Question

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians​

Answer 1

Answer:

Let the two similar triangles ABC and DEF in which AP and DQ are the medians respectively.

we have to prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

If ΔABC~ΔDEF

⇒ \frac{ar(BCA)}{ar(EFD)}=\frac{AB^2}{DE^2} → (1)

As ΔABC~ΔDEF

\frac{AB}{DE}=\frac{BC}{EF}=\frac{2BP}{2EQ}

Hence, \frac{AB}{DE}=\frac{BP}{EQ}

In ΔABP and ΔDEQ

\frac{AB}{DE}=\frac{BP}{EQ}

∠B=∠E (∵ΔABC~ΔDEF)

By SAS rule, ΔABP~ΔDEQ

⇒ \frac{AB}{DE}=\frac{AP}{DQ}

Squaring, we get

\frac{AB^2}{DE^2}=\frac{AP^2}{DQ^2} → (2)

Comparing (1) and (2), we get

\frac{ar(BCA)}{ar(EFD)}=\frac{AP^2}{DQ^2}

Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

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Answer 2

Step-by-step explanation:

Step 1:

Given Data: Δ ABC ~ Δ PQR

To Prove: (ΔABC) / (ΔPQR) =

Step 2:

Draw AM ⊥ BC, PN ⊥ QR

(ΔABC) / (ΔPQR) = (½ × BC × AM) / (½ × QR × PN)

= BC/QR × AM/PN........................................... [I]

In Δ ABM and Δ PQN,

Step 3:

∠B = ∠Q (Δ ABC ~ Δ PQR)

∠M = ∠N (each 90°)

Step 4:

So, Δ ABM ~ Δ PQN  

AM/PN = AB/PQ ... ………………. [ii]

AB/PQ = BC/QR = CA/RP (Δ ABC ~ Δ PQR)..................... [iii]

Step 5:

Therefore Equation (i)

(ΔABC) / (ΔPQR) = BC/QR × AM/PN

= AB/PQ × AB/PQ [From Equation (ii) and Equation (iii)]

Step 6:

Using Equation (iii)

(ABC) / (PQR) =>

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