Question

Each of the equal sides of an isosceles triangles measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle.

Answer 1

Assume a triangle ABC in which side AB = AC = x+2 cm,as height AD which meets BC at D = x cm

also BD = DC = 6 cm as in isosceles triangle perpndicular frm vertex to base bisects base

in triangle ADC by Pythagoras theorm

$(x+2{)}^2=(6{)}^2+x^2$

$x^2+4+4x=36+x^2$

$x=\frac{32}{4}=8cm$

AB = AC = 8+2 =10 cm

$area=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times 12\times 8=48c{m}^2$