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Here we will learn how to solve different types of problems on finding area of triangle and parallelogram.
1. In the figure, XQ ∥ SY, PS ∥ QR, XS ⊥ SY, QY ⊥ SY and QY = 3 cm. Find the areas of ∆MSR and parallelogram PQRS.
Solution:
ar(∆MSR) = 12 × ar(rectangle of SR of height QY)
= 12 × SR × QY
= 12 × 6 × 3 cm2
= 9 cm2.
Also, ar(∆MSR) = 12 × ar(parallelogram PQRS).
Therefore, 9 cm2 = 12 × ar(parallelogram PQRS).
Therefore, ar(parallelogram PQRS) = 9 × 2 cm2 = 18 cm2.
2. In the figure, PQRS is a parallelogram, M is a point on QR such that QM : MR = 1 : 2.SM produced meets PQ produced at N. If the area of the triangle RMN = 20 cm2, calculate the areas of the parallelogram PQRS and ∆RSM.
Solution:
Draw NO ∥ QR which cuts SR produced at O. Then RONQ is a parallelogram. Join RN.
Now, ar(∆QMN)ar(∆RMN) = QMMR; (since both traingles have equal altitudes).
Therefore, ar(∆QMN)20cm2 = 12.
Therefore, ar(∆QMN) = 10 cm2.
Therefore, ar(∆QRN) = ar(∆QMN) + ar(∆RMN)
= 10 cm2 + 20 cm2
= 30 cm2.
Therefore, ar(parallelogram QRON) = 2ar(∆QRN) = 2 × 30 cm2 = 60 cm2 .................... (i)
Now, ar(parallelogramPQRS)ar(parallelogramQRON) = BaseSR×HeightBaseRO×Height = SRRO; (Since, both the parallelograms have have the same height)
Therefore, ar(parallelogramPQRS)ar(parallelogramQRON) = SRQN ................... (ii)
In ∆MQN and ∆MRS,
∠MQN = ∠MRS and ∠QNM= ∠MSR (Since, QN ∥ SR).
Therefore, ∆MQN ∼ ∆MRS (By AA axiom of similarity).
Therefore, corresponding sides are proportional.
So, MQMR = QNSR ................... (iii)
From (ii) and (iii),
ar(parallelogramPQRS)ar(parallelogramQRON) = MRMQ = 21
Therefore, ar(parallelogram PQRS) = 2 × 60 cm2 [From (i)]
= 120 cm2.
Now, ar(∆RSN) = 12 × ar(parallelogram PQRS)
= 12 × 120 cm2
= 60 cm2.
Therefore, ar(∆RSM) = ar(∆RSN) – ar(∆RMN)
= 60 cm2 - 20 cm2
= 40 cm2.
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