Question

ABC is a right angled triangle with B = ${90}^{\circ }$ M is the mid point of AC and BM = $\sqrt{117}$cm

Sum of the lengths of the sides AB and BC is $30cm$ The area of the triangle is

• A. 96 $c{m}^2$
• B. 108 $c{m}^2$
• C. 114 $c{m}^2$
• D. 125 $c{m}^2$

Answer 1

The correct option is B 108 $c{m}^2$

R.E.F image

Midpoint of hypotenuse of right - angle triangle is also

the circumference

Also the angle subscribed by triangle in

a semicircle is right -angle

$\therefore$ In this circle,

$AM=MC=MB=$ radius ...(1)

Given, $BM=\sqrt{177}cmAB+BC=30cm$

BY Pythagoras theorem, In $\Delta ABC$

$A{C}^2=A{B}^2+B{C}^2$

$(AM+MC{)}^2=(AB+BC{)}^2-2AB.BC$ ($\because AC=AM+MC=MB+MB)$

$(2MB{)}^2=(30{)}^2-2AB.BC$

$(2\sqrt{117}{)}^2-(30{)}^2=-2AB.BC$

$468-900=-2AB.BC$

$AB.BC=\frac{432}{2}=216$

In $\Delta ABC,$ BC is Base & AB is height

$\therefore A\left(\Delta ABC\right)=\frac{1}{2}BC\times AB$

$=\frac{1}{2}\times 216$

$A\left(\Delta ABC\right)=108c{m}^2$