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) = \(\frac{1}{2}\) × ar(rectangle of SR of height QY)
= \(\frac{1}{2}\) × SR × QY
= \(\frac{1}{2}\) × 6 × 3 cm\(^{2}\)
= 9 cm\(^{2}\).
Also, ar(∆MSR) = \(\frac{1}{2}\) × ar(parallelogram PQRS).
Therefore, 9 cm\(^{2}\) = \(\frac{1}{2}\) × ar(parallelogram PQRS).
Therefore, ar(parallelogram PQRS) = 9 × 2 cm\(^{2}\) = 18 cm\(^{2}\).
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 cm\(^{2}\), 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, \(\frac{ ar(∆QMN)}{ ar(∆RMN)}\) = \(\frac{QM}{MR}\); (since both traingles have equal altitudes).
Therefore, \(\frac{ ar(∆QMN) }{20 cm^{2}}\) = \(\frac{1}{2}\).
Therefore, ar(∆QMN) = 10 cm\(^{2}\).
Therefore, ar(∆QRN) = ar(∆QMN) + ar(∆RMN)
= 10 cm\(^{2}\) + 20 cm\(^{2}\)
= 30 cm\(^{2}\).
Therefore, ar(parallelogram QRON) = 2ar(∆QRN) = 2 × 30 cm\(^{2}\) = 60 cm\(^{2}\) .................... (i)
Now, \(\frac{ar(parallelogram PQRS)}{ar(parallelogram QRON)}\) = \(\frac{Base SR × Height}{ Base RO × Height}\) = \(\frac{SR}{RO}\); (Since, both the parallelograms have have the same height)
Therefore, \(\frac{ar(parallelogram PQRS)}{ar(parallelogram QRON)}\) = \(\frac{SR}{QN}\) ................... (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, \(\frac{MQ}{MR}\) = \(\frac{QN}{SR}\) ................... (iii)
From (ii) and (iii),
\(\frac{ar(parallelogram PQRS)}{ar(parallelogram QRON)}\) = \(\frac{MR}{MQ}\) = \(\frac{2}{1}\)
Therefore, ar(parallelogram PQRS) = 2 × 60 cm\(^{2}\) [From (i)]
= 120 cm\(^{2}\).
Now, ar(∆RSN) = \(\frac{1}{2}\) × ar(parallelogram PQRS)
= \(\frac{1}{2}\) × 120 cm\(^{2}\)
= 60 cm\(^{2}\).
Therefore, ar(∆RSM) = ar(∆RSN) – ar(∆RMN)
= 60 cm\(^{2}\) - 20 cm\(^{2}\)
= 40 cm\(^{2}\).
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