Name: Proof of Angle in the Same Segment of a Circle
Uploaded: 2019-07-19
Duration: 1 min 26 s
Description: The most important and simple theorem from the chapter circle that angle in the same segment of the circle are equal. Let's derive this theorem.

Question

Two chords

A B

and

C D

of lengths

5 c m

and

11 c m

respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between

A B

and

C D

is

6 c m

, find the radius of the circle.

Accepted Answer

Given: $A B ∥ C D$ , $A B = 5 c m$ , $C D = 11 c m$ and $O P = 4 c m$ .

Join $O A$ and $O C$ .

Let the radius of the circle be $r c m$ and $O$ be the centre.

Draw $O P ⊥ A B$ and $O Q ⊥ C D .$

We know, $O Q ⊥ C D, O P ⊥ A B and A B ∥ C D .$

Therefore, points $P, O and Q are collinear . So P Q = 6 c m .$

Let $O P = x .$

Then $O Q = (6 - x) c m$

and $O A = O C = r$

Also, $A P = P B = 2.5 c m$ and $C Q = Q D = 5.5 c m .$

In right triangles $Q A P$ and $O C Q,$ we have

$O A^{2} = O P^{2} + A P^{2} and O C^{2} = O Q^{2} + C Q^{2}$

$∴ r^{2} = x^{2} + {(2.5)}^{2} . . . . . . (1)$

$and r^{2} = {(6 - x)}^{2} + {(5.5)}^{2} . . . . . . (2)$

$\Rightarrow x^{2} + {(2.5)}^{2} = {(6 - x)}^{2} + {(5.5)}^{2}$

$\Rightarrow x^{2} + 6.25 = 35 + x^{2} - 12 x + 30.25$

$\Rightarrow 12 x = 60$

$\Rightarrow x = 5$

$Putting x = 5 in equation (1) we get$

$r^{2} = 25 + {(2.5)}^{2}$

$\Rightarrow r^{2} = 25 + 6.25$

$\Rightarrow r^{2} = 31.25$

$\Rightarrow r = 5.6 c m$

Gujarat Board Solutions for Chapter: Circles, Exercise 6: EXERCISE (Optional)*

Gujarat Board Mathematics Solutions for Exercise - Gujarat Board Solutions for Chapter: Circles, Exercise 6: EXERCISE (Optional)*

Questions from Gujarat Board Solutions for Chapter: Circles, Exercise 6: EXERCISE (Optional)* with Hints & Solutions

Learn and Prepare for any exam you want !