Question

prove that the area of an equilateral triangle described on one side of a square is equal of half the area of the equilateral triangle described on one of its diagonal.

Answer 1

Hey

let square ABCD and triangle ABE is described on side AB and triangle ACF is described on diagonal AC you can take other sides also.

Now

triangle ABE is similar to triangle ACF because both are equilateral triangle

By area theorem

Ar of ABE/ Ar of CEF = (AB / AC)^2
......eq 1

By Pythagoras theorem in ABC

AB^2 + AC^2 = AC^2

sides of sq are equal

2 AB^2 = AC ^2

(AB/AC )^2 = 1/2

now putting this value in first eq

Ar of ABE/ Ar of CEF =1/2

2 ar of ABE= Ar of CEF

HOPE IT HELPS

Answer 2

C O N C E P T :

• In the above question, a triangle with all sides equal and angles equal is an equilateral triangle and two equilateral triangles are always similar. the length of side of square is a then the length of diagonal is √ 2

• One must read the question properly and have to draw the diagram clearly and then we have to prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

S O L U T I O N :

➯ Here ABCD is a square and △AEC and △AFB are two equilateral triangles described on diagonal and side of a square respectively.

➯ Let us analyze the above question through a diagram

★ We have to prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of it’s diagonal.

➯ That means, Area ( △ AFB ) / Area ( △ AEC ) = 1 / 2

★ We know that two equilateral triangles are similar.

➯ So, In △AEC and △AFB, by SSS congruency.

➯ AE / AF = EC / FB = CA / BA

★ We know that area of an equilateral triangle is √3 / 4 a ²

➯ In △AEC, the area of triangle can be expressed as

➯ √ 3 / 4 ( AE )² , √ 3 / 4 ( EC ) ² , √ 3 / 4 ( CA ) ²

➯ and in △AFB, the area of triangle can be expressed as

➯ √ 3 / 4 ( BF ) ², √ 3 / 4 ( AF ) ² , √3 / 4 ( AB ) ²

★ For now we will consider area of △AEC is

➯ √ 3 / 4 ( AE ) ² and Area of ∆ AFB is √ 3 / 4 ( BF ) ²

Now,

➯ Area ( ∆ AFB ) / Area ( ∆ AEC ) = ( √3 / 4 ( AE ) ² / √ 3 / 4 ( BF ) ² = AE ² / BF ²

★ If the length of the side of the square is ‘a’. Then by pythagoras theorem length of diagonal of square = √ a ² + a² = √ 2 a² = √ 2 a

★ The length of side AE is ‘a’ as the △AFB is described on the side of a square. So, AE=a and the length of side BF is √ 2 a because the △AEC is described on the diagonal of square. Now substituting AE =a and BF

= √ 2 a in equation (1) we will get,

➯ Area ( ∆ AFB ) / Area ( ∆ AEC ) = AE ² / BF ² = ( AF / BF ) ² = ( a / √ 2 a )

❮ So, we proved that Area ( ∆ AFB ) / Area ( ∆ AEC ) = 1 / 2 ❯

★ Hence, we proved that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal