Area problems
- 1. What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high ?
Options- A. 13 m
- B. 14 m
- C. 15 m
- D. 16 m Discuss
Correct Answer: 13 m
Explanation:
The length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high is
=> mts.
- 2. The diagonals of two squares are in the ratio of 2:5. Find the ratio of their areas.
Options- A. 2:25
- B. 3:25
- C. 4:25
- D. 6:25 Discuss
Correct Answer: 4:25
Explanation:
let the diagonals of the squares be 2x and 5x respectively
ratio of their areas = = 4:25
- 3. An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square?
Options- A. 1:2
- B. 1:3
- C. 2:3
- D. 3^(1/2) :2 Discuss
Correct Answer: 3^(1/2) :2
Explanation:
area of a square = a² sq cm
length of the diagonal = cm
area of equilateral triangle with side
=
required ratio =
- 4. The circumference of a circle, whose area is 24.64 sq.m
Options- A. 14.64
- B. 16.36
- C. 17.60
- D. 18.40 Discuss
Correct Answer: 17.60
Explanation:
- 5. In an old Palace there are 50 stairs to go on top floor. Each is 20 cm high, 30 cm wide and 60 cm long. If all the stairs are re-plastered at the rate of Rs. 4 per 100 sq.cm. What is the expenditure ?
Options- A. Rs. 4680
- B. Rs. 7600
- C. Rs. 6000
- D. Rs. 5480 Discuss
Correct Answer: Rs. 6000
Explanation:
Total are of all stairs = 50 (20x60 + 30x60)
= 50 (1200 + 1800)
= 50 x 3000
= 150000 sq cm.
Given rate of 100 sq cm = Rs. 4
Then for 150000 sq. cm = ?
=> 4 x 150000/100 = Rs. 6000
- 6. The length of a rectangle is 10 cm more than the breadth. Find the diagonal length of rectangle if perimeter is 44 cm ?
Options- A. 12.50 cm
- B. 14.21 cm
- C. 15.98 cm
- D. 17.08 cm Discuss
Correct Answer: 17.08 cm
Explanation:
Let the length of the rectangle = L cm
Then the breadth of the rectangle = (L - 10) cm
Perimeter of a rectangle = 2(L + B) cm
=> 44 = 2(L + L - 10)
=> 44 = 4L - 20
=> 4L = 64
=> L = 16 cm
=> B = L - 10 = 6 cm
Diagonal = = = 17.08 cm
- 7. The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.
Options- A. 24
- B. 48
- C. 60
- D. 72 Discuss
Correct Answer: 48
Explanation:
Let ABC be the isosceles triangle and AD be the altitude
Let AB = AC = x. Then, BC = (32 - 2x).
Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC,
BC = (32- 2x) = (32 - 20) cm = 12 cm.
Hence, required area =
- 8. In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements.
Options- A. .7%
- B. 0.8%
- C. 0.9%
- D. 0.3% Discuss
Correct Answer: 0.8%
Explanation:
let x and y be the sides of the rectangle then
correct area =
Error% = %
- 9. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is
Options- A. 12
- B. 32
- C. 42
- D. 52 Discuss
Correct Answer: 42
Explanation:
required area = [14 x 14 - (4 x 1/4 x 22/7 x 7 x 7) ]sq.cm
= 196 - 154 = 42 sq.cm.
- 10. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
Options- A. 315
- B. 385
- C. 415
- D. 485 Discuss
Correct Answer: 315
Explanation:
Let the areas of the two parts be x and (700-x) hectares
therefore,
So, the two parts are 385 and 315.
Hence, Area of the smaller = 315 hectares
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