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


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.



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.



Share this page: What’s this?

Recent Articles

  1. Rupees and Paise | Paise Coins | Rupee Coins | Rupee Notes

    Dec 04, 23 02:14 PM

    Different types of Indian Coins
    Money consists of rupees and paise; we require money to purchase things. 100 paise make one rupee. List of paise and rupees in the shape of coins and notes:

    Read More

  2. Months of the Year | List of 12 Months of the Year |Jan, Feb, Mar, Apr

    Dec 04, 23 01:50 PM

    Months of the Year
    There are 12 months in a year. The months are January, February, march, April, May, June, July, August, September, October, November and December. The year begins with the January month. December is t…

    Read More

  3. The Story about Seasons | Spring | Summer | Autumn | Winter

    Dec 04, 23 01:49 PM

    The Four Seasons
    Kids let’s enjoy the story about seasons. Here we will discuss about the four seasons and the duration. Some months are too hot and some are too cold. The period of hot months is called the hot

    Read More