Question

help plz cant solve​

Answer 1

Hi✌️✌️

Let the congruent side of the isosceles triangle be a and the height be h.

Given $a=h+4$

Since the height of the triangle, divides it into two right angled triangles,

$a^{2} =h^{2}+(\frac{b}{2} ) => (h+4) =h^{2}+12^{2}$

$h^{2}+16+8h=h^{2}+144$

$8h=128$

$h=16cm$

$a=20cm$

Perimeter of the triangle = Sum of all sides =$a+a+b =20+20+24=64cm$

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