Question 1
The area of the square garden is 625 sq.m.What is the area of a path of width 2.5 m around it if the path is out side the garden?
 A 286 sq. m B 275 sq. m C 250 sq. m D 300 sq. m
Mensuration 2D
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 Question 2
What is the area of a triangular field of dimensions 3 4 and 5m?
 A 6 sq. m B 8 sq. m C 14 sq. m D 18 sq. m
Mensuration 2D
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Question 2 Explanation:
given is a right angled triangle
½ * 3* 4 = 6 m2
 Question 3
The area of the rectangle is 2880 sq.m. The ratio of the sides is 5:4, the perimeter is:
 A 216 m B 206 m C 180 m D 160 m
Mensuration 2D
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Question 3 Explanation:
5x * 4x = 2880
x2 = 144
x= 12
perimeter = 60*2+48*2 = 216
 Question 4
If diagonal of a square is 12. What is the area of the square?
 A 68 sq. m B 83 sq . m C 72 sq. m D 80 sq. m
Mensuration 2D
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Question 4 Explanation:
2a2 = 122 a2 = 72
 Question 5
The diagonals of rhombus are 16 cm and 12 cm. Its perimeter is:
 A 30 B 34 C 48 D 40
Mensuration 2D
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 Question 6
The sides of rectangular garden are 25 m x 49 m. What is the perimeter of a square garden with same area?
 A 120 B 108 C 140 D 160
Mensuration 2D
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Question 6 Explanation:
area of square L2= 25 m x 49 m
L = 5*7 = 35
Perimeter of square = 35*4 = 140 meters
 Question 7
If an error of 10% excess is made in calculating the side of a square, the % error in its area is
 A 21 B 121 C 40 D 60
Mensuration 2D
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Question 7 Explanation:
Area of square = L2
=(1.1L)2 = 1.21 L2
=21 %
 Question 8
What is the radius of a circular field of area 616 sq.m?
 A 42 B 40 C 35 D 30
Mensuration 2D
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 Question 9
The perimeter of a field of length 100 m.and breadth is 50 m is:
 A 500 m B 400 m C 300 m D 200 m
Mensuration 2D
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 Question 10
If the radius of a circle is increased by 6%, then the area is increased by:
 A 14.96 B 11.24 C 13.64 D 12.36
Mensuration 2D
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Question 10 Explanation:
New Area of the Circle = Pi * (R + 6% of R)^2 = Pi * (R + 0.06R)^2 = Pi * (1.06R)^2 = Pi * R^2 * (1.06)^2. Therefore, percentage increase in Area = [ Pi * R^2 * (1.06)^2 - Pi * R^2 ]/ (Pi * R^2) = 1.06^2 - 1 = 0.1236 = 12.36%
 Question 11
If an area enclosed by a circle or a square or an equilateral triangle is the same, then the maximum perimeter is possessed by:
 A circle B square C equilateral triangle D triangle and square have equal perimeters greater than that of circle
Mensuration 2D
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Question 11 Explanation:
Let the area be a. Then, Radius of the circle = (a/Pi)^0.5 Side of square = a^0.5 Side of equilateral triangle = (4a/3^0.5)^0.5 Therefore, Perimeter of circle = 2*Pi*(a/Pi)^0.5 = 2*(Pi*a)^0.5 = 2*(3.14*a)^0.5 = 3.54a^0.5 Perimeter of square = 4a^0.5 Perimeter of equilateral triangle = 3(4a/3^0.5)^0.5 = 3(4a/1.732)^0.5 = 3(2.31a)^0.5 = 3*1.52a^0.5 = 4.56a^0.5 Therefore, perimeter of the equilateral triangle is the highest.
 Question 12
The area of the largest triangle that can be inscribed in a semi-circle of radius r, is:
 A r^2 B 2r^2 C r^3 D 2r^3
Mensuration 2D
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Question 12 Explanation:
The largest triangle that can be inscribed is the right-angled triangle with its base as the diameter of the semi-circle. Area = 1/2 * base * height = 1/2 * 2r * r = r^2
 Question 13
An equilateral triangle, a square and a circle have equal perimeters. If T denotes the area of the triangle, S denotes the area of the square, and C denotes the area of the circle, then:
 A S < T < C B T < C < S C T < S < C D S < C < T
Mensuration 2D
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Question 13 Explanation:
Let p be the perimeter. Then, Side of equilateral triangle = p/3 Side of square = p/4 Side of circle = p/(2*Pi) Now, Area of equilateral triangle, T = √3/4a² = 1.732/4 * (p/3)² = 0.0481 p² Area of square, S = (p/4)² = 0.0625 p² Area of circle, C = Pi * [p/(2*Pi)]² = Pi/4Pi² p² = 1/4Pi p² = 0.0796 p² Clearly, T < S < C.
 Question 14
The sides of a triangle are 6 cm, 11 cm and 15 cm. The radius of its in-circle is:
 A 3*(2)^0.5 B (4/5)*(2)^0.5 C (5/4)*(2)^0.5 D 6*(2)^0.5
Mensuration 2D
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Question 14 Explanation:
Let a, b, c be the sides of the triangle. Then, we know Semi-perimeter = 1/2 * (a + b + c) = 1/2 * (6 + 11 + 15) = 16 cm. Area = [s*(s-a)*(s-b)*(s-c)]^0.5 = [16*10*5*1]^0.5 = 20(2)^0.5 cm^2. Therefore, Radius of in-circle = Area/Semi-perimeter = 20(2)^0.5 / 16 = (5/4)*(2)^0.5 cm.
 Question 15
The circumference of a circle is 100 cm. The side of a square inscribed in the circle is:
 A 50(2^0.5) B 100/Pi C [50(2^0.5)]/Pi D [100(2^0.5)]/Pi
Mensuration 2D
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Question 15 Explanation:
Given, Circumference = 100cm 2*Pi*r = 100 r = 50/Pi. The diagonal of the square inscribed, D = 2r = 100/Pi. Area of the square = 1/2 * D^2 = 1/2 * (100/Pi)^2 Therefore, Side of the square = [ 1/2 * (100/Pi)^2 ]^0.5 = 1/(2^0.5) * 100/Pi = 50(2^0.5)/Pi.
There are 15 questions to complete.

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