Answer
Verified
330.7k+ 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
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
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
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
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
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
What organs are located on the left side of your body class 11 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE