Question

In ∆ABC, PQ||BC; P:AB and Q:AC. PB=AQ, AP=9 units, QC=4 units, PB=?​

Answer 1

Step-by-step explanation:

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)

Answer 2

On the basis of question we draw a figure of a ∆ABC in which a straight line parallel to BC intersects AB and AC at points P and Q respectively.



Given , AP = QC, PB= 4 units and AQ = 9 units

PQ || BC

∴AP=AQABAC[By Basic Proportionality Theorem ]

⇒AB=ACAPAQ

⇒AP + PB=AQ + QCAPAQ

⇒PB=QC=APAPAQAQ

⇒ AP² = PB. AQ = 4 × 9 = 36 

∴ AP = 6 units