# Problems on Finding Area of Triangle and Parallelogram

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