The area of a triangle whose vertices are $(a, c + a), (a, c)$ and $(- a, c - a)$ are

a^{2}

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b^{2}

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c^{2}

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a^{2} + c^{2}

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Solution

The correct option is A $a^{2}$
Area of a triangle with vertices $(x_{1}, y_{1})$ ; $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $∣ ∣ \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{2} ∣ ∣$
Hence, area of the triangle with given vertices is $∣ ∣ \frac{a (c - c + a) + a (c - a - c - a) - a (c + a - c)}{2} ∣ ∣ = ∣ ∣ \frac{a^{2} - 2 a^{2} - a^{2}}{2} ∣ ∣ = a^{2}$

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Similar questions

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\sqrt{b^{2} + c^{2}}, \sqrt{c^{2} + a^{2}}, \sqrt{a^{2} + b^{2}}

a, b, c > 0

△_{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^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

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

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