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}).

9th Grade Math

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.






Source link

Notice: compact(): Undefined variable: limits in /customers/6/d/3/sciencetells.co.uk/httpd.www/wp-includes/class-wp-comment-query.php on line 853 Notice: compact(): Undefined variable: groupby in /customers/6/d/3/sciencetells.co.uk/httpd.www/wp-includes/class-wp-comment-query.php on line 853