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- Question
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
- Correct Answer
- 4:25
Explanationlet the diagonals of the squares be 2x and 5x respectively
ratio of their areas = = 4:25
- 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
- 2. The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter
Options- A. Rs.16500
- B. Rs.15500
- C. Rs.17500
- D. Rs.18500 Discuss
- 3. A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle.
Options- A. 1cm
- B. 2cm
- C. 3cm
- D. 4cm Discuss
- 4. What is the area of an obtuse angled triangle given below with two sides are 8 and 12 and the angle included between two sides is 150 deg?
Options- A. 48 sq units
- B. 24 sq units
- C. 12 sq units
- D. 6 sq units Discuss
- 5. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is
Options- A. 200m
- B. 300m
- C. 400m
- D. 100m Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Area problems
Search Results
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.
Correct Answer: Rs.16500
Explanation:
l=5.5m w=3.75m
area of the floor = 5.5 x 3.75 = 20.625 sq m
cost of paving = 800 x 20.625 = Rs. 16500
Correct Answer: 3cm
Explanation:
r =3
Correct Answer: 24 sq units
Explanation:
We know that,
The area of a triangle with two sides given and included angle
A = 1/2 x product of sides x Sin(angle)
Here the two sides are 8 & 12
Angle = 150
Area = 1/2 x 8 x 12 x sin150
Sin(150) = sin(90+60) = cos(60) = 1/2
A = 48 x 1/2 = 24
Area of the given triangle = 24 sq units.
Correct Answer: 200m
Explanation:
Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
Area of triangle =
Area of rectangle = b*h2
As per Given,
h1=200
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 =
Correct Answer: 17.60
Explanation:
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
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
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 =
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