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Q:

Find the length of the altitude of an equilateral triangle of side $3 \sqrt{3}$ cm.

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 = $a \frac{\sqrt{3}}{2}$ = $3 \sqrt{3} \times \frac{\sqrt{3}}{2}$ = 4.5

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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, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

$\Rightarrow x^{2} = {(8)}^{2} + {(16 - x)}^{2} \Rightarrow x = 10$

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

Hence, required area = $\frac{1}{2} * B C * A D = \frac{1}{2} * 12 * 8 = 48 c m^{2}$

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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
$13.5 * 10000 m^{2} = 135000 m^{2}$
Let altitude = x metres and base = 3x metres.
Then, $\frac{1}{2} * 3 x * x = 135000 \Rightarrow x^{2} = 90000 \Rightarrow x = 300$

Base = 900 m and Altitude = 300 m.

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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 = $\sqrt{(13^{2} - 12^{2})} = \sqrt{25}$ = 5 cm.

Its area = $\frac{1}{2} * b a s e * h e i g h t$ = $\frac{1}{2} * 12 * 5$ = $30 c m^{2}$ .

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Q:

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

A) 105555	B) 106665
C) 106656	D) 108333

Answer & Explanation Answer: C) 106656

Explanation:

The number of arrangements of 4 different digits taken 4 at a time is given by $4 P_{4}$ = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

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Q:

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.

A) 84	B) 64
C) 44	D) 22

Answer & Explanation Answer: A) 84

Explanation:

Let a = 13, b = 14 and c = 15. Then, $s = \frac{1}{2} (a + b + c)$ =21

(s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = $\sqrt{[s (s - a) (s - b) (s - c)]}$ = $\sqrt{(21 \times 8 \times 7 \times 6)}$ = 84 sq.cm

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Q:

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

A) b=6; l=18; H=6	B) b=5; l=6; H=18
C) l=6; b=18; H=15	D) l=5; b=18; H=18

Answer & Explanation Answer: A) b=6; l=18; H=6

Explanation:

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m

$x \times \frac{3 x}{2} = 54 \Rightarrow x^{2} = 54 \times 2 \Rightarrow x = 6$

So, breadth = 6 m and length = $\frac{3 (6)}{2}$ = 9 m.

Now, papered area = (1720/10) = 172 sq.m

Area of 1 door and 2 windows = 8 sq.m

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m

$2 \times (9 + 6) H = 180 \Rightarrow H = 6$

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Q:

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A) 20	B) 30
C) 40	D) 50

Answer & Explanation Answer: D) 50

Explanation:

Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

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