Please tell how to solve Q 8 Share with your friends Share 0 Gursheen Kaur answered this Dear Student, Solution) In ∆ABC, we have P and R as the mid point of sides AB and AC, thenPR∥BC and PR = BC2 Mid point theorem⇒PR∥BQIn ∆ABC, we have Q and R as the mid points of sides BC and AC, thenQR∥BA and QR = BA2 Mid point theorem⇒QR∥BPIn quad PBQR, PR∥BQ proved aboveQR∥BP proved above⇒PBQR is a ∥gmWe know that a diagonal of a ∥gm divides it into 2 congruent ∆'s.Now, PQ is the diagonal of ∥gm PBQR, then ∆PBQ≅∆QRP⇒ar∆PBQ = ar∆QRP Congruent ∆'s have equal areas .......1In ∆ABC, we have P and R as the mid point of sides AB and AC, thenPR∥BC and PR = BC2 Mid point theorem⇒PR∥QCIn ∆ABC, we have P and Q as the mid points of sides AB and BC, thenPQ∥AC and PQ = AC2 Mid point theorem⇒PQ∥RCIn quad PQCR, PR∥QC proved abovePQ∥RC proved above⇒PQRC is a ∥gmWe know that a diagonal of a ∥gm divides it into 2 congruent ∆'s.Now,QR is the diagonal of ∥gm PQRC, then ∆RQC≅∆QRP⇒ar∆RQC = ar∆QRP Congruent ∆'s have equal areas .......2In ∆ABC, we have Q and R as the mid point of sides BC and AC, thenQR∥BA and QR = BA2 Mid point theorem⇒QR∥PAIn ∆ABC, we have P and Q as the mid points of sides AB and BC, thenPQ∥AC and PQ = AC2 Mid point theorem⇒PQ∥ARIn quad PQRA, QR∥PA proved abovePQ∥AR proved above⇒PQRA is a ∥gmWe know that a diagonal of a ∥gm divides it into 2 congruent ∆'s.Now,PR is the diagonal of ∥gm PQRA, then ∆QRP≅∆APR⇒ar∆QRP = ar∆ARP Congruent ∆'s have equal areas .......3From 1, 2 and 3, we getar∆PBQ = ar∆QRP = ar∆ARP = ar∆RQC .........4Now, ar∆ABC = ar∆PBQ + ar∆QRP + ar∆ARP + ar∆RQC⇒ar∆ABC = ar∆PBQ + ar∆PBQ + ar∆PBQ + ar∆PBQ Using 4⇒ar∆PBQ = 14ar∆ABC ........5In ∆PBQ, S is the mid point of PQ, then BS is median to PQ of ∆PBQ.We know that median of a ∆ divides it into 2 ∆'s of equal areas, thenar∆QSB = ar∆PSB = 12ar∆PBQ⇒ar∆PBQ = 2ar∆QSBNow, from 5, we get2ar∆QSB = 14ar∆ABC⇒ar∆QSB = 18ar∆ABC Or area of traingle ABC = 8 × area of triangle QSB Hence proved. Regards! 1 View Full Answer
