Question

In Figure AP=4.2,AQ=6.3,AR=1.8,AS=1.2
Prove that PSllQR
plzz answer it ​

Answer 1

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MATHS

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP=1cm, PB=3cm, AQ=1.5cm, QC=4.5cm, prove that area of ΔAPQ is 1/16 of the area of ΔABC.

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ANSWER

In ΔABC, PQ is a line which meets AB in P and AC in Q.

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

Now,

PB

AP

=

3

1

and

QC

AQ

=

4.5

1.5

=

3

1

⇒

PB

AP

=

QC

AQ

From figure: AB=AP+PB=1+3=4cm

AC=AQ+QC=1.5+4.5=6cm

In ΔAPQ and ΔABC,

AB

AP

=

AC

AQ

angle A (common)

ΔAPQ and ΔABC are similar triangles.

Now,

area of (ΔABC)

area of (ΔAPQ)

=

AB

2

AP

2

=

(4)

2

(1)

2

=

16

1

Which implies,

area of ΔAPQ=

16

1

of the area of ΔABC

Hence Proved.

Answer 2

Given:-

AP=4.2

AQ=6.3

AR=1.8

AS=1.2

TO proove:-

PS || QR

proof:-

angle,

PAS=RAQ (vertically opposite)-----(1)

and,

$\frac{4.2}{1.2} = \frac{6.3}{1.8}$

$3.5 = 3.5$

Ratio of two side of trinagle are same so,

$\frac{AP}{AS} = \frac{AQ}{AR} - - (2)$

by eq. 1 and two

ΔPAS ~ ΔRAQ

and,

angle,

P=Q (triangle are similar)

if interior angle is same

so,

the line

PS AND QR is parallel

hence prooved