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.
Answer
290.5k+ views
Hint: 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 $\sqrt{2}a.$
Complete step-by-step answer:
Let us analyze the above question through a diagram-
.
Here ABCD is a square and $\vartriangle AEC$ and $\vartriangle AFB$ are two equilateral triangles described on diagonal and side of a square respectively.
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, $\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{1}{2}$.
We know that two equilateral triangles are similar.
So, In $\vartriangle AEC$ and $\vartriangle AFB$, by SSS congruency.
$\dfrac{AE}{AF}=\dfrac{EC}{FB}=\dfrac{CA}{BA}$
We know that area of an equilateral triangle is $\dfrac{\sqrt{3}}{4}{{a}^{2}}$.
In $\vartriangle AEC$, the area of triangle can be expressed as $\dfrac{\sqrt{3}}{4}{{(AE)}^{2}},\dfrac{\sqrt{3}}{4}{{(EC)}^{2}},\dfrac{\sqrt{3}}{4}{{(CA)}^{2}}$.
and in $\vartriangle AFB$, the area of triangle can be expressed as $\dfrac{\sqrt{3}}{4}{{(BF)}^{2}},\dfrac{\sqrt{3}}{4}{{(AF)}^{2}},\dfrac{\sqrt{3}}{4}{{(AB)}^{2}}$ .
For now we will consider area of $\vartriangle AEC$ is $\dfrac{\sqrt{3}}{4}{{(AE)}^{2}}$ and the area of $\vartriangle AFB$ is $\dfrac{\sqrt{3}}{4}{{(BF)}^{2}}$.
Now, $\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{\left( \dfrac{\sqrt{3}}{4}{{(AE)}^{2}} \right)}{\left( \dfrac{\sqrt{3}}{4}{{(BF)}^{2}} \right)}=\dfrac{A{{E}^{2}}}{B{{F}^{2}}}\cdot \cdot \cdot \cdot \cdot (1)$
If the length of the side of the square is ‘a’. Then by pythagoras theorem length of diagonal of square=$\sqrt{{{a}^{2}}+{{a}^{2}}}=\sqrt{2{{a}^{2}}}=\sqrt{2}a$.
The length of side AE is ‘a’ as the $\vartriangle AFB$ is described on the side of a square. So, AE=a and the length of side BF is $\sqrt{2}a$ because the $\vartriangle AEC$ is described on the diagonal of square. Now substituting AE=a and BF=$\sqrt{2}a$ in equation (1) we will get,
$\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{A{{E}^{2}}}{B{{F}^{2}}}={{\left( \dfrac{AE}{BF} \right)}^{2}}={{\left( \dfrac{a}{\sqrt{2}a} \right)}^{2}}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=\dfrac{1}{2}$.
So, we proved that $\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{1}{2}$.
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.
Note: 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.
Complete step-by-step answer:
Let us analyze the above question through a diagram-

Here ABCD is a square and $\vartriangle AEC$ and $\vartriangle AFB$ are two equilateral triangles described on diagonal and side of a square respectively.
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, $\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{1}{2}$.
We know that two equilateral triangles are similar.
So, In $\vartriangle AEC$ and $\vartriangle AFB$, by SSS congruency.
$\dfrac{AE}{AF}=\dfrac{EC}{FB}=\dfrac{CA}{BA}$
We know that area of an equilateral triangle is $\dfrac{\sqrt{3}}{4}{{a}^{2}}$.
In $\vartriangle AEC$, the area of triangle can be expressed as $\dfrac{\sqrt{3}}{4}{{(AE)}^{2}},\dfrac{\sqrt{3}}{4}{{(EC)}^{2}},\dfrac{\sqrt{3}}{4}{{(CA)}^{2}}$.
and in $\vartriangle AFB$, the area of triangle can be expressed as $\dfrac{\sqrt{3}}{4}{{(BF)}^{2}},\dfrac{\sqrt{3}}{4}{{(AF)}^{2}},\dfrac{\sqrt{3}}{4}{{(AB)}^{2}}$ .
For now we will consider area of $\vartriangle AEC$ is $\dfrac{\sqrt{3}}{4}{{(AE)}^{2}}$ and the area of $\vartriangle AFB$ is $\dfrac{\sqrt{3}}{4}{{(BF)}^{2}}$.
Now, $\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{\left( \dfrac{\sqrt{3}}{4}{{(AE)}^{2}} \right)}{\left( \dfrac{\sqrt{3}}{4}{{(BF)}^{2}} \right)}=\dfrac{A{{E}^{2}}}{B{{F}^{2}}}\cdot \cdot \cdot \cdot \cdot (1)$
If the length of the side of the square is ‘a’. Then by pythagoras theorem length of diagonal of square=$\sqrt{{{a}^{2}}+{{a}^{2}}}=\sqrt{2{{a}^{2}}}=\sqrt{2}a$.
The length of side AE is ‘a’ as the $\vartriangle AFB$ is described on the side of a square. So, AE=a and the length of side BF is $\sqrt{2}a$ because the $\vartriangle AEC$ is described on the diagonal of square. Now substituting AE=a and BF=$\sqrt{2}a$ in equation (1) we will get,
$\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{A{{E}^{2}}}{B{{F}^{2}}}={{\left( \dfrac{AE}{BF} \right)}^{2}}={{\left( \dfrac{a}{\sqrt{2}a} \right)}^{2}}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=\dfrac{1}{2}$.
So, we proved that $\dfrac{Area(\vartriangle AFB)}{Area(\vartriangle AEC)}=\dfrac{1}{2}$.
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.
Note: 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.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
Change the following sentences into negative and interrogative class 10 english CBSE

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

What is 1 divided by 0 class 8 maths CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Convert compound sentence to simple sentence He is class 10 english CBSE

India lies between latitudes and longitudes class 12 social science CBSE

Why are rivers important for the countrys economy class 12 social science CBSE

Distinguish between Khadar and Bhangar class 9 social science CBSE
