Question

pls ans ques no 56 full solution pls

Answer 1

Hey Mate !!!!

In the given figure,

ΔRPQ is a right angle triangle where ∠RPQ = 90°

i.e. By Pythagoras Theorem,

QP² + RP² = QR²

24² + 7² = RQ²

i.e. RQ = √24² + 7²

i.e. RQ = √576 + 49

i.e. RQ  = √625

i.e. RQ = 25 cm

Now, From the given figure,

Area of Shaded Region = Area of the semicircle - Area of the right Δ

Area of the Semicircle = 1/2 πr² (here, π = 3.14 and r = RQ/2 = 25/2 = 12.5)

                                     = 1/2 X 3.14 X (12.5)²

                                     = 1/2 X 3.14 X 156.25

                                     = 245.3125 cm²

Area of the right Δ = 1/2 X base X height (base=RP=7cm,height=QP=24cm)

                               = 1/2 X 7 X 24 = 84 cm²                        

  ∴ Area of Shaded Region = Area of the semicircle - Area of the right Δ

  Area of Shaded Region =245.3125 cm² - 84 cm² = 161.3125 cm²

Hence , the area of the shaded region is 161.3125 cm² .