**Practice Questions on Circles and Cyclic Quadrilateral :**

Here we are going to see some example problems on circles and cyclic quadrilaterals.

**Question 1 :**

In the given figure, AB and CD are the parallel chords of a circle with centre O. Such that AB = 8cm and CD = 6cm. If OM ⊥ AB and OL ⊥ CD distance between LM is 7cm. Find the radius of the circle.

**Solution :**

Let us join the points OC, OD, OA and OB.

OA = OB = OC and OD are equal in length and it is radius.

Let OL = x

In triangle OLD,

CL = CD = 3 cm

OL^{2} + LD^{2} = OD^{2}

x^{2} + 3^{2} = OD^{2}

OD^{2} = x^{2} + 9 ------------(1)

In triangle OAB,

OM = 7 - x

AM = BM = 4 cm

OM^{2} + MB^{2} = OB^{2}

(7 - x)^{2} + 4^{2} = OB^{2}

OB^{2} = (7-x)^{2} + 16 ------------(2)

(1) = (2)

(7-x)^{2} + 16 = x^{2} + 9

49 - 14x + x^{2} + 16 - x^{2} - 9 = 0

-14x + 56 = 0

x = 56/14 = 4

By applying the value of x in (1), we get

OD^{2} = 4^{2} + 9

OD^{2} = 16 + 9 = 25

OD = 5 cm

**Question 2 :**

The arch of a bridge has dimensions as shown, where the arch measure 2 m at its highest point and its width is 6 m. What is the radius of the circle that contains the arch?

**Solution :**

DB = 2 m and DB = 3 m

CE = CB = AC (radius)

Let DC = x

In triangle CDB,

CB^{2} = CD^{2} + DB^{2}

CB^{2} = x^{2} + 3^{2 } ---(1)

CE = x + 2

By applying the value of CE = x + 2 (CB) in (1), we get

(x + 2)^{2} = x^{2} + 9

x^{2} + 4x + 4 - x^{2} - 9 = 0

4x - 5 = 0

x = 5/4

x = 1.25

CE = 1.25 + 2 = 3.25 m

**Question 3 :**

In figure, <ABC =120 , where A,B and C are points on the circle with centre O. Find <OAC ?

**Solution :**

**Question 4 :**

A school wants to conduct tree plantation programme. For this a teacher allotted a circle of radius 6m ground to nineth standard students for planting sapplings. Four students plant trees at the points A,B,C and D as shown in figure. Here AB = 8m, CD = 10m and AB is perpendicular to CD. If another student places a flower pot at the point P, the intersection of AB and CD, then find the distance from the centre to P.

**Solution :**

AM = (1/2) AB

AM = (1/2)8 = 4

CN = (1/2) 10 = 5

OA = 6

OA^{2} = OM^{2} + MO^{2}

6^{2} = OM^{2} + 4^{2}

36 - 16 = OM^{2}

OM = √20 = 2√5

OD^{2} = ON^{2} + ND^{2}

6^{2} = ON^{2} + 5^{2}

36 - 25 = ON^{2}

ON = √11

OMPN is a rectangle and OP is a diagonal.

OP^{2} = ON^{2} + PN^{2}

OP^{2} = √11^{2} + (√20)^{2}

OP^{2} = 11 + 20

OP = √31 = 5.56 (approximately)

Question 5 :

In the given figure, <POQ = 100 and <PQR = 30, then find <RPO .

Solution :

After having gone through the stuff given above, we hope that the students would have understood, "Practice Questions on Circles and Cyclic Quadrilateral"

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