CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
465472_9b70d7be9cca437abeff969573eb26f4.png


Solution

Given:
$$\triangle ABC$$ is an Equilateral Triangle.
$$AB=BC=AC$$
$$CD$$ Perpendicular to $$AB$$
To Prove:
$$3AC^2=4CD^2$$
Proof:

In $$\triangle ADC$$
By Pythagoras Theorem,
$$AC^2=AD^2+DC^2\,.....(1)$$
In $$\triangle BDC$$
By Pythagoras Theorem,
$$BC^2=DC^2+BD^2\,.....(2)$$
Adding 1) and 2) We get,
$$2AC^2=2DC^2+AD^2+BD^2$$ $$(Since\,\, AC=BC)$$
Since, $$AD=BD=\dfrac{1}{2}AC$$ $$(Since\,\,AC=AB)$$
$$\therefore$$ $$2AC^2=2DC^2+\dfrac{1}{2}AC^2$$

$$\therefore$$ $$2AC^2-\dfrac{1}{2}AC^2=2DC^2$$

$$\therefore$$ $$3AC^2=4CD^2$$


495464_465472_ans_638f31473e464f6f8f6559304073d0fc.png

Mathematics
RS Agarwal
Standard X

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image