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,

AC² = AD² + CD²

BC² = AD² + (BC/2)² [Since AC = BC and CD = BC/2]

BC² = AD² + BC²/4

BC² - BC²/4 = AD²

3BC²/4 = AD²

3BC² = 4AD²

☛ 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 3BC² = 4AD²

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