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- Question
The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?
Options- A. 50
- B. 60
- C. 70
- D. 80
- Correct Answer
- 60
Explanationlet ABC be the isosceles triangle, the AD be the altitude
Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base]
since in an isoceles triange the altitude bisects the base so
BD = DC = 16-x
In a triangle ADC,
BC = 32-2x = 32-20 = 12 cm
Hence, required area = = = 60 sq cm
- 1. One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?
Options- A. 110
- B. 120
- C. 130
- D. 140 Discuss
- 2. The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is
Options- A. 49
- B. 154
- C. 378
- D. 1078 Discuss
- 3. The ratio of the area of a square to that of the square drawn on its diagonal is
Options- A. 1:2
- B. 2;3
- C. 3:1
- D. 4:1 Discuss
- 4. A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be
Options- A. 7744
- B. 8844
- C. 5544
- D. 4444 Discuss
- 5. Find the ratio of the areas of the incircle and circumcircle of a square.
Options- A. 1:1
- B. 1:2
- C. 1:3
- D. 1:4 Discuss
- 6. If the length of the diagonal of a square is 20cm,then its perimeter must be a. 402 cm b. 302 cm c. 10cm d. 15 2 cm
Options- A. a
- B. b
- C. c
- D. d Discuss
- 7. The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750 sq.m. The length of the hall is
Options- A. 20
- B. 25
- C. 30
- D. 35 Discuss
- 8. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 9. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
Options- A. B=900;H=300
- B. B=300;H=900
- C. B=600;H=700
- D. B=500;H=900 Discuss
- 10. A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is
Options- A. Rs. 258
- B. Rs. 358
- C. Rs. 458
- D. Rs. 558 Discuss
Area problems
Search Results
Correct Answer: 120
Explanation:
Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m
Correct Answer: 154
Explanation:
The diameter is equal to the shortest side of the rectangle.
So radius= 14/2 = 7cm.
Therefore,
Correct Answer: 1:2
Explanation:
ratio =
Correct Answer: 7744
Explanation:
length of wire =
= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm
Correct Answer: 1:2
Explanation:
Let the side of the square be x. Then, its diagonal =
Radius of incircle =
Radius of circum circle=
Required ratio =
Correct Answer: a
Explanation:
We know that d=?2s
Given diagonal = 20 cm
=> s = 20/ cm
Therefore, perimeter of the square is 4s = 4 x 20/ = 40 cm.
Correct Answer: 30
Explanation:
Let breadth = x m
Then, length = (x+5)m
Area of a rectangle = Length x Breadth
x(x+5) = 750
x² + 5x - 750= 0
(x+30)(x-25)= 0
x = 25 or x = -30
Hence, breadth x = 25m
=> Length = x + 5 = 25 + 5 = 30m.
Correct Answer: 3
Explanation:
Area of the park = (60 x 40) = 2400
Area of the lawn = 2109
Area of the crossroads = (2400 - 2109) = 291
Let the width of the road be x metres. Then,
(x - 97)(x - 3) = 0
x = 3.
Correct Answer: B=900;H=300
Explanation:
Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
Let altitude = x metres and base = 3x metres.
Then,
Base = 900 m and Altitude = 300 m.
Correct Answer: Rs. 558
Explanation:
Area to be plastered =
=
Cost of plastering =
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