Question

In the given figure, P is a point on the side BC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5PQ, find the area of ∆AQC : area of ∆ABC.​

Answer 1

Answer:

Step-by-step explanation:

Given:

From figure,

in △ABC , P is a point on the side BC such that PC=2BP, and Q is a point on AP such that QA=5PQ

To Find:

area of ΔAQC : area of △ABC.

Now, It is given that: PC=2BP

PC/2=BP

We know that, BC=BP+PC

Now substitute the values, we get

BC = BP + PC

      =PC/2+PC

      =(PC+2PC)/2

      =3PC/2

       =PC

=2/3 ar (ΔABC)……(1)

​

 

It is given that, QA=5PQ

QA/5=PQ

We know that, QA=QA+PQ

So, QA=5/6 AP ar(ΔAQC)=5/6ar(ΔAPC)

=5/6(2/3ar(ΔABC))[From(1)]

ar (ΔAQC)=5/9 ar (ΔABC)

ar (ΔAQC)/ar(ΔAQC)=5/9

Hence proved.

Answer 2

Answer:

∆abc= ∆cqp is answer

Step-by-step explanation:

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