Question

Two isocles triangles have eual vertical angles and their aras are in the ratio 16:25. Find the ratio of thire corresponding heights.​

Answer 1

Correct question:-

• Two isosceles triangles have equal vertical angles and their areas are in the ratio 16:25. Find the ratio of their corresponding heights.

Solution:-

Let, ∆ABC and ∆DEF be the two isosceles triangles in which AB = AC and DE = DF, ∠A = ∠D.

and Area ∆ABC/Area ∆DEF = 16/25

Draw,Al ⊥ BC

And, DM ⊥ EF.

Now, AB = AC, DE = DF

$\sf \cfrac{AB}{AC} = 1$

$\sf \cfrac{DE}{DF} = 1$

Both are equal to 1. So,

$\implies \sf \cfrac{AB}{AC} = \cfrac{DE}{DF}$

$\implies \sf \cfrac{AB}{DE} = \cfrac{AC}{DF}$

In triangles ABC and DEF,

$\sf \cfrac{AB}{DE} = \cfrac{AC}{DF}$ and ∠A = ∠D

So, By SAS congruency,

∆ABC ∼ ∆DEF

Area ∆ABC/∆DEF = AL²/DM²

$\implies \sf \cfrac{16}{25} = \cfrac{AL^{2}}{DM^{2}}$

$\implies \sf \cfrac{(4)^{2}}{(5)^{2}} = \cfrac{AL^{2}}{DM^{2}}$

$\implies \sf \cfrac{4}{5} = \cfrac{AL}{DM}$

Or, $\implies \sf \cfrac{AL}{DM} = \cfrac{4}{5}$

AL:DM = 4:5

Therefore,

Ratio of corresponding heights is 4:5.