Question

The perimeter of an isosceles tri- angle is 544 cm and each of the equal sides is $\frac{5}{6}$ times the base . What is the area (in $cm^2$) of the triangle ?

• A. 38172
• B. 18372
• C. 31872
• D. 13872

Answer 1

Answer:

D. 13872

Let AD be the altitude.

Base = x cm

Each equal side $=\frac{5x}{6}cm$

$x+2\times{\frac{5x}{6}}=544$

$\frac{3x+5x}{3}=544$

$8x=544\times{3}$

$x=\frac{544\times{3}}{8}$

$x=BC=204$

$BD=\frac{204}{2} = 102 cm$

$AB=\frac{5x}{6}=\frac{5\times{204}}{6}=170cm$

and $AD=\sqrt{AB^2-BD^2}$

$=\sqrt{170^2-102^2}$

$=\sqrt{(170+102)(170-102)}$

$=\sqrt{272\times{68}}=136cm$

$\triangle{ABC}=\frac{1}{2}BC\times{AD}$

$=\frac{1}{2}\times{204}\times{136}$

$=13872$ sq.cm