If a plane cuts off intercepts $O A = a, O B = b, O C = c$ from the co-ordinate axes, then the area of the triangle $A B C =$

\frac{1}{2} \sqrt{b^{2} c^{2} + c^{2} a^{2} + a^{2} b^{2}}

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\frac{1}{2} (b c + c a + a b)

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\frac{1}{2} a b c

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Solution

The correct option is A $\frac{1}{2} \sqrt{b^{2} c^{2} + c^{2} a^{2} + a^{2} b^{2}}$
Given that,
$O A = a$
$O B = b$
and $O C = c$
Area of triangle $= \frac{1}{2} \sqrt{b^{2} c^{2} + c^{2} a^{2} + a^{2} b^{2}}$
Square part $(A)$ is correct answer.

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Similar questions

If a plane cuts off intercepts OA = a, OB = b, OC = c from the co-ordinate axes, then the area of the triangle ABC=

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

P_{1} : c y + b z = x

P_{2} : a z + c x = y

P_{3} : b x + a y = z .

P_{1}, P_{2}

P_{3}

△_{1}, △_{2}

(b, c), (c, a), (a, b)

(a c - b^{2}, a b - c^{2}), (b a - c^{2}, b c - a^{2}), (c b - a^{2}, c a - b^{2})

\frac{△_{1}}{△_{2}} = (a + b + c)^{2}

a + b + c = a b c

\frac{a (b^{2} c^{2} - 1)}{b c + 1} + \frac{b (c^{2} a^{2} - 1)}{c a + 1} + \frac{c (a^{2} b^{2} - 1)}{a b + 1} = 2 a b c

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