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AD is an altitude of an equilateral triangle ABC. On AD as the base, another equilateral triangle ADE is constructed, then area (Triangle ADE): area (Triangle ABC) = 3:4.
(a) True
(b) False

Last updated date: 16th Jul 2024
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Hint: As we have both the triangles as equilateral triangles, so they are similar. We know that if triangle ABC is congruent to triangle PQR, then, \[\dfrac{Ar\left( \Delta ABC \right)}{Ar\left( \Delta PQR \right)}={{\left( \dfrac{AB}{PQ} \right)}^{2}}={{\left( \dfrac{BC}{QR} \right)}^{2}}={{\left( \dfrac{AC}{PR} \right)}^{2}}.\] So, we compare the sides to find the ratio of the area between triangle ADE and triangle ABC.

Complete step-by-step solution:
We are given that ABC is an equilateral triangle. Let us consider its side as ‘a’. So, we get,
\[AB=BC=AC=a......\left( i \right)\]
AD is the altitude through while we will construct another triangle ADE.
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We know that in triangle ABC, altitude is the perpendicular bisector. So, we have,
\[BD=DC......\left( ii \right)\]
From (ii), we have BD = DC, so we get,
\[\Rightarrow BC=BD+BD\]
\[\Rightarrow BC=2BD\]
So, we get,
\[\Rightarrow \dfrac{BC}{2}=BD\]
Noe from (i), we get,
\[BD=\dfrac{a}{2}....\left( iii \right)\]
As AD is a perpendicular bisector, we get triangle ABD as right triangle. Now, applying Pythagoras theorem in triangle ABD which says
From here, we get,
Using the value of (i) and (iii), we get,
\[\Rightarrow A{{D}^{2}}=\dfrac{4{{a}^{2}}-{{a}^{2}}}{4}\]
\[\Rightarrow A{{D}^{2}}=\dfrac{3{{a}^{2}}}{4}\]
Now, canceling square on both the sides, we get,
\[AD=\dfrac{\sqrt{3}a}{2}.....\left( iv \right)\]
So, we get the length of the sides of the triangle ADC as \[AD=\dfrac{\sqrt{3}a}{2}.\]
As triangle ABC and triangle ADE are both equilateral triangles, this means that both the triangles have all the angles as \[{{60}^{\circ }}.\] Therefore, by AAA similarity criteria, they are similar.
\[\dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}={{\left( \dfrac{AD}{AB} \right)}^{2}}\]
Using (i) and (iv), we get,
\[\Rightarrow \dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}=\dfrac{{{\left( \dfrac{\sqrt{3}{a}}{2} \right)}^{2}}}{{{a}^{2}}}\]
\[\Rightarrow \dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}=\dfrac{\dfrac{3{{a}^{2}}}{4}}{{{a}^{2}}}\]
\[\Rightarrow \dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}=\dfrac{3{{a}^{2}}}{4{{a}^{2}}}\]
\[\Rightarrow \dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}=\dfrac{3}{4}\]
So, we get,
\[\dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}=\dfrac{3}{4}\]
Hence, option (a) is the right answer.

Note: The alternate method to find AD, i.e the altitude is as follows. We know that the area of the equilateral triangle is also given as
\[\text{Area}=\dfrac{\sqrt{3}}{4}\times {{\left( \text{side} \right)}^{2}}\]
In triangle ABC with base BC and altitude AD, we get,
\[\text{Area of }\Delta \text{ABC}=\dfrac{1}{2}\times BC\times AD\]
\[\text{Area of }\Delta ABC=\dfrac{\sqrt{3}}{4}{{\left( \text{side} \right)}^{2}}\]
\[\Rightarrow \text{Area of }\Delta ABC=\dfrac{\sqrt{3}}{4}\times B{{C}^{2}}\]
So comparing both, we get,
\[\dfrac{1}{2}\times BC\times AD=\dfrac{\sqrt{3}}{4}\times B{{C}^{2}}\]
Solving for AD, we get,
\[\Rightarrow AD=\dfrac{\sqrt{3}}{4}\times B{{C}^{2}}\times \dfrac{2}{BC}\]
\[\Rightarrow AD=\dfrac{\sqrt{3}}{2}\times BC\]
As BC = a, we can write as,
\[\Rightarrow AD=\dfrac{\sqrt{3}}{2}a\]