Question

Find the area of a right angled triangle, if the radius of its circumcircle is 3 cm and altitude drawn to the hypotenuse is 2 cm.

Answer 1

$\huge \sf \red{Solution.}$

Let ABC be a right angled triangle with ∠A=90°.

Let O be the centre of the circumcirlce of Δ ABC

(Circle passing through three vertices of ∆ ABC.)

Because ∠ A = 90°, therefore Hypotenuse BC of

right triangle ABC is diameter of circumcirlce.

( ∵ Angle in a semi-circle is 90°)

∴ BC = 2 OB = 2r

= 2 × 3 = 6 cm

AD = 2 cm

∴ Area of the Δ ABC

$\frac{1}{2} \times Base \times Height$

$frac{1}{2} \times BC \times AD$

$\frac{1}{2} \times 6 \times 2 = 6 \: cm {}^{2}$

Answer 2

$\\ \boxed{ \rm \: let \: \triangle \: abc \: be \: right \: angled \: at \: B}$

$\textsf{Then hypotenus \: AC = diameter \: of \: its \: circumcircle}$

$\textsf{ = (3× 2) = 6cm}$

$\textsf{Let BL⏊AC .Then,BL=2cm}$

$\textsf{Area of triangle ABC}= \rm \huge {\frac{1}{2}} \small ×AC×BL$

$\rm = \large \big( \frac{1}{2} \times 6 \times 2 \big)cm {}^{2} = 6cm {}^{2}$