Question

In right triangle ABC ,b=90 and AB=BC and hypotenuse is 2√2 . Find the area of the triangle ABC.​

Answer 1

△ABC is a right angled triangle

∠ABC=90

o

A circle is drawn with diameter intersecting AC in P, PQ is the tangent to circle which intersects BC at Q.

Join BP

PQ and BQ are tangents drawn from an external point Q

∴ PQ=BQ....(1) (length of tangents drawn from an external point to the circle )

∠PBQ=∠BPQ (In a triangle, angle opposite to equal sides are given that AB is the diamter

∠APB=90

o

(angle in a semicircleis a right angle)

∠APB+∠BPC=180

o

(linear pair)

∴ ∠BPC=180

o

−∠APB=180

o

−90

o

=90

o

Consider △BPC

∠BPC+∠PBC+∠PCB=180

o

(angle sum)

∴ ∠PBC+∠PCB=180−∠BPC=180=90=90

o

....(2)

∠BPC=90

o

∴∠BPQ+∠CPQ=90

o

...(3)

from 2 and 3

∠PBC+∠PCB=∠BPQ+∠CPQ

∠PCQ=∠CPQ (Since ∠BPQ=∠PBQ)

Consider ∠PQC

∠PCQ=∠CPQ

∴ PQ=CQ...(4)

From (1) and (4) we get

BQ=QC

∴ tangent at P bisect the side BC