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Area Problems

Area Problems
The area of an equilateral triangle whose side is 8 cms is?

16?3 cm2
21.3 cm2
4?3 cm2
64 cm2

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Req. Area = ?3/4 x a2 = ?3/4 x 82 cm2= 16?3 cm2

Area Problems
The length of a room is 5.5 m and width is 3.75 m. Find the cost of paying the floor by slabs at the rate of Rs. 800 per sq.metre.

Rs.15500
Rs.16500
Rs.17500
Rs.18500

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area = 5.5 × 3.75 sq. metre. Cost for 1 sq. metre. = Rs. 800 Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

Area Problems
The Area of a square is 50 sq. units. Then the area of the circle drawn on its diagonal is?

100? sq. units
25? sq. units
50? sq. units
None of theas

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area = (Diagonal)2 / 2 = 50? Diagonal = 10 units ? Radius of required circle = 5 unitsIts area = [? x (5)2 ] cm2= 25? units2

Area Problems
The length of a rectangular plot is twice of its width. If the length of a diagonal is 9?5 meters, the perimeter of the rectangular is?

54 m
27 m
81 m
None of these

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let breadth = y meters,Then, length = 2y meters? Diagonal = ?y2 + (2y)2 = ?5y2 metersSo, ?5y2 = 9 ?5? y= 9Thus, breadth = 9 m and length = 18 m? Perimeter = 2 (18 + 9) m = 54m.

Area Problems
The diameter of the driving wheel of a bus is 140 cm. How many revolution, per minute must the wheel make in order to keep a speed of 66 kmph ?

550
250
350
150

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Circumference = No.of revolutions * Distance covered   Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.Number of revolutions per min. =(1100/4.4) = 250.

Area Problems
The length of each side of an equilateral triangle having an area of 4?3 cm 2 is?

4/?3 cm
3 cm
4 cm
?3/4 cm

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area of equilateral triangle = ?3/4 a2 = 4?3.? a2 = 16? a = 4 cm

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