Question

ABC is an isosceles triangle whose equal sides AB and AC are 10 cm each and base BC = 8cm. AD perpendicular to BC and angle BPC = 90°. Find the area of the shaded portion is following Figure​

Answer 1

Answer:

IN PICTURE

Step-by-step explanation:

answer is looking big but it is small

Answer 2

Perpendicular drawn on side of isosceless triangle = AD

Thus, it will bisect the side BC.

= BD = CD = 8/2 =4 cm

In △ABD , Using the  Pythagoras theorem

AB² = AD² + BD²

10² = AD² + 4²

100−16 = AD²

AD² = 84

AD = 2√21

In △PDB and △PDC

BD = CD (Proved)

PD = PD  (Common)

∠PDB =∠PDC (Each 90°)

Thus, by SAS congruency c△PDB ≅ △PDC

In △PBC , Using Pythagoras theorem

PB² + PC² = BC²

PC² + PC² = 8²    

2PC² = 64PC²

Pc = 4√2

In △ABC

= 1/2 × Base × Height

= 1/2 × BC × AD

= 1/2 × 8 × 2√21

= 8√21

Similarly,

Area of △PBC

= 1/2 × Base × Height

= 1/2 × PB × PC

= 1/2 × 4√2 × 4√2

= 16 cm²

Thus,

8√21 - 16

= 8 × 4.5 - 16

= 20.6cm²