AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

In △ ABD and △ ACD,

⇒ ∠ADB = ∠ADC (Both equal to 90°)

⇒ AD = AD (Common side)

⇒ AB = AC (Since, ABC is an equilateral triangle)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{10}{2}$ = 5 cm.

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ 10² = AD² + 5²

⇒ AD² = 10² - 5²

⇒ AD² = 100 - 25

⇒ AD² = 75

⇒ AD = $\sqrt{75}$ = 8.7 cm.

Hence, AD = 8.7 cm.