Question

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

Answer 1

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

Answer 2

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