Let-f-be-a-real-valued-function-defined-on-the-inte-rval-1-1-If-the-area-of-the-equilateral-triangle-with-0-0-and-x-f-x-as-two-vertices-is-3-4-then-f-x-is-equal-to-A-1-x-2-

Question Number 119807 by Ar Brandon last updated on 27/Oct/20

Let f be a real - valued function defined on the inte -

rval [- 1, 1] . If the area of the equilateral triangle with

(0, 0) and (x, f (x)) as two vertices is \sqrt{3} / 4, then f (x)

is equal to

(A) \sqrt{1 - x^{2}} (B) \sqrt{1 + x^{2}}

(C) - \sqrt{1 - x^{2}} (D) - \sqrt{1 + x^{2}}

Answered by TANMAY PANACEA last updated on 28/Oct/20

s i d e o f e q u i l a t e r a [t r i a n g l e = a = \sqrt{x^{2} + f^{2} (x)}

\frac{\sqrt{3}}{4} \times a^{2} = \frac{\sqrt{3}}{4}

a^{2} = 1

x^{2} + f^{2} (x) = 1

f (x) = \sqrt{1 - x^{2}}

Commented by Ar Brandon last updated on 28/Oct/20

Thank you Sir

Commented by Ar Brandon last updated on 28/Oct/20

Thanks Sir

Commented by Ar Brandon last updated on 03/Nov/20

Sir how come \frac{\sqrt{3}}{4} \times a^{2} ? Is it by using

\frac{1}{2} a \times a \sin 60 ° ? 60 ° being the angle

between two adjacent sides .

Commented by TANMAY PANACEA last updated on 28/Oct/20

a r e a = \sqrt{s (s - a) (s - b) (s - c)}

h e r e a = b = c

s = \frac{a + b + c}{2} = \frac{a + a + a}{2} = \frac{3 a}{2}

(s - a) = \frac{3 a}{2} - a = \frac{a}{2}

a r e a = \sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}} = \frac{\sqrt{3}}{4} a^{2}

Commented by TANMAY PANACEA last updated on 28/Oct/20

o r m d t h o d a r e a = \frac{1}{2} \times a \times h

s i n 60^{o} = \frac{h}{a} \to h = \frac{\sqrt{3}}{2} a

a r e a = \frac{\sqrt{3}}{2} \times a \times \frac{1}{2} \times a = \frac{\sqrt{3}}{4} a^{2}

Commented by TANMAY PANACEA last updated on 28/Oct/20

y e s s i r