Mensuration 2D

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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