Question

A circle has radius root2 cm. It is divided into two segments by a chord of length 2cm. Prove that the angle subtended by the chord a point in major segment is 45 degree.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Hence proved that the angle subtended by the chord a point in major segment $\angle QPR=45^{\circ}$

To prove:

The angle subtended in a major segment by a chord will be 45 degrees.

Solution:

Given

$\begin{array} { l } { \mathrm { OR } = \mathrm { OQ } = \sqrt { 2 } \mathrm { cm } } \\\\ { \mathrm { QR } = 2 \mathrm { cm } } \end{array}$

In Triangle OQR

$\begin{array} { l } { \mathrm { OQ } ^ { 2 } + \mathrm { OR } ^ { 2 } = ( \sqrt { 2 } ) ^ { 2 } + ( \sqrt { 2 } ) ^ { 2 } } \\\\ { \mathrm { OQ } ^ { 2 } + \mathrm { OR } ^ { 2 } = 2 + 2 } \\\\ { \mathrm { OQ } ^ { 2 } + \mathrm { OR } ^ { 2 } = 4 } \\\\ { \mathrm { QR } ^ { 2 } = 4 } \\ \\{ \therefore \mathrm { OQ } ^ { 2 } + \mathrm { OR } ^ { 2 } = \mathrm { QR } ^ { 2 } } \end{array}$

Therefore, Triangle OQR is a right angled triangle.

Angles subtended at the centre will be twice the angle at the circumference.

$\begin{array} { c } { \angle O Q R = 2 ( \angle Q P R ) } \\\\ { 90 ^ { \circ } = 2 \angle Q P R } \\\\ { \angle Q P R = \frac { 90 ^ { \circ } } { 2 } } \\\\ { \angle Q P R = 45 ^ { \circ } } \end{array}$

Hence proved