Question

AD is an altitude of an equilateral triangle ABC. on AD as base another equilateral triangle ade is constructed prove that area of triangle A D ratio area of triangle ABC is equal to 3 ratio 4

Answer 1

Answer:

Step-by-step explanation:

Since,ABC is a equilateral triangle

So,AB=BC=CA=a

Area of give triangle=√3/4a^2

And height is √3a/2

Answer 2

Given: AD is an altitude of an equilateral triangle ABC.

On AD as base another equilateral triangle ADE is constructed.

To prove: $\dfrac{\text{Area of triangle AED}}{\text{Area of triangle ABC}}=\dfrac{3}{4}$

Proof:

Please see the attachment for graph.

Let the side of equilateral triangle ABC be a

The length of altitude of AD $=\dfrac{\sqrt{3}a}{2}$

Thus, The side the equilateral triangle AED $=\dfrac{\sqrt{3}a}{2}$

Area of equilateral triangle $=\dfrac{\sqrt{3}(side)^2}{4}$

Area of triangle ABC  $=\dfrac{\sqrt{3}a^2}{4}$

Area of triangle AED  $=\dfrac{\sqrt{3}(\dfrac{\sqrt{3}a}{2})^2}{4}=\dfrac{\sqrt{3}}{4}\cdot \dfrac{3a^2}{4}$

The ratio of ar(AED) to ar(ABC) $=\dfrac{\dfrac{\sqrt{3}}{4}\cdot \dfrac{3a^2}{4}}{\dfrac{\sqrt{3}}{4}a^2}=\dfrac{3}{4}$

Ratio of ar(AED) to ar(ABC) = 3:4

Hence proved