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Area & Perimeter of Triangle for CLAT - Practice Questions & MCQ

Edited By admin | Updated on Oct 06, 2023 06:35 PM | #CLAT

Concepts Covered - 1

Area & Perimeter of Triangle

You have studied in previous classes about different figures such as squares, rectangles, triangles and quadrilaterals. You also know how to calculated perimeters and the areas of some of these figures like rectangle, square etc.

So, if your classroom floor is a shape of rectangular with length 12 m and width 10 m, its perimeter would be 2(12 m + 10 m) = 44 m and its area would be 12 m × 10 m, i.e., 120 m² .

Unit of measurement for length or breadth is taken as metre (m) or centimetre (cm) etc. Unit of measurement for area of any plane figure is taken as square metre (m² ) or square centimetre (cm² ) etc.

You also studied in previous classes that how to find the area of a triangle.

$\text{Area of a triangle }=\frac{1}{2} \times base \times height$

Let us recall how we find area of right angle triangle, equilateral triangle and isosceles triangle.

When the triangle is right angled, we can directly apply the formula by using two sides containing the right angle as base and height. For example, suppose that the sides of a right triangle ABC are 3 cm, 4 cm and 5 cm; we take base as 4 cm and height as 3 cm.

Then the area of ∆ ABC is given by

$\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times 4 \times 3 \mathrm{cm}^{2}, \text { i.e., } 6 \mathrm{cm}^{2}$

Note that we could also take 3 cm as the base and 4 cm as height.

Equilateral triangle

Now suppose an equilateral triangle ABC with side 10cm is given. To find its area we need its height. We can find the height of triangle using Pythagoras Theorem.

Take the mid-point of BC as D and join it to A. We know that ADB is a right triangle. Therefore, by using Pythagoras Theorem, we can find the length AD as shown below:

AB² = AD² + BD²

i.e. (10)² = AD² + (5)², since, BD = DC

Therefore, we have AD² = 75²

i.e. AD = $\sqrt{75}\;cm=5\sqrt3\;cm$

Then area of ∆ ABC = $\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times 10 \times 5 \sqrt{3} \mathrm{cm}^{2}=25 \sqrt{3} \mathrm{cm}^{2}$

If each side of an equilateral triangle is a unit. Then its area is $\frac{\sqrt3}{4}\left ( a^2 \right )$ sq. unit and and height = $\frac{\sqrt3}{2}\left ( a \right )$ .units.

Isosceles triangle

Let a triangle ABC with two equal sides AB and AC as 5 cm each and unequal side BC as 8 cm is given.

In this case also, we want to know the height of the triangle. So, from A we draw a perpendicular AD to side BC. You can see that this perpendicular AD divides the base BC of the triangle in two equal parts.

Therefore, BD = DC = 1/2(BC) = 4 cm

Then, by using Pythagoras theorem, we get

AD² = AB² - BD²

= 5² - 4²= 25 - 16 = 9

So, AD = 3

Now, area of ∆ ABC = $\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times 8 \times 3 \mathrm{cm}^{2}=12 \mathrm{cm}^{2}$

Till now, we have studied that how to find area of right angle, equilateral, isosceles triangle. In these triamgles, height is given or can be found using Pythagoras Theorem.

Now suppose that we know the lengths of the sides of a scalene triangle and not the height. Can you still find its area? If you want to apply the formula, you will have to calculate its height. But we do not have a clue to calculate the height.

Let's see a formula, where we don't have to calculate height to find the area of any triangle.

Heron’s Formula

The formula given by Heron about the area of a triangle, is also known as Hero’s formula. It is stated as:

$\text{Area of a triangle }=\sqrt{s(s-a)(s-b)(s-c)}$ , where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the perimeter of the triangle = $\frac{a+b+c}{2}$ .

This formula is helpful where it is not possible to find the height of the triangle easily.

TIP:-

Heron was born in about 10AD possibly in Alexandria in Egypt. He worked in applied mathematics. His works on mathematical and physical subjects are so numerous and varied that he is considered to be an encyclopedic writer in these fields. His geometrical works deal largely with problems on mensuration written in three books. Book I deals with the area of squares, rectangles, triangles, trapezoids (trapezia), various other specialised quadrilaterals, the regular polygons, circles, surfaces of cylinders, cones, spheres etc. In this book, Heron has derived the famous formula for the area of a triangle in terms of its three sides.

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Brahmagupta's formula provides the area A of a cyclic quadrilateral (i.e., a simple quadrilateral that is inscribed in a circle) with sides of length a, b, c, and d as

$\mathrm{A=\sqrt{(s-a)(s-b)(s-c)(s-d)}}$

where s is the semiperimeter

$s=\frac{a+b+c+d}{2}$ .

Proof:

Let the quadrilateral be ABCD, with AB = a, BC = b, CD = c and AD = d. Extend AD and BC to meet at E, outside the circumcircle:

(If AD∥BC then considers the other pair of the opposite sides. If those two are also parallel, the quadrilateral is a rectangle, and Brahmagupta's formula reduces to the standard formula for the area of a rectangle.)

Let CE = x and DE = y.

Then area of ∆CDE (using Heron's Formula)

$\\\mathrm{Ar[CDE]=\sqrt{\left (\frac{x+y+c}{2} \right )(x+y-c)(x-y+c)(-x+y+c)}}\\\mathrm{4Ar[CDE]=\sqrt{(x+y+c)(x+y-c)(x-y+c)(-x+y+c)}}\;\;\;\;\;\ldots(1)$

But triangles ABE and CDE are similar, implying:

$\frac{Ar[A B E]}{Ar[C D E]}=\frac{a^{2}}{c^{2}}$

from which

$\\\frac{Ar[C D E]-Ar[A B E]}{Ar[C D E]}=\frac{c^2-a^{2}}{c^{2}}\\\mathrm{Let,\;A=Ar{[CDE]}-Ar[ABE]}\\\Rightarrow A=Ar[CDE]\times\left ( \frac{c^2-a^{2}}{c^{2}} \right )\;\;\;\;\;\;\;\;\;\;\;\;\ldots(2)\\\text{A is area of the Quadrilateral}$

We also have the proportions

$\\ \frac{x}{c}=\frac{y-d}{a} \\ \frac{y}{c}=\frac{x-b}{a}$

Adding the two and solving for (x+y) gives

$x+y=c \left (\frac{b+d}{c-a} \right )$

Similarly, subtracting one from the other and solving for x−y we obtain

$x-y=c \left (\frac{b-d}{c+a} \right )$

from which we can find all the terms in Heron's formula. For example,

$\begin{aligned} x+y+c &=c \left (\frac{b+d}{c-a} \right )+c=c \left (\frac{b+d+c-a}{c-a} \right )=2 c \left (\frac{s-a}{c-a} \right )\\ x+y-c &=c\left ( \frac{b+d}{c-a} \right )-c=c \left (\frac{b+d-c+a}{c-a} \right )=2 c\left ( \frac{s-c}{c-a} \right ) \\ x-y+c &=c \left (\frac{b-d}{c+a} \right )+c=c\left ( \frac{b-d+c+a}{c+a} \right )=2 c \left (\frac{s-d}{c+a} \right ), \text { and } \\ -x+y+c &=-c \left (\frac{b-d}{c+a} \right )+c=c\left ( \frac{-b+d+c+a}{c+a} \right )=2 c \left (\frac{s-b}{c+a} \right ) \end{aligned}$