In the given figure PA/AQ=PB/BR=3. IF the area of triangle pqr is 32cm^2.then find the area of the quadrilateral aqrb
Answers
Answered by
49
AB∥QR,PQisatransversal,
∠PAB=∠PQR
(Correspondingangles)
SinceAB∥QR,PRisatransversal,
∠PBA=∠PRQ
(Correspondingangles)
In∆PABand∆PQR,
∠PAB=∠PQR[provedabove]
∠PBA=∠PRQ[provedabove]
so,∆PAB~∆PQR(AAsimilarity)
⇒
PA
PQ
=
PB
PR
=
AB
QR
[Correspondingsidesofsimilar∆'sareproportional]
Now,
PA
AQ
=3
⇒
AQ
PA
=
1
3
Adding'1'bothsides,weget
AQ
PA
+1=
1
3
+1
⇒
AQ+PA
PA
=
4
3
⇒
PQ
PA
=
4
3
⇒
PA
PQ
=
3
4
Weknowthatratioofareasof2
similar∆'sisequaltothesquaresof
theratioofcorrespondingsides.
so,
ar(∆PAB)
ar(∆PQR)
=(
PA
PQ
)2
⇒
ar(∆PQR)−ar(AQRB)
ar(∆PQR)
=(
3
4
)2=
9
16
⇒
32−ar(AQRB)
32
=
9
16
⇒512−16ar(AQRB)=288
⇒16ar(AQRB)=512−288=224
⇒ar(AQRB)=
224
16
=14cm2
Answered by
15
AB∥QR,PQisatransversal, ∠PAB=∠PQR (Correspondingangles) SinceAB∥QR,PRisatransversal, ∠PBA=∠PRQ (Correspondingangles) In∆PABand∆PQR, ∠PAB=∠PQR[provedabove] ∠PBA=∠PRQ[provedabove] so,∆PAB~∆PQR(AAsimilarity) ⇒ PA PQ = PB PR = AB QR [Correspondingsidesofsimilar∆'sareproportional] Now, PA AQ =3 ⇒ AQ PA = 1 3 Adding'1'bothsides,weget AQ PA +1= 1 3 +1 ⇒ AQ+PA PA = 4 3 ⇒ PQ PA = 4 3 ⇒ PA PQ = 3 4 Weknowthatratioofareasof2 similar∆'sisequaltothesquaresof theratioofcorrespondingsides. so, ar(∆PAB) ar(∆PQR) =( PA PQ )2 ⇒ ar(∆PQR)−ar(AQRB) ar(∆PQR) =( 3 4 )2= 9 16 ⇒ 32−ar(AQRB) 32 = 9 16 ⇒512−16ar(AQRB)=288 ⇒16ar(AQRB)=512−288=224 ⇒ar(AQRB)= 224 16 =14cm2
Similar questions