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(a) True

(b) False

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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.

We know that in triangle ABC, altitude is the perpendicular bisector. So, we have,

\[BD=DC......\left( ii \right)\]

\[BC=BD+DC\]

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

\[A{{D}^{2}}+B{{D}^{2}}=A{{B}^{2}}\]

From here, we get,

\[A{{D}^{2}}=A{{B}^{2}}-B{{D}^{2}}\]

Using the value of (i) and (iii), we get,

\[A{{D}^{2}}={{a}^{2}}-\dfrac{{{a}^{2}}}{4}\]

\[\Rightarrow A{{D}^{2}}=\dfrac{4{{a}^{2}}-{{a}^{2}}}{4}\]

\[\Rightarrow A{{D}^{2}}=\dfrac{3{{a}^{2}}}{4}\]

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.

So,

\[\dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}={{\left( \dfrac{AD}{AB} \right)}^{2}}\]

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,

\[\dfrac{Ar\left( \Delta ADE \right)}{Ar\left( \Delta ABC \right)}=\dfrac{3}{4}\]

\[\text{Area}=\dfrac{\sqrt{3}}{4}\times {{\left( \text{side} \right)}^{2}}\]

In triangle ABC with base BC and altitude AD, we get,

\[\text{Area of }\Delta \text{ABC}=\dfrac{1}{2}\times BC\times AD\]

Also,

\[\text{Area of }\Delta ABC=\dfrac{\sqrt{3}}{4}{{\left( \text{side} \right)}^{2}}\]

\[\Rightarrow \text{Area of }\Delta ABC=\dfrac{\sqrt{3}}{4}\times B{{C}^{2}}\]

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,

\[\Rightarrow AD=\dfrac{\sqrt{3}}{2}a\]