Question

Area of the triangle formed by (a, b + c), (b, c + a), (c, a + b) in sq.units​

Answer 1

ANSWER:

Given:

• Coordinates of vertices of a triangle = A(a, b + c), B(b, c + a), C(c, a + b)

To Find:

• Area of ΔABC

Diagram:

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A(a,b+c) }\put(0.5,-0.3){ \bf C(c,a+b) }\put(5,-0.3){ \bf B(b,c+a) }\put(4.5,1.5){ \bf Area=? }\end{picture}$

Solution:

$\text{We know that,}\\\\:\longrightarrow\text{Area of a triangle}=\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|\\\\\text{Here,}\\\\:\hookrightarrow x_1=a, y_1=(b+c), x_2=b, y_2=(c+a), x_3=c, y_3=(a+b)\\\\\text{So,}\\\\:\implies\text{Area of \Delta ABC}\\\\:\implies\dfrac{1}{2}\bigg|a(c+a\!\!\!/-a\!\!\!/-b)+b(a+b\!\!\!/-b\!\!\!/-c)+c(b+c\!\!\!/-c\!\!\!/-a)\bigg|\\\\:\implies\dfrac{1}{2}\bigg|a(c-b)+b(a-c)+c(b-a)\bigg|$

$:\implies\dfrac{1}{2}\bigg|ac\!\!\!\!/\,-ab\!\!\!\!/\,+ab\!\!\!\!/\,-bc\!\!\!\!/\,+bc\!\!\!\!/\,-ac\!\!\!\!/\,\bigg|\\\\:\implies\dfrac{1}{2}\times0\\\\\text{So,}\\\\\bf{:\implies0\:sq.\:units.}\\\\\text{\bf{The area of \Delta ABC is 0 sq. units.}}$

Formula Used:

• Area of a triangle = 1/2×|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.