The ratio of the areas of the in-circle and the circum-circle of a square is :
(A) 1 : 2
(B) $$\sqrt 2 :1$$
(C) $$1:\sqrt 2 $$
(D) 2 : 1
Solution:
Let r1 and r2 be the radii of the in-circle and circum-circle of a square respectively and let each side of the square be a. Then, $$\eqalign{ & {r_1} = \frac{a}{2} \cr & {r_2} = \frac{1}{2} \times {\text{diagonal of the sequence}} \cr & {r_2} = \frac{1}{2} \times \sqrt 2 a \cr & {r_2} = \frac{{\sqrt 2 a}}{2}cm \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{\pi \times {{\left( {\frac{a}{2}} \right)}^2}}}{{\pi \times {{\left( {\frac{{\sqrt 2 a}}{2}} \right)}^2}}} \cr & = 1:2 \cr} $$
62.
One side of a right-angled triangle is twice the other, and the hypotenuse is 10 cm. The area of the triangle is :
(A) 20 cm2
(B) $$33\frac{1}{3}$$ cm2
(C) 40 cm2
(D) 50 cm2
Solution:
Let the sides be a cm and 2a cm Then, $$\eqalign{ & {a^2} + {\left( {2a} \right)^2} = {\left( {10} \right)^2} \cr & \Rightarrow 5{a^2} = 100 \cr & \Rightarrow {a^2} = 20 \cr} $$ ∴ Area : $$\eqalign{ & = \left( {\frac{1}{2} \times a \times 2a} \right) \cr & = {a^2} \cr & = 20\,c{m^2} \cr} $$
63.
A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:
The area of a circular field is equal to the area of a rectangular field. The ratio of the length and the breadth of the rectangular field is 14 : 11 respectively and perimeter is 100 metres. What is the diameter of the circular field ?
(A) 14 m
(B) 22 m
(C) 24 m
(D) 28 m
Solution:
$$\eqalign{ & 2\left( {14x + 11x} \right) = 100 \cr & \Rightarrow 25x = 50 \cr & \Rightarrow x = 2 \cr} $$ So, length and breadth of the rectangular field are 28 m and 22 m respectively Area of the circle = Area of the rectangular field = (28 × 22) m2 = 616 m2 Let the radius of the circle be r metres Then, $$\eqalign{ & \frac{{22}}{7} \times {r^2} = 616 \cr & \Rightarrow {r^2} = \frac{{616 \times 7}}{{22}} \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14\,m \cr} $$ ∴ Diameter : $$\eqalign{ & = \left( {2 \times 14} \right)\,m \cr & = 28\,m \cr} $$
65.
ABCD is a rectangle and E and F are the mid-points of AD and DC respectively. Then the ratio of the areas of EDF and AEFC would be :
(A) 1 : 2
(B) 1 : 3
(C) 1 : 4
(D) 2 : 3
Solution:
Let AD = x and DC = y Then, $$AE = ED = \frac{x}{2}{\text{ and}}$$ $$DE = FC = \frac{y}{2}$$ $${\text{Area }}\left( {\vartriangle EDF} \right) = \left( {\frac{1}{2} \times \frac{x}{2} \times \frac{x}{2}} \right)$$ $$ = \frac{{xy}}{8}$$ $${\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = $$ $${\text{Area }}\left( {\vartriangle ADC} \right) - $$ $${\text{Area}}\left( {\vartriangle EDF} \right)$$ $$\eqalign{ & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{xy}}{2} - \frac{{xy}}{8} \cr & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{3xy}}{8} \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{xy}}{8}:\frac{{3xy}}{8} \cr & = 1:3 \cr} $$
66.
The ratio of bases of two triangles is x : y and that of their areas is a : b. Then the ratio of their corresponding altitudes will be :
The perimeter of a rectangle is 60 metres. If its length is twice its breadth, then its area is :
(A) 160 m2
(B) 180 m2
(C) 200 m2
(D) 220 m2
Solution:
Let the breadth of the rectangle be x metres Then, length of the rectangle = 2x metres ⇒ 2(2x + x) = 60 ⇒ 6x = 60 ⇒ x = 10 So, length = 20 m, breadth = 10 m ∴ Area = (20 × 10) m2 = 200 m2
68.
The breadth of a rectangular field is $$\frac{3}{4}$$ of its length and its area is 300 sq. metres. What will be the area (in sq. metres) of the garden of breath 1.5 metres developed around the field ?
(A) 96 m2
(B) 105 m2
(C) 114 m2
(D) Cannot be determined
Solution:
Let the length of the field be x metres Then, breadth of the field = $$\frac{3x}{4}$$ metres $$\eqalign{ & x \times \frac{{3x}}{4} = 300 \cr & \Rightarrow {x^2} = 300 \times \frac{4}{3} \cr & \Rightarrow {x^2} = 400 \cr & \Rightarrow x = 20 \cr} $$ So, length = 20 m, breadth = 15 m ∴ Area of the garden : $$\eqalign{ & = \left[ {\left\{ {\left( {20 + 3} \right) \times \left( {15 + 3} \right)} \right\} - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left[ {\left( {23 \times 18} \right) - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left( {414 - 300} \right){m^2} \cr & = 114\,{m^2} \cr} $$
69.
The area of the four walls of a room is 120 m2 and the length is twice the breadth. If the height of the room is 4 m, then the area of the floor is :
(A) 48 m2
(B) 49 m2
(C) 50 m2
(D) 52 m2
Solution:
Let the breadth = x metres and length = (2x) metres Area of 4 walls = [2(2x + x)× 4] m2 = (24x) m2 ∴ 24x = 120 ⇒ x = 5 So, length = 10 m, and breadth = 5 m Area of the floor = (10 × 5) m2 = 50 m2
70.
The diameter of a circle is equal to the perimeter of a square whose area is 3136 cm2 . What is the circumference of the circle ?
(A) 352 cm
(B) 704 cm
(C) 39424 cm
(D) 1024 cm
Solution:
Area of square = 3136 cm2 Side of squared = $$\sqrt {3136} $$ = 56 cm Perimeter of square : = 4a = (4 × 56) cm = 224 cm = diameter of circle ∴ Circumference of circle : $$\eqalign{ & = \pi d \cr & = \frac{{22}}{7} \times 224 \cr & = 704{\text{ cm}} \cr} $$