#everydayquiz #10Question #advancemaths
We are providing you a quiz on Mensuration -I Area and Perimeter(2-Dimensional Plane Figures) of Geometry Section from the Advanced Maths. that can be helpful for all the SSC aspirants exam.
1.The height of an equilateral triangle whose perimeter is 27 cm. is :
(a) 9/2 √3 cm
(b) 9/2 cm
(c) 9 cm
(d) none of these
2.Find the distance between the two parallel sides of a trapezium if the area of the trapezium is 250 sq.m and the two parallel sides are equal to 15 m and 10 m respectively.
(a) 15 m
(b) 20 m
(c) 25 m
(d) 22 m
3.The length of a rectangle is increased by 60%. By what percent would the width be decreased so as to maintain the same area?
(a) 37 1/2%
(b) 60%
(c) 75%
(d) 120%
4.The diagram represents the area swept by the wiper of a car. With the dimensions given in the figure, calculate the shaded area swept by the wiper.
(a) 102.67 cm^2
(b) 205.34 cm^2
(c) 51. 33 cm^2
(d) 208.16 cm^2
5.The ratio between the length and width of the rectangular field is 3 : 2. If only length is increased by 5 m. The new area of the field is 2600 m^2. What is the width of the rectangular field?
(a) 60 m
(b) 50 m
(c) 40 m
(d) 65 m
6.In the given figure, when the outer circles all have radii ‘R’ then the radius of the inner circle will be :
(a) 2/(√2+1)R
(b) 1/√2 R
(c)(√2-1)R
(d)√2 R
7.Find the area of the shaded region if the radius of each of the circle is 1 cm.
(a) 2 - π/3
(b) √3- π
(c)√3-π/2
(d)√3-π/4
8.The perimeter of a rhombus is 40 cm and the measure of an angle is 60, then the area of it is:
(a) 1000 √3 cm^2
(b) 50 √3 cm^2
(c) 160 √3 cm^2
(d) 100 cm^2
9.A right angled isosceles triangle is inscribed in a semi-circle of radius 7cm. The area enclosed by the semi-circle but exterior to the triangle is :
(a) 14 cm^2
(b) 28 cm^2
(c) 44 cm^2
(d) 68 cm^2
10.The area of the square with AC as a side is 128 cm^2. What is the sum of the areas of semicircles drawn on AB and AC as diameters, given ABC is an isosceles right angled triangle and AC is its hypotenuse.
(a) 32 π cm^2
(b) 16π cm^2
(c) 16 cm^2
(d) 32 cm^2
ANSWERS ANS SOLUTIONS :
1.(a) 3a = 27
a = 9cm
height = √3/2 a = 9/2 √3
2.(b) Area = 1/2(sum of parallel sides) * (distance between them)
=> 250 = 1/2(15 + 10) * h
=> h = 20m
3.(a) Let length = width = 100m
If length = 160m, then let width = x m
s.t. 160x = 10000
=> x = 10000/160 = 1000/16 = 62 1/2
so width is reduced to 37 1/2 %
4.(a) Larger Radius (R) = 14 + 7 = 21 cm
Smaller Radius (S) = 7 cm
so Area of shaded portion = πR^2 x θ/(360°)- πr^2 x θ/(360°)
= π (30°)/(360°) (21 * 21 – 7 * 7)
= 22/7 * 1/12 * 7 * 7(8)
= 102.67 cm^2
5.(c) Let length = 3x, then width = 2x
so (3x +5)2x = 2600 ⇒(3x + 5)x = 1300
We, go through the options
Option (c) 2x = 40 ⇒x = 20 which satisfy the above equation
so Width = 2x = 40m
6.(c)
7.(c)ABC is an equilateral triangle with sides = 2cm
so Area of shaded region = Area of ∆ABC – Area of 3 quadrants.
√3/4 (2)^2 - 3(πr^2 θ/(360°))
[θ=60° ∵ ∆ABC is an equilateral triangle]
= √3/4 × 4-3 (π*1*1/6)
= √3-π/2
8.(b)
9.(b)
10.(b)
1.(a) 3a = 27
a = 9cm
height = √3/2 a = 9/2 √3
2.(b) Area = 1/2(sum of parallel sides) * (distance between them)
=> 250 = 1/2(15 + 10) * h
=> h = 20m
3.(a) Let length = width = 100m
If length = 160m, then let width = x m
s.t. 160x = 10000
=> x = 10000/160 = 1000/16 = 62 1/2
so width is reduced to 37 1/2 %
4.(a) Larger Radius (R) = 14 + 7 = 21 cm
Smaller Radius (S) = 7 cm
so Area of shaded portion = πR^2 x θ/(360°)- πr^2 x θ/(360°)
= π (30°)/(360°) (21 * 21 – 7 * 7)
= 22/7 * 1/12 * 7 * 7(8)
= 102.67 cm^2
5.(c) Let length = 3x, then width = 2x
so (3x +5)2x = 2600 ⇒(3x + 5)x = 1300
We, go through the options
Option (c) 2x = 40 ⇒x = 20 which satisfy the above equation
so Width = 2x = 40m
6.(c)
7.(c)ABC is an equilateral triangle with sides = 2cm
so Area of shaded region = Area of ∆ABC – Area of 3 quadrants.
√3/4 (2)^2 - 3(πr^2 θ/(360°))
[θ=60° ∵ ∆ABC is an equilateral triangle]
= √3/4 × 4-3 (π*1*1/6)
= √3-π/2
8.(b)
9.(b)
10.(b)
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