Find tbes ofthe shdedo in Fig 12.19.PO-34 on PR-7 m and O is the centr

1.	Find tbes ofthe shdedo in Fig 12.19.PO-34 on PR-7 m and O is the centre of th
Answer» Given : PQ= 24 cm ,PR = 7 cm We know that any angle made by the diameter QR in the semicircle is 90°. ∠RPQ = 90° In right angled ∆RPQ RQ² = PQ² + PR² [By pythagoras theorem] RQ² = 24² + 7² RQ² = 576 + 49 RQ² = 625 RQ = √625cm RQ= 25 cm radius of the circle (OQ)= 25 / 2 cm Area of right ∆ RPQ= ½ × Base × height Area of right ∆ RPQ= ½ × RP × PQ Area of right ∆ RPQ = ½ × 7 × 24 = 7 × 12 = 84 cm² Area of right ∆ RPQ = 84 cm² Area of semicircle= πr²/2 = (22/7) × (25/2)² / 2 = (22 × 25 × 25)/ (7× 2 × 2 × 2) = 11 × 625 /28 = 6875/28 cm² Area of semicircle = 6875/28 cm² Area of the shaded region = Area of semicircle - Area of right ∆ RPQ = (6875/28 - 84 )cm² = (6875 - 2532)/ 28 Area of the shaded region = 4523 / 28= 161.54 cm² Hence, the area of the shaded region = 161.54 cm² Like my answer if you find it useful!

Find tbes ofthe shdedo in Fig 12.19.PO-34 on PR-7 m and O is the centre of th

Answer»

Given :

PQ= 24 cm ,PR = 7 cm

We know that any angle made by the diameter QR in the semicircle is 90°.

∠RPQ = 90°

In right angled ∆RPQ

RQ² = PQ² + PR²

[By pythagoras theorem]

RQ² = 24² + 7²

RQ² = 576 + 49

RQ² = 625

RQ = √625cm

RQ= 25 cm

radius of the circle (OQ)= 25 / 2 cm

Area of right ∆ RPQ= ½ × Base × height

Area of right ∆ RPQ= ½ × RP × PQ

Area of right ∆ RPQ = ½ × 7 × 24 = 7 × 12 = 84 cm²

Area of right ∆ RPQ = 84 cm²

Area of semicircle= πr²/2

= (22/7) × (25/2)² / 2

= (22 × 25 × 25)/ (7× 2 × 2 × 2)

= 11 × 625 /28 = 6875/28 cm²

Area of semicircle = 6875/28 cm²

Area of the shaded region = Area of semicircle - Area of right ∆ RPQ

= (6875/28 - 84 )cm²

= (6875 - 2532)/ 28

Area of the shaded region = 4523 / 28= 161.54 cm²

Hence, the area of the shaded region = 161.54 cm²

Like my answer if you find it useful!