Question

In the given figure, prove that ar (∆AQC) = ar (PBR)​

Answer 1

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\purple{Solution:-}$

⏩We observe that triangles ABQ and PBQ are on the same base BQ and between the same parallels AP and BQ.

Therefore,

ar(∆ABQ)= ar(∆PBQ) .....→(i)

Similarly, triangles BCQ and BRQ are on the same base BQ and between the same parallels BQ and CR.

Therefore,

ar(∆BCQ) = ar(∆BRQ).......→ (ii)

Now,

On adding (i) and (ii), we get,

ar (∆ABQ) + ar (∆BCQ) = ar (∆PBQ) + ar(∆BRQ)

=> ar (∆AQC) = ar (∆PBR)

Hence, proved...:-)

✔✔✔✔✔✔✔✔✔

Hope it helps...❣❣❣

⭐♥✨❤⭐♥✨❤⭐

Be Brainly...✌✌✌

♣ WALKER ♠

Answer 2

Answer:

Step-by-step explanation:

Pls mark my answer as brainliest