Question 1

PQRS is a square. The ∠ SRP is equal to :

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Question 2

The two sides of a right triangle containing the right angle measure 3 cm and 4 cm. The radius of the in circle of the triangle is :

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Question 3

If one of the diagonals of a rhombus is equal to its side, then the diagonals of the rhombus are in the ratio :

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Let ABCD is a rhombus in which AB = BC = CD = DA = diagonal CA. In triangle ABC , AB = BC = CA , therefore ∠B = 60° ∠D = ∠B = 60° ∠A=180 - ∠B = 180° - 60° =120° ∠C = ∠A=120° Hence ,The ratio of diagonals = √2 : 1

Question 4

We have an angle of 2 1°/ 2 .How big it will look through a glass that magnifies things three times?

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Question 5

Two circles of unit radius touch each other and each of them touches internally a circle of radius two, as shown in the following figure. The radius of the circle which thouches all the three circles:

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Question 6

If, in the following figure, PA = 8 cm, PD = 4 cm, CD equal to 3 cm, then AB is :

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Question 7

In the given figure, what is the length of AD in terms of b and c :

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Question 8

In a quadrilateral ABCD, ∠B = 90° and AD 2 = AB 2 + BC 2 + CD 2, then ∠ACD is equal to:

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In a quadrilateral ABCD, Given :- ∠B = 90° AB2 + BC2 + CD2 = AC2 + CD2 = AD2 { ∴ In triangle ABC , AB2 + BC2 = AC2 } So, in triangle ACD, angle opposite to AD = 90° ( by Converse of Pythagoras theorem) ⇒ ∠ACD = 90°

Question 9

If P and Q are the mid points of the sides CA and GB respectively of a triangle ABC, right-angled at C. Then the value of 4 ( AQ 2 + BP 2) is equal to :

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Question 10

In a triangle ABC, ∠A = x°, ∠B = y° and ∠C = (y + 20)°. If 4x – y = 10, then the triangle is :

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Given :- ∠A = x° , ∠B = y° , ∠C = ( y + 20 )° and 4x – y = 10 We know that , ∠A + ∠B + ∠C = 180° ⇒ x + y + ( y + 20 ) = 180° ⇒ x + 2y = 180 - 20 = 160° ........... ( ⅰ ) 4x – y = 10 ........... ( ⅱ ) From ( ⅰ ) and ( ⅱ ) , we get ⇒ y = 70 , x = 20 ∴ The angles of the triangle are 20° , 70° , 90° . Hence , the triangle will be Right-angle triangle .

Question 11

In a circular lawn, there is a 16 m long path in the form of a chord. If the path is 6 m away from the center of the lawn, then find the radius of the circular lawn.

Accepted Answer

From given figure , we can see that Let AB = 16 m , OM = 6 m , AM = BM = 8 m and OA = r In triangle AMO , By pythagoras theorem , OA2 = AM2 + OM2 OA2 = 82 + 62 = 64 + 36 OA2 = 100 ⇒ OA = √ 100 = 10 ∴ OA = r = 10 m The radius of the circular lawn = 10 m

Question 12

Of all the chords of a circle passing through a given point in it, the smallest is that which :

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Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle. Let CD is another chord passes through point M. We have to prove that AB < CD. Now join OM and draw OL perpendicular to CD. In right angle triangle OLM, OM is the hypotenuse. So OM > OL ⇒ chord CD is nearer to O in comparison to AB. ⇒ CD > AB ⇒ AB < CD So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point. Hence required answer will be option D .

Question 13

ABCD is a parallelogram and X, Y are the mid-points of sides AB and CD respectively. Then quadrilateral AXCY is a :

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Question 14

In the figure, BD and CD are angle bisectors of ∠ ABC and ∠ ACE, respectively. Then ∠ BDC is equal to :

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Question 15

In a Δ ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of ∠ OBC and ∠ QCB intersect at a point O. Then ∠ BOC is equal to:

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Question 16

In a Δ ABC, the bisectors of ∠ B and ∠ C intersect each other at a point O. Then ∠ BOC is equal to :

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Question 17

In the fig. XY || AC and XY divides triangular region ABC into two part equal in area. Then AX is equal to : AB

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Question 18

Two poles of ht. a and b meters are p meters apart ( b > a). The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is :

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Plane Geometry Questions