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CAT Area of Equilateral and Isosceles Triangle - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Quick Facts

5 Questions around this concept.

Solve by difficulty

Each side of an equilateral triangle measures a unit. Then the area of the triangle is $\frac{\sqrt3}{4}a^2\;$ sq. unit.

True

False

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The height of an equilateral triangle is 10 cm. Its area is (in cm²)

$50$

$100$

$50\sqrt3$

$\frac{100}{\sqrt3}$

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Concepts Covered - 1

Area of Equilateral and Isosceles Triangle

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 = $\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times 8 \times 3 \mathrm{cm}^{2}=12 \mathrm{cm}^{2}$