Question

ABC is a truangle and PQ is a straight line meeting AB in P and AC in Q.if AP=1cm and BP=3cm,AQ=1.5 and QC=4.5.prove that area of apq=1/16(area of ABC)​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

AP=1cm,PB=3cm,AQ=1.5cm,QC=4.5cm [ Given ]

Then, AB=AP+PB=1+3=4cm

In △APQ and △ABC,

∠A=∠A [ Common angle ]

AB

AP

=

AC

AQ

[ Each equal to

4

1

]

∴ △APQ∼△ABC [ By SAS similarity ]

∴

area(△ABC)

area(△APQ)

=

AB

2

AP

2

[ By area of similar triangle theorem ]

∴

area(△ABC)

area(△APQ)

=

4

2

1

2

∴

area(△ABC)

area(△APQ)

=

16

1

∴ area(△APQ)=

16

1

×area(△ABC)