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Q:

The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.

A) 2808 B) 3808
C) 4808 D) 5808
 
Answer & Explanation Answer: D) 5808

Explanation:

Area = (13.86 x 10000) sq.m = 138600 sq.m

 

πR2=138600R2=138600×722R=210m 

 

Circumference = 2πR2=2×227×210=1320m 

 

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

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23 24870
Q:

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A) 27 and 23 B) 24 and 23
C) 25 and 23 D) 22 and 23
 
Answer & Explanation Answer: A) 27 and 23

Explanation:

Let the two parallel sides of the trapezium be a cm and b cm. 

Then, a - b = 4 

And, 12×a+b×19=475=>a+b=50 

Solving (i) and (ii), we get: a = 27, b = 23. 

So, the two parallel sides are 27 cm and 23 cm.

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23 23581
Q:

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq. cm, find its height

A) 6 B) 7
C) 8 D) 9
 
Answer & Explanation Answer: A) 6

Explanation:

Let the height of the parallelogram be x. cm. Then, base = (2x) cm.

2x×x=72=>2x2=72=>x=6 

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32 22160
Q:

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.

A) 13:9 B) 16:9
C) 15:9 D) 14:9
 
Answer & Explanation Answer: B) 16:9

Explanation:

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then, 12×x×3h12×y×4h=43xy=43×43=169 

 

Required ratio = 16 : 9.

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23 18367
Q:

Find the length of the altitude of an equilateral triangle of side  33cm.

A) 4.5 B) 3.5
C) 2.5 D) 6.5
 
Answer & Explanation Answer: A) 4.5

Explanation:

Let the length of the altitude be h.
Then, h = a32   = 33 × 32 = 4.5

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3 7323
Q:

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.

A) 24 B) 48
C) 60 D) 72
 
Answer & Explanation Answer: B) 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,AC2=AD2+DC2 

x2=82+16-x2x=10 

BC = (32- 2x) = (32 - 20) cm = 12 cm. 

Hence, required area =  12*BC*AD=12*12*8=48cm2

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11 13567
Q:

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.

A) B=900;H=300 B) B=300;H=900
C) B=600;H=700 D) B=500;H=900
 
Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
13.5*10000m2=135000m2 
Let altitude = x metres  and  base = 3x metres.
Then, 12*3x*x=135000x2=90000x=300 

Base = 900 m and Altitude = 300 m.

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62 27630
Q:

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm.

A) 30 B) 40
C) 50 D) 60
 
Answer & Explanation Answer: A) 30

Explanation:

Height of the triangle = 132-122=25 = 5 cm.

 

Its area =12*base*height12*12*530cm2.

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48 24647