Answer: B
A: The area of a triangle with sides 6, 8, 10
By this information we can deduce that it is a right side traingle (6^2 + 8^2 = 10^2)
So it’s area equals: 6 * 8 / 2 = 24

B: The area of an equilateral triangle with side 8.
We know the base of triangle is 8, for calculating it’s area we need to know it’s height. If we draw the shape we will see:
h^2 + (8/2)^2 = 8^2 ->
h^2 = 64 - 16 = 48 ->
h = √48 = 4√3
Area = 1/2 * h* base = 1/2 * 4√3 * 8 = 16√3

Now we need to compare 24 and 16√3, Which 16√3 is bigger. The answer is B

** In general height of a equilateral traingle which it’s sides are “a", is √3*a/2
h^2 + (a/2)^2 = a^2 ->
h^2 = a^2 - a^2/4 ->
h^2 = 3a^2/4
h = √3*a/2

I was confused by A!

Col. A: \(\frac{1}{2}(6)(8) = 24\)
Col. B: \(\frac{\sqrt{3}}{4}(8^2) = 16\sqrt{3}\)

Dividing by \(16\) both sides;

Col. A: \(1.5\)
Col. B: \(1.73\)

Hence, option B

QA don't state that 6,8,10 is right angle triangle? it can be scalen triangle....

Even if doesn't say it is a right triangle, but it will always be!

Apply Cosine rule:

Angle C (opposite to side 10) = \(Cos^-1 [\frac{6^2 + 8^2 - 10^2}{(2)(6)(8)}]\)

C = \(Cos^-1 [\frac{0}{90}]\)

C = \(Cos^-1 (0) = 90\)

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