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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