Question

Example 16 In the adjoining figure, ABC is an
isosceles triangle with base BC = 8 cm and
AB = AC = 12 cm. AD is perpendicular to BC and
O is a point on AD such that ZBOC = 90°. Find
the area of the shaded region.​

Answer 1

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