If two similar triangle have same area , then find the ratio of their

1.	If two similar triangle have same area , then find the ratio of their corresponding sides
Answer» 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 ⇒ $\frac{arΔABC}{arΔPQR}=(\frac{AB}{PQ})^{2}=(\frac{BC}{QR})^{2}=(\frac{CA}{RP})^{2}$

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

Answer»

Given:-

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

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