Question

solve this question????

Answer 1

Answer:

Step-by-step explanation:

Consider two triangles ABC and PQR

Given : ΔABC  similar to ΔPQR

∴ AB/PQ = BC/QR = AC/PR [ sides of similar triangles are proportional]

Also ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R [CPST] --- ( i )

and ar(ABC) = ar(PQR)

To Prove: ΔABC ≅ ΔPQR

Draw AD ⊥ BC and PM ⊥ QR

You can observe two right triangles ADC and PMR

In Δ ADC and Δ PMR

∠ADC = ∠PMR  [90° Each]

∠ACD = ∠PRM  [from ( i ) ]

∴ Δ ADC is similar to Δ PMR   [ By AA Similarity Criteria]

∴AD/PM = AC/PR  [ Sides of similar triangle are proportional]

∴ AB/PQ = BC/QR = AC/PR = AD/PM   --- ( ii )

Now,

$\frac{ar(ABC)}{ar(PQR)}=\frac{1/2 \times BC \times AD}{1/2 \times QR \times PM}$

$\frac{ar(ABC)}{ar(PQR)}=\frac{BC \times AD}{QR \times PM}$

Putting AD/PM = BC/QR from eq (ii)

$\frac{ar(ABC)}{ar(PQR)}=\frac{BC \times BC}{QR \times QR}$

$\frac{ar(ABC)}{ar(PQR)}=\frac{BC^2}{QR^2}$

Given that ar(ABC) = ar(PQR) ∴r(ABC) : ar(PQR) = 1 : 1

$\frac{1}{1}=\frac{BC^2}{QR^2}$

BC/QR = 1  or BC = QR

∴ AB/PQ = 1 or AB = PQ   [∵BC/QR = AB/PQ]

Similarly AC = RP

Now, In ΔABC and ΔPQR

AB = PQ [Proved above]

BC = QR [Proved above]

CA = RP [ Proved above]

∴ ΔABC ≅ ΔPQR [By SSS]