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Find the altitude of an equilateral triangle of side 8 cm.
CBSE | Class 10 | Exercise 6.3 | Maths | NCERT Exemplar | Triangles
Find the altitude of an equilateral triangle of side 8 cm.

Solution:

Let an equilateral triangle of side 8 cm be ABC.

AB\text{ }=\text{ }BC\text{ }=\text{ }CA\text{ }=\text{ }8\text{ }cm. (all sides of an equilateral triangle is equal)

NCERT Exemplar Solutions Class 10 Maths Chapter 6 Ex. 6.3-10

Construct an altitude AD perpendicular to BC.

Now, D is the mid-point of BC.

\therefore ~BD\text{ }=\text{ }CD\text{ }={\scriptscriptstyle 1\!/\!{ }_2}

BC\text{ }=\text{ }8/2\text{ }=\text{ }4\text{ }cm

Now, by using the Pythagoras theorem

A{{B}^{2}}~=\text{ }A{{D}^{2}}~+\text{ }B{{D}^{2}}

\Rightarrow ~{{\left( 8 \right)}^{2}}~=\text{ }A{{D}^{2}}~+\text{ }{{\left( 4 \right)}^{2}}

\Rightarrow ~64\text{ }=\text{ }A{{D}^{2}}~+\text{ }16

\Rightarrow ~AD\text{ }=\surd 48\text{ }=\text{ }4\surd 3\text{ }cm.

As a result, altitude of an equilateral triangle is 4√3 cm.

OR

By Heron’ Formula
area = √[s(s-a)(s-b)(s-c)]
=√[12×4×4×4]
=16√3 cm²

As we know that, area = (1/2)×base×altitude
⇒16√3 = (1/2)×8×altitude
⇒altitude = 4√3 cm

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