The area of a nbsp-triangle having vertices -a- b- c- -b- c - a- and -c- a - b- is - 0-1-a - b - c-a-2 - b-2 - c-2-

The correct option is A 0 Area of a nbsp;triangle with vertices (x_1, y_1), (x_2, y_2) and (x_3, y_3) is given by Area=1/2× |(x_1)(y_2 y_3)+(x_2)(y_3 y_1)+(x ...

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Byju's Answer

Standard X

Mathematics

Area of a Triangle Using Coordinate of Vertices

The area of a...

Question

The area of a triangle having vertices (a, b+ c), (b, c + a) and (c, a + b) is ___.

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a + b + c

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$a^{2}$ + $b^{2}$ + $c^{2}$

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Solution

The correct option is A
0

Area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3})$ is given by
$A r e a = \frac{1}{2} \times | (x_{1}) (y_{2} - y_{3}) + (x_{2}) (y_{3} - y_{1}) + (x_{3}) (y_{1} - y_{2}) |$

Thus, $Area of the given triangle$
$= \frac{1}{2} | a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a) |$
$= \frac{1}{2} | a c - a b + a b - b c + b c - a c | = 0$

Suggest Corrections

Similar questions

Q. Prove that :

|\begin{matrix} a^{2} & a^{2} - {(b - c)}^{2} & b c \\ b^{2} & b^{2} - {(c - a)}^{2} & c a \\ c^{2} & c^{2} - {(a - b)}^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})

The area of a triangle having vertices (a, b+ c), (b, c + a) and (c, a + b) is ___.

Q. If

△_{1}, △_{2}

be the areas of two triangles with vertices

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

, and

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

, then

\frac{△_{1}}{△_{2}} = (a + b + c)^{2}

Q. Using factor theorem, show that

a - b, b - c

and

c - a

are the factors of

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

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

Q. If

a, b, c

are the sides of a triangle then the range of

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

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The area of a triangle having vertices (a, b+ c), (b, c + a) and (c, a + b) is ___.

The correct option is A 0 Area of a triangle with vertices (x1,y1),(x2,y2) and (x3,y3) is given by Area=12×|(x1)(y2−y3)+(x2)(y3−y1)+(x3)(y1−y2)| Thus, Area of the given triangle =12|a(c+a−a−b)+b(a+b−b−c)+c(b+c−c−a)| =12|ac−ab+ab−bc+bc−ac|=0