Question

ABC is a triangle and a line PQ is
parallel to BC if AP=1cm, PB=3 cm
QC=9 cm then the length of AC is​

Answer 1

Step-by-step explanation:

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)