# Prove that the area of the 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.

Answer

Verified

362.7k+ views

Hint: Consider any square whose edge length is \[a\]. Find the length of the diagonal of the square using hypotenuse property. Calculate the area of both the triangles constructed on edge of square and diagonal of square using the formula \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\] and then compare them to get the desired result.

Complete step-by-step answer:

We have a square \[ABCD\] whose length of each side is \[a\]. Two triangles are constructed on this square, one on side \[AB\], the triangle \[\vartriangle ABE\] and other on diagonal \[AC\], the triangle \[\vartriangle AFC\], as shown in the figure.

We will calculate the length of diagonal \[AC\]. We know that all angles of a square are equal to \[{{90}^{\circ }}\]. Thus, \[\vartriangle ABC\] is a right angled triangle, right angled at \[B\] and the length of the sides \[BC\] and \[AB\] are equal to \[a\]. We have to find the length of the side \[AC\].

As \[\vartriangle ABC\] is a right angled triangle, it follows Hypotenuse Property. Thus, we have \[{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}\].

Substituting the value of length of sides, we have \[{{a}^{2}}+{{a}^{2}}={{\left( AC \right)}^{2}}\].

\[\begin{align}

& \Rightarrow {{\left( AC \right)}^{2}}=2{{a}^{2}} \\

& \Rightarrow AC=\sqrt{2}a \\

\end{align}\]

Thus, the length of side \[AC\] is \[\sqrt{2}a\]. As \[\vartriangle AFC\] is an equilateral triangle, each of its sides is equal and has length \[\sqrt{2}a\].

We will now calculate the area of both the triangles. We know that the area of the equilateral triangle whose length of sides is \[a\] is \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\].

We will firstly calculate the area of \[\vartriangle ABE\]. As the length of each side of the triangle is \[a\] is \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\].

We will now calculate the area of \[\vartriangle AFC\]. Each side of this triangle has length \[\sqrt{2}a\]. So, the area of triangle is \[\dfrac{\sqrt{3}}{4}{{\left( \sqrt{2}a \right)}^{2}}=\dfrac{\sqrt{3}}{4}\left( 2{{a}^{2}} \right)=\dfrac{\sqrt{3}}{2}{{a}^{2}}\].

We will now find the ratio of area of both triangles. Thus, \[\dfrac{area\left( \vartriangle ABE \right)}{area\left( \vartriangle AFC \right)}=\dfrac{\dfrac{\sqrt{3}}{4}{{a}^{2}}}{\dfrac{\sqrt{3}}{2}{{a}^{2}}}=\dfrac{2}{4}=\dfrac{1}{2}\].

Hence, we have proved that the area of the 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.

Note: One must clearly know the meaning of terms equilateral triangle and the Hypotenuse Property. Equilateral triangle is a triangle in which the measure of all the sides is equal. Hypotenuse Property states that in a right angled triangle, the sum of squares of length of base and perpendicular is equal to the square of the length of hypotenuse. One should also observe that as all the sides of the square are equal, the length of both the diagonals will be equal as well. So, it doesn’t matter which diagonal we are considering to draw the equilateral triangle.

Complete step-by-step answer:

We have a square \[ABCD\] whose length of each side is \[a\]. Two triangles are constructed on this square, one on side \[AB\], the triangle \[\vartriangle ABE\] and other on diagonal \[AC\], the triangle \[\vartriangle AFC\], as shown in the figure.

We will calculate the length of diagonal \[AC\]. We know that all angles of a square are equal to \[{{90}^{\circ }}\]. Thus, \[\vartriangle ABC\] is a right angled triangle, right angled at \[B\] and the length of the sides \[BC\] and \[AB\] are equal to \[a\]. We have to find the length of the side \[AC\].

As \[\vartriangle ABC\] is a right angled triangle, it follows Hypotenuse Property. Thus, we have \[{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}\].

Substituting the value of length of sides, we have \[{{a}^{2}}+{{a}^{2}}={{\left( AC \right)}^{2}}\].

\[\begin{align}

& \Rightarrow {{\left( AC \right)}^{2}}=2{{a}^{2}} \\

& \Rightarrow AC=\sqrt{2}a \\

\end{align}\]

Thus, the length of side \[AC\] is \[\sqrt{2}a\]. As \[\vartriangle AFC\] is an equilateral triangle, each of its sides is equal and has length \[\sqrt{2}a\].

We will now calculate the area of both the triangles. We know that the area of the equilateral triangle whose length of sides is \[a\] is \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\].

We will firstly calculate the area of \[\vartriangle ABE\]. As the length of each side of the triangle is \[a\] is \[\dfrac{\sqrt{3}}{4}{{a}^{2}}\].

We will now calculate the area of \[\vartriangle AFC\]. Each side of this triangle has length \[\sqrt{2}a\]. So, the area of triangle is \[\dfrac{\sqrt{3}}{4}{{\left( \sqrt{2}a \right)}^{2}}=\dfrac{\sqrt{3}}{4}\left( 2{{a}^{2}} \right)=\dfrac{\sqrt{3}}{2}{{a}^{2}}\].

We will now find the ratio of area of both triangles. Thus, \[\dfrac{area\left( \vartriangle ABE \right)}{area\left( \vartriangle AFC \right)}=\dfrac{\dfrac{\sqrt{3}}{4}{{a}^{2}}}{\dfrac{\sqrt{3}}{2}{{a}^{2}}}=\dfrac{2}{4}=\dfrac{1}{2}\].

Hence, we have proved that the area of the 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.

Note: One must clearly know the meaning of terms equilateral triangle and the Hypotenuse Property. Equilateral triangle is a triangle in which the measure of all the sides is equal. Hypotenuse Property states that in a right angled triangle, the sum of squares of length of base and perpendicular is equal to the square of the length of hypotenuse. One should also observe that as all the sides of the square are equal, the length of both the diagonals will be equal as well. So, it doesn’t matter which diagonal we are considering to draw the equilateral triangle.

Last updated date: 02nd Oct 2023

•

Total views: 362.7k

•

Views today: 4.62k

Recently Updated Pages

What do you mean by public facilities

Please Write an Essay on Disaster Management

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers