Question

In a ΔABC, AB = BC = CA = 2 a and AD⊥BC. Prove that
(i) AD = a√3 (ii) Area (ΔABC) =√3a²

Answer 1

(i)

ΔABC is a equilateral triangle so BD=CD=a.

By The Pythagorean Theorem

In ΔABD, ∠D=90°

AB²=BD²+AD²

We know AB=2a and BD=a

4a²=a²+AD²

AD²=3a²

∴ AD=(√3)a

(ii)

ΔABC=(1/2)*AB*AD=(1/2)*2a*(√3)a=√3a

2 gets erased. Hence shown.

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Answer 2

Step-by-step explanation:

GIVEN:AB = BC = CA = 2a

AD perpendicular BC

TO PROVE:AD = √3a and √3a²

AB = BC = CA. Therefore it is an equilateral triangle.

AD PERPENDICULAR BC

In ∆ABD

(AB)² = (BD)² + (AD)²

(2a)²=(a)² + (AD)² { AD PERPENDICULAR BC.So,

BD = BC/2

BD = 2a/2

BD = a }

4a² = a² + (AD)²

4a² -a² = (AD)²

3a² = (AD)²

AD = √3a²

AD = a√3

AREA OF AN EQUILATERAL TRIANGLE = √3/4a²

= √3/4(2a)²

= √3/4(4a²)

= √3a²

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