Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals

Solution:

As we know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

ΔABE is described on the side AB of the square ABCD

ΔDBF is described on the diagonal BD of the square ABCD

Since ΔABE and ΔDBF are equilateral triangles

ΔABE ∼ ΔDBF [each angle in an equilateral triangle is 60 degree]

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Area of ΔABE / Area of ΔDBF = ( AB)² /(DB)²

Area of ΔABE / Area of ΔDBF = (AB)² / (√2AB)²  [diagonal of a square is 2 × side]

Area of ΔABE / Area of ΔDBF = AB² / 2AB²

Area of ΔABE / Area of ΔDBF = 1/2

Area of ΔABE = 1/2 × Area of ΔDBF

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

Video Solution:

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals
 

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.4 Question 7

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals

Hence it is proved that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals

☛ Related Questions:

• Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm^2 and 121 cm^2. If EF = 15.4 cm, find BC.
• Diagonals of a trapezium ABCD with AB || DC intersect each other at the point If AB = 2 CD,find the ratio of the areas of triangles AOB and COD.
• In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area( ABC) / area(DBC) = AO/DO
• If the areas of two similar triangles are equal, prove that they are congruent.