Question

In the given figure, O is the Centre of the circle with radius 5 cm. OP is perpendicular to CD, AB || CD,

AB= 6 cm and CD = 8 cm. Determine PQ.​

Answer 1

Question

In figure, O is the centre of the circle id radius 5 cm. OP is perpendicular to AB, OQ is perpendicular to CD, AB || CD , AB = 6 cm and CD = 8 cm. Determine PQ

Given

PQ = OP + OQ

r = 5cm = OA = OC

AB = 6cm CD = 8cm

Solution

2AP = AB ; 2CQ = CD

AP = 3cm ; CQ = 4 cm

By pythagorus theorem

In Triangle OPA

$\sf {OA}^{2} = {OP}^{2} + {AP}^{2} \\ \\ \sf \: (5 {)}^{2} = (OP {)}^{2} + (3 {)}^{2} \\ \\ \sf {OP}^{2} = 16 \: \implies \: OP = 4cm$

In Triangle OQC

$\\ \sf {OC}^{2} = {OQ}^{2} + {CQ}^{2} \\ \\ \sf(5 {)}^{2} = {OQ}^{2} + (4 {)}^{2} \\ \\ \sf{OQ}^{2} = 9 \implies \: OQ = 3cm$

PQ = OK + OQ = 4 + 3 = 7 cm

Hence , PQ = 7cm

Answer 2

Answer:

$\huge\underline\mathfrak\red{❥Answer}$

Step-by-step explanation:

Since the perpendicular from the center of the circle to a chord bisects the chord.

P and Q are the mid points of AB and CD In right triangles

OAP and OCQ,

we have

OA2 = OP2 + AP2andOC2 = OQ2 + CQ2 52 = OP2 + 32and52 = OQ2 + 42 OP2 = 52 - 32andOQ2 = 52 - 42 OP2 = 16andOQ2 = 9 OP = 4andOQ = 3 PQ = OP+ OQ = 4 + 3 = 7 cm