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Area Problems

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

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

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

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

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 ?

350
150
250
550

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 three sides of a triangle are 15, 25 and x units. Which one of the following is correct?

10 ? x
10
10
10? x ? 40

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

In a triangleSum of two sides is always greater than 3rd side i.e., x < 25 + 15 = 40 .....(i)Difference of two sides is always less than 3rd side i.e., 25 - 15 = 10 < x ...(ii) From Eqs. (i) and (ii) , we get 10 < x < 40

Area Problems
In a circle of radius 21 cm an arc subtends an angle of 72° at the centre. The length of the arc is?

21.6 cm
19.8 cm
13.2 cm
26.4 cm

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Arc length = 2?r (?° / 360°) = 2 x (22/7) x 21 x (72° / 360°) cm= 26.4 cm

Area Problems
A veranda 40 meters long 15 meters broad is to paved with stones each measuring 6 dm by 5 dm. the number of stones required is?

3000
2000
None of these
1000

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Length = (40 x 10 ) dm = 400 dm.Breadth = (15 x 10 ) dm = 150 dm.Area of veranda = (400 x 150 ) dm2Area of one stone = (6 x 5 ) dm2? Required number of stones = (400 x 150) /(6 x 5) = 2000

Area Problems
If the base of a rectangular is increased by 10% and the area is unchanged , then the corresponding altitude must to be decreased by?

111/9 %
10%
11%
91/11 %

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %

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