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Q:

In the given figure, PQRS is a square of side 8 cm. PQO = 60 deg. What is the area (in sq.cm) of the triangle POQ?

A) 32√3 B) 24[(√3) – 1]
C) 48[(√3) – 1] D) 16[3 – (√3)]
 
Answer & Explanation Answer: D) 16[3 – (√3)]

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Q:

In the given figure, in a right angle triangle ABC, AB = 12 cm and AC = 15 cm. A square is inscribed in the triangle. One of the vertices of square coincides with the vertex of triangle. What is the maximum possible area (in cm.sq) of the square?

A) 1296/49 B) 25
C) 1225/36 D) 1225/64
 
Answer & Explanation Answer: A) 1296/49

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Q:

In the given figure, E and F are the centers of two identical circles. What is the ratio of area of triangle AOB to the area of triangle DOC?

A) 1 : 3 B) 1 : 9
C) 1 : 8 D) 1 : 4
 
Answer & Explanation Answer: B) 1 : 9

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Q:

In the given figure, from the point P two tangents PA and PB are drawn to a circle with centre Oand radius 5 cm. From the point O, OC and OD are drawn parallel to PA and PB respectively. If the length of the chord ABis 5 cm, then what is the value (in deg) of COD?

A) 90 B) 120
C) 150 D) 135
 
Answer & Explanation Answer: B) 120

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Q:

In the given figure, a circle touches the sides of the quadrilateral PQRS. The radius of the circle is 9 cm. RSP = SRQ = 60 deg and PQR = QPS = 120 deg. What is the perimeter (in cm) of the quadrilateral?

A) 36√3 B) 24√3
C) 48√3 D) 32
 
Answer & Explanation Answer: C) 48√3

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Q:

In the given figure, if  QR/XY = 14/9 and PY = 18 cm, then what is the value (in cm) of PQ?

 

 

A) 28 B) 18
C) 21 D) 24
 
Answer & Explanation Answer: A) 28

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Q:

In the given figure, OX, OY and OZ are perpendicular bisectors of the three sides of the triangle. If QPR = 65 deg and PQR = 60 deg, then what is the value (in deg) of QOR + POR ?

A) 250 B) 180
C) 210 D) 125
 
Answer & Explanation Answer: A) 250

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Q:

Can You Guess The Missing Numbers In The Given Math Puzzle?

A) B)
C) D)
 
Answer & Explanation Answer: A)

Explanation:

Let the missing numbers be x, y, a and b

x + y = 8 ---- (1)
a - b = 6 ---- (2)
x + a = 13 --- (3)
y + b = 8 ---- (4)

2x + a + y = 21 --- From (1 & 3)
a + y = 14 ----  From (2 & 4)

Substitute (2 & 4) in (1 & 3)

2x + 14 = 21
2x = 7
x = 3.5

From (1) ---> 3.5 + y = 8
=> y = 8 - 3.5 = 4.5

a + y = 14 => a = 14 - y = 14 - 4.5 => a = 9.5
From (4) ---- y + b = 8 => b = 8 - y = 8 - 4.5 = 3.5

Hence,
x = 3.5
y = 4.5
a = 9.5
b = 3.5

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