Sign in
Please select an account to continue using cracku.in
↓ →
In the given figure. Triangle ABC is drawn such that AB is tangent to a circle at A, and BC passes through the centre of the circle. Point C lies on the circle. If BC = 36 cm and AB = 24 cm . Find the area (in $$cm^2$$) of triangle ABC.
OAB will be a right-angled triangle where OA is a radius and it is perpendicular to AB.
We know AB = 24 cm and BC = 26 cm .
Let the radius of circle be "r" .
In right angle triangle OAB : $$OB=r+36-2r=36-r$$
$$OA^2+AB^2=OB^2$$
$$r^2+24^2=(36-r)^2$$
$$24^2=36^2-72r$$
$$r=10cm$$.
Triangle OAC is isosceles triangle.
If $$\angle\ OAC=\theta\ $$ $$\rightarrow$$ $$\angle\ AOB=2\theta\ $$ and $$\angle\ ABO=90-2\theta\ $$ .
Hence, from right angled triangle : $$\cos\left(90-2\theta\right)=\sin2\theta\ =\dfrac{\ 24}{36-r}=\dfrac{\ 24}{26}=\dfrac{\ 12}{13}$$
Hence, area of triangle ABC = area(OAB) + area(OCA)
$$=\ \dfrac{\ 1}{2}\left(r\right)\left(24\right)+\ \dfrac{\ 1}{2}\left(r\right)\left(r\right)\sin\left(180-2\theta\ \right)$$
$$=12\left(10\right)+50\ \left(\dfrac{\ 12}{13}\right)$$
= $$120+46.15$$
$$=166.15\ sq.cm$$.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get: