Question

area theorem class 10​

Answer 1

Answer:

Now, area of ΔABC = 12 × BC × AD

area of ΔPQR = 12 × QR × PE

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

area of ΔABCarea of ΔPQR = 12×BC×AD12×QR×PE

⇒ area of ΔABCarea of ΔPQR = BC × ADQR × PE ……………. (1)

Now in ∆ABD and ∆PQE, it can be seen that:

∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)

∠ADB = ∠PEQ (Since both the angles are 90°)

From AA criterion of similarity ∆ADB ~ ∆PEQ

⇒ ADPE = ABPQ  …………….(2)

Since it is known that ΔABC~ ΔPQR,

ABPQ = BCQR = ACPR …………….(3)

Substituting this value in equation (1), we get

area of ΔABCarea of ΔPQR = ABPQ × ADPE

Using equation (2), we can write

area of ΔABCarea of ΔPQR = ABPQ × ADPE

⇒area of ΔABCarea of ΔPQR =(ABPQ)2

Also from equation (3),

area of ΔABCarea of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2

This proves that the ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

Step-by-step explanation:

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