Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

In ∆ABC,P is a point on AB ,PR is drawn parallel to BC and PR=BC.PR curd AC at Qprove that ar(∆AQR)=ar(∆BPQ).

In ∆ABC,P is a point on AB ,PR is drawn parallel to BC and PR=BC.PR curd AC at Qprove that ar(∆AQR)=ar(∆BPQ).

Grade:12th pass

1 Answers

Amit jha
38 Points
3 years ago
Since,PR।।BC and AP = BP =>AQ=QC Now, triangles PQB and PQC are on the same base and b/w same parallels PQ and BC => ar(PQB) = ar(PQC)Now,In triangles PQC and ARQ PQ. = RQ AnglesPQC= ARQ [vertically opposite angles] QA=QC Hence, by SAS congruency rule => Triangles PQC and AQR are congruent to each other Since,ar(PQC) = ar (PQB) and also ar(PQC) = ar(ARQ) => ar(PQB) = ar(ARQ)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free