Question

Two isosceles triangles have equal bases and their areas in the ratio of 16 is to 49 find the ratio of their corresponding altitude

Answer 1

Let the two triangles be ABC AND DEF. Let AE be altitude of ABC triangle and DG be altitude of DEF triangle .

BC = EF

1/2× BC × AE /1/2× EF× DG = 16/ 49

= AE / DG = 16 / 49

i couldn' t attach answer as it was showing loading error.

Mark my ANSWER AS BRAINLIEST

Answer 2

══════════════════════════

$\it\huge\mathfrak\red{Answer:-}$]

Let ,

The triangles be ABC AND DEF.

Let ,

Seg AE be altitude of ABC triangle

Seg DG be altitude of DEF triangle .

SegBC =Seg EF

=$\frac{1}{2}$× BC × AE ×$\frac{1}{2}$× EF× DG =$\frac{16}{49}$

= AE/DG =$\frac{16}{49}$

══════════════════════════

$\it\huge\mathfrak\red{Thanks}$]