071
Hence, the side is 46 m.

99. Find the value of the sum of interior angles of a regular hexagon. Also find the value of each interior angle.
  A.  4π,
2 / 3
π
  B.  2π,
3 / 5
π
  C.  4π,
3 / 8
π
  D.  6π,
4 / 7
π
     
   
View Answer

Shortcut:
There is a regular polygon of 'n' sides. If length of each side is 'z' metre, then the sum of the interior angles is given by (n − 2)π, where, n = > 3 and the value of each interior angle is [
n − 2 / n
π
Here, n = 6
Using this value in the shortcut, we get:
Sum of the interior angles = (6 − 2)π = 4π
value of each interior angle =
6 − 2 / 6
π
=
4 / 6
π
=
2 / 3
π

100. There is a regular polygon of 8 sides. Find the sum of the exterior angles and the value of each exterior angle.
  A.  Sum: 4π; Each exterior angle:
π / 5
  B.  Sum: 3π; Each exterior angle:
π / 3
  C.  Sum: 2π; Each exterior angle:
π / 4
  D.  Sum: 5π; Each exterior angle:
π / 6
     
   
View Answer

Shortcut:
There is a regular polygon of 'n' sides. If length of each side is 'z' metre, then the sum of the exterior angles is given by 2π and the value of each exterior angle is
2π / n

Here, n = 8
Using this value in the shortcut, we get:
sum of the exterior angles is = 2π
Value of each exterior angle =
2π / 8
=
π / 4

Copyright © 2020-2022. All rights reserved. Designed, Developed and content provided by Anjula Graphics & Web Desigining .