Question

Revisit
Given that in APOR, POR = 90°, SPQ= 4 SPR PL-3, LQ-1, PM - 2, MR-3. Let A denote
area of quadrilateral MNSR, find
75
XA
2
P
2.
M M
3
L
1
\N
3
Qt
R
S

Answer 1

Given :- in ∆POR, ∠POR = 90°, ∠SPQ = ∠SPR , PL = 3, LQ = 1, PM = 2, MR = 3.

To Find :- (75/2) A , where A is area of quadrilateral MNSR ?

Solution :-

In right ∆PQR,

→ QR = √[PR² - PQ²] { By pythagoras theorem. }

→ QR = √[5² - 4²] = 3 .

now,

→ Area ∆PLM / Area ∆PQR = (1/2) * 3 * 2 * sin QPR / (1/2) * 4 * 5 * sin QPR = (3*2)/(4*5) = (3/10) --------- Eqn.(1)

similarly,

→ Area PLN / Area PNM = (3/2) { since PN is angle bisector .}

→ Area PNM / Area PLM = (2/5)

→ Area PNM = (2/5) Area PLM

using Eqn.(1),

→ Area PNM = (2/5) * (3/10) * (Area PQR) = (3/25) Area PQR ------- Eqn.(2)

now,

→ PQ / PR = QS / SR { By angle bisector theorem.}

→ (4/5) = QS / SR

→ QS = (4/5) SR

→ QS = (4/9) QR

→ QS = (4/9) * 3

→ QS = (4/3) unit.

then,

→ Area ∆PQS = (1/2) * Base * Height = (1/2) * (4/3) * 4 = (8/3) cm² --------- Eqn.(3)

therefore,

→ Area ⌷ MNSR = Area ∆PQR - Area ∆PQS - Area PNM

putting values from Eqn.(2) and Eqn.(3) ,

→ Area ⌷ MNSR = (1/2) * Base * Height - (8/3) - (3/25) Area PQR

→ Area ⌷ MNSR = (1/2) * 4 * 3 - (8/3) - (3/25) * {(1/2) * 4 * 3}

→ Area ⌷ MNSR = 6 - (8/3) - (3/25) * 6

→ Area ⌷ MNSR = 6 - (8/3) - (18/25)

→ Area ⌷ MNSR = (450 - 200 - 54)/75

→ Area ⌷ MNSR = (196/75) unit² = A .

hence,

→ (75/2) A = (75/2) * (196/75) = 98 unit² (Ans.)

{ Excellent Question. }

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