Question

Find the area of $△ABC$ with vertices A(0,-1), B(2,1) and C(0,3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio ofyou have to be careful how you use less-than signs, ampersands, and other HTML special the area of two triangles is 4: 1.

Answer 1

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Let​​​​Let A(0, -1), B(2, 1) and C(0, 3) be the vertices of ΔABC. Let D, E, F be the mid points of sides BC, CA and AB respectively. Then, the coordinates of D, E and F are (1, 2), (0, 1) and (1, 0) respectively.

Now,

Area of ΔABC = $\frac{1}{2}$(x1 (y2 - y3 ) + x2 (y3 - y1 ) + x3 (y1 - y2 ))

⇒ Area of ΔABC = $\frac{1}{2}$(0(1 - 3) + 2(3 - (-1)) + 0(0 - 1))

⇒ Area of ΔABC =$\frac{1}{2}$ (0 + 8 + 0)= 4 sq. units.

Area of ΔDEF =$\frac{1}{2}$ (x1 (y2 - y3 ) + x2 (y3 - y1 ) + x3 (y1 - y2 ))

⇒ Area of ΔDEF = $\frac{1}{2}$ (1(1 - 0) + 0(0 - 2) + 1(2 - 1))

⇒ Area of ΔDEF = $\frac{1}{2}$ (1 + 1) = 1 sq. units.

∴ Area of ΔDEF : Area of ΔABC = 1 : 4