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\(^{2}\).
From Problems on Finding Area of Triangle and Parallelogram to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Dec 04, 23 02:14 PM
Dec 04, 23 01:50 PM
Dec 04, 23 01:49 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.