Question

in the adjoining figure triangle ABC right angled atA.Semi cirlces are drawn on the sides of the triangle ABC.Prove that area of shaded region is equal to area of trainlgeABC.

Answer 1

area of semi circle on AB
is (1/2) π (AB/2)^2
$\frac{\pi \: {ab}^{2} }{8}$
similarly area of semi circle on AC
is (1/2) π (AC/2)^2
$\frac{\pi \: {ac}^{2} }{8}$
area of shaded region is
area(∆ABC)+area(semicircleAB)+area(semicircleAC)-area(semicircleBC)...(*)


area(semicircleBC) is
$\frac{\pi \: {bc}^{2} }{8}$
as (ac)^2 + (ab)^2 = (bc)^2
Pythagorean theorom..
multiplying π/8 both sides

(π/8)(ac)^2 + (π/8)(ab)^2 = (π/8)(bc)^2
area(semicircleAC)+area(semicircleAB)=area(semicircleBC)
so filling in equation (*)
we get
area(shaded.part) = area(∆ABC)

Answer 2

As, ∠ACB = 90° ( angle in semicircle is always 90°)

And ∠PCB = 90° ( as ACB = 90°)

Now, in ∆ABC and ∆PCB, using pythagoras therem

AB2 = AC2 +BC2

And BP2 = BC2 +CP2

So AB2 = AC2 + ( BP2 - PC2 )

Or AB2 = AC2 -PC2 +BP2

Or AB2 = (AC-PC)(AC+PC) + BP2

As [ AC + PC = AP]

So AB2 = ( AC - PC) ×AP + BP2

AB2 = AC × AP- PC ×AP + BP2

Using secant formula, PA ×PC = PB×PD

So AB2 = AC × AP- PB×PD + BP2

Or AB2 = AC × AP +BP ( BP - PD)

Or AB2 = AC × AP + BP × BD (hence proved)