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
Answers
Answered by
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)
Similar questions