# In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes

**Solution:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [Pythagoras theorem]

In ΔABC,

AB = BC = CA (sides of the triangle), AD is the altitude

AD ⊥ BC

We know that in an equilateral triangle perpendicular drawn from vertex to opposite side bisects the side

Thus, BD = CD = BC/2

Now in ΔADC,

AC^{2} = AD^{2} + CD^{2}

BC^{2} = AD^{2} + (BC/2)^{2} [Since AC = BC and CD = BC/2]

BC^{2} = AD^{2} + BC^{2}/4

BC^{2} - BC^{2}/4 = AD^{2}

3BC^{2}/4 = AD^{2}

3BC^{2} = 4AD^{2}

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 16

**Summary:**

In an equilateral triangle, we have proved that three times the square of one side is equal to four times the square of one of its altitudes. Hence proved 3BC^{2} = 4AD^{2}

**☛ Related Questions:**

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