Question

Plz Solve this Question. ASAP Pls.​

Answer 1

$\large\bold{\underline{\underline{Answer:-}}}$

$\sf{ \green{ \underline{ \underline{GIVEN : }}}}$

ABC is an equilateral triangle with side 'a'.

AD is the Altitude of the triangle.

$\sf{ \green{ \underline{ \underline{TO \: \: FIND : }}}}$

Length of the Altitude.

$\sf{ \green{ \underline{ \underline{SOLUTION : }}}}$

As, AD is the Altitude of the triangle ABC. So it bisects the base BC perpendicularly, because we know altitude of divides an equilateral triangle in two equal halves.

Therefore, BD = CD and Ang. ADB = Ang. ADC = 90°.

$\boxed{ \therefore \: \sf \: We \: got \: AD = CD = \frac{a}{2}}$

Now, from right angle triangle ADB, using Pythagoras Theorem

$\boxed{\sf{\: {AB}^{2} = {AD}^{2} + {BD}^{2} }}$

On putting the values of AB, AD and BD we get,

$\implies \: \sf{\: {a}^{2} = {AD}^{2} + {( \frac{a}{2}) }^{2} }$

$\implies \: \sf{\: {a}^{2} = {AD}^{2} + \frac{ {a}^{2} }{4} }$

$\implies \: \sf{\: {a}^{2} = \frac{ { {4AD}^{2} \: + \: a}^{2} }{4} }$

[On cross multiplication,]

$\implies \: \sf{\: {4a}^{2} = { {4AD}^{2} \: + \: a}^{2}}$

$\implies \: \sf{\: {4a}^{2} - {a}^{2} = { {4AD}^{2}}}$

$\implies \: \sf{\: {4AD}^{2}} = {3a}^{2}$

$\implies \: \sf{\: {AD}^{2}} = \frac{3}{4} {a}^{2}$

$\implies \: \sf{\: \sqrt{{AD}^{2}}} = \sqrt{ \frac{3 {a}^{2} }{4} }$

$\boxed{\therefore \: \sf{\: {AD} = \frac{ \sqrt{3}a }{2}} \: unit}$

Therefore, the length of the Altitude AD is $\sf {\frac{ \sqrt{3}a}{2} \: unit }$.

Answer 2

Answer:

The length of altitude if side is a then altitude is (root3/2)a

Derivation is a²- a²/4 = 3/4a² By Pythagoras theorem in any of tye smaller triangle firmed by altitude.

Root of 3a²/4 = root3a/2

Explanation:

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