If the length of the diagonal of a square is 20cm, | Area Questions & Answers

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.

$2 x \times x = 72 = > 2 x^{2} = 72 = > x = 6$

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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, $\frac{\frac{1}{2} \times x \times 3 h}{\frac{1}{2} \times y \times 4 h} = \frac{4}{3} \Rightarrow \frac{x}{y} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$

Required ratio = 16 : 9.

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23 18380

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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48 24669

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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A) a	B) b
C) c	D) d

If the length of the diagonal of a square is 20cm,then its perimeter must be

a. $40 \sqrt{2}$ cm b. $30 \sqrt{2}$ cm c. 10cm d. 15 $\sqrt{2}$ cm

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If the length of the diagonal of a square is 20cm,then its perimeter must be a. 402cm b. 302cm c. 10cm d. 152cm

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If the length of the diagonal of a square is 20cm,then its perimeter must be

a. $40 \sqrt{2}$ cm b. $30 \sqrt{2}$ cm c. 10cm d. 15 $\sqrt{2}$ cm