Question

if two similar triangle have same area , then find the ratio of their corresponding sides​

Answer 1

Answer:-

Given:-

• The two similar triangle have same area

Solution:-

If two similar triangle have same area,  then the ratio of their corresponding sides.

To prove this theorem, consider two similar triangles ΔABC and ΔPQR

$\frac{arΔABC}{arΔPQR}=(\frac{AB}{PQ})^{2}=(\frac{BC}{QR})^{2}=(\frac{CA}{RP})^{2}$

Since area of triangle = 1/2 × base × altitude

To find the area of ΔABC and ΔPQR draw the altitudes AD and PE from the vertex A and P of ΔABC and ΔPQR

Now, area of ΔABC = 1/2×BC×AD

area of ΔPQR = 1/2×QR×PE

The ratio of the areas of both the triangles can now be given as:-

$\frac{arΔABC}{arΔPQR}=\frac{\frac{1}{2}×BC×AD}{\frac{1}{2}×QR×PE}$

$\frac{arΔABC}{arΔPQR}=\frac{BC×AD}{QR×PE}$

Now, in ΔABD and ΔPQE it can be seen

∠ABC=∠PQR (since ΔABC ≅ ΔPQR)

∠ABD=∠PEQ (since both the angles are 90°)

From AA criterion of similarity ΔADB≅ΔPEQ

$\frac{AB}{PE}=\frac{AB}{PQ}$

since it is known that ΔABC≅ΔPQR

$\frac{AB}{PE}=\frac{BC}{QR}=\frac{CA}{RP}$

Substituting this value in equation, we get

$\frac{arΔABC}{arΔPQR}=\frac{AB}{PQ}×\frac{AD}{PE}$

We can write

$\frac{arΔABC}{arΔPQR}=(\frac{AB}{PQ})^{2}$

Similarly we can prove

⇒$\frac{arΔABC}{arΔPQR}=(\frac{AB}{PQ})^{2}=(\frac{BC}{QR})^{2}=(\frac{CA}{RP})^{2}$