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?

286 sq. m | |

275 sq. m | |

250 sq. m | |

300 sq. m |

**Mensuration 2D**

**Discuss it**

Question 2 |

What is the area of a triangular field of dimensions 3 4 and 5m?

6 sq. m | |

8 sq. m | |

14 sq. m | |

18 sq. m |

**Mensuration 2D**

**Discuss it**

Question 2 Explanation:

given is a right angled triangle

½ * 3* 4 = 6 m2

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:

216 m | |

206 m | |

180 m | |

160 m |

**Mensuration 2D**

**Discuss it**

Question 3 Explanation:

5x * 4x = 2880

x2 = 144

x= 12

perimeter = 60*2+48*2 = 216

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?

68 sq. m | |

83 sq . m | |

72 sq. m | |

80 sq. m |

**Mensuration 2D**

**Discuss it**

Question 4 Explanation:

2a

^{2}= 12^{2}a^{2}= 72Question 5 |

The diagonals of rhombus are 16 cm and 12 cm. Its perimeter is:

30 | |

34 | |

48 | |

40 |

**Mensuration 2D**

**Discuss it**

Question 6 |

The sides of rectangular garden are 25 m x 49 m. What is the perimeter of a square garden with same area?

120 | |

108 | |

140 | |

160 |

**Mensuration 2D**

**Discuss it**

Question 6 Explanation:

area of square L2= 25 m x 49 m

L = 5*7 = 35

Perimeter of square = 35*4 = 140 meters

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

21 | |

121 | |

40 | |

60 |

**Mensuration 2D**

**Discuss it**

Question 7 Explanation:

Area of square = L2

=(1.1L)2 = 1.21 L2

=21 %

=(1.1L)2 = 1.21 L2

=21 %

Question 9 |

The perimeter of a field of length 100 m.and breadth is 50 m is:

500 m | |

400 m | |

300 m | |

200 m |

**Mensuration 2D**

**Discuss it**

Question 10 |

If the radius of a circle is increased by 6%, then the area is increased by:

14.96 | |

11.24 | |

13.64 | |

12.36 |

**Mensuration 2D**

**Discuss it**

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:

circle | |

square | |

equilateral triangle | |

triangle and square have equal perimeters greater than that of circle |

**Mensuration 2D**

**Discuss it**

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:

r^2 | |

2r^2 | |

r^3 | |

2r^3 |

**Mensuration 2D**

**Discuss it**

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:

S < T < C | |

T < C < S | |

T < S < C | |

S < C < T |

**Mensuration 2D**

**Discuss it**

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:

3*(2)^0.5 | |

(4/5)*(2)^0.5 | |

(5/4)*(2)^0.5 | |

6*(2)^0.5 |

**Mensuration 2D**

**Discuss it**

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:

50(2^0.5) | |

100/Pi | |

[50(2^0.5)]/Pi | |

[100(2^0.5)]/Pi |

**Mensuration 2D**

**Discuss it**

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.