Math, asked by Tony9011, 10 months ago

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

Answered by amitnrw
8

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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