Question

In APQR; PO = PR. A is a point in PQ and B is a point in PR, so that OR = RA = AB = BP (i) Show that : ZP: ZR= 1:3 (ii) Find the value of ZQ.​

Answer 1

Step-by-step explanation:

Given: △PQR, Q Q=P R, A and B are points such that, QR=R A=A B=BP

Now, In △PAB

A B=PB

Thus, ∠P=∠PAB=P (isosceles triangle property)

and ∠ABR=∠P+∠PAB

∠ABR=2P (Exterior angle property)

Now, In △ABR

A B=R A

∠ABR=∠ARB=2P (Isosceles triangle property)

Sum of angles of the triangle = 180

∠ABR+∠ARB+∠BAR=180

2P+2P+∠BAR=180

∠BAR=180−4P

Sum of angles on a straight line = 180

∠PAB+∠BAR+∠RAB=180

P+180−4P+∠RAB=180

∠RAB=3 P

In △QRA

QR=R A

∠Q=∠QAR=3 P (Isosceles triangle property)

But , Q Q=P R

∠Q=∠R=3 P

hence, ∠P:∠R=1:3

Answer 2

Answer:

Ok Bro

Step-by-step explanation:

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In ΔPQR;PQ=PR and A is a point in PQ and B is a point in PR, so that QR=RA=AB=BP Then: ∠P:∠R

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Correct option is

C

1:3

Given: △PQR, PQ=PR, A and B are points such that, QR=RA=AB=BP

Now, In △PAB

AB=PB

Thus, ∠P=∠PAB=P (isosceles triangle property)

and ∠ABR=∠P+∠PAB

∠ABR=2P (Exterior angle property)

Now, In △ABR

AB=RA

∠ABR=∠ARB=2P (Isosceles triangle property)

Sum of angles of the triangle = 180

∠ABR+∠ARB+∠BAR=180

2P+2P+∠BAR=180

∠BAR=180−4P

Sum of angles on a straight line = 180

∠PAB+∠BAR+∠RAB=180

P+180−4P+∠RAB=180

∠RAB=3P

In △QRA

QR=RA

∠Q=∠QAR=3P (Isosceles triangle property)

But , PQ=PR

∠Q=∠R=3P

hence, ∠P:∠R=1:3