In a figure, PQR is a right angled triangle and from Q, three lines are drawn which intersect PR at A, B and C. if PA= AB=BC= CR=26cm, then find the value of QA sq. +QB sq. +Qc sq.
Olympiad IOM silverzone Class 9,2019
Answers
QA² + QB² + QC² = 9464 cm² if PQR is a right angled triangle from Q, three lines are drawn which intersect PR at A, B and C such that PA= AB=BC= CR=26cm
Step-by-step explanation:
PA= AB=BC= CR=26cm
=> PR = PA + AB + BC + CR = 104 cm
Lets draw perpendicular from A , B & C at PQ
as L , M , N respectively
ΔPLA ≈ Δ PQR
=> PA/PR = PL /PQ = LA/QR
=> 1/4 = PL /PQ = LA/QR
=> PL = PQ/4 & LA = QR/4
QL = PQ - QL = PQ - PQ/4 = 3PQ/4
=> QA² = LA² + QL²
=> QA² = (QR/4)² + (3PQ/4)²
=> QA² = QR²/16 + 9PQ²/16
=> QA² = (QR² + 9PQ²)/16
Similarly ΔPMB ≈ Δ PQR
=> PB/PR = PM /PQ = MB/QR
=> 1/2 = PM /PQ = MB/QR
=> PM = PQ/2 => QM = PQ - PQ/2 = PQ/2
MB = QR/2
QB² = MB² + QM²
=> QB² = (QR/2)² + (PQ/2)²
=> QB² = (QR² + PQ²)/4
ΔPNC ≈ Δ PQR
=> PC/PR = PN /PQ = NC/QR
=> 3/4 = PN /PQ = NC/QR
=> PN = 3PQ/4 => QN = PQ/4
NC = 3QR/4
QC² = (PQ/4)² + (3QR/4)²
=> QC² = (PQ² + 9QR²)/16
QA² + QB² + QC² = (QR² + 9PQ²)/16 + (QR² + PQ²)/4 + (PQ² + 9QR²)/16
= (14QR² + 14PQ²)/16
= 14 * PR² / 16
= 14 * 104² /16
= 9464 cm²
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