Question

Find the area and match the pair.

• A. 47.5 square units
• B. 72 square units
• C. 10.5 square units

Answer 1

Area of the triangle with coordinates $(x_1,y_1),(x_2,y_2)\text{and}(x_3,y_3)$
= $\frac{1}{2}\left(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$


$\therefore$ Area of the above shown triangle =$\frac{1}{2}\left((-5)\left(-2-5\right)+(-4)\left(5-6\right)+(7\left)\right(6-(-2))\right]$
= $\frac{1}{2}(35+4+56)$
= 47.5 square units.



By joining B to D, we get two triangles ABD and BCD.

Area of the $\Delta$ ABD = $\frac{1}{2}\left((-5)\left(-5-5\right)+(-4)\left(5-7\right)+4\left(7+5\right)\right]$
= $\frac{1}{2}(50+8+48)$
= 53 square units

Area of the $\Delta$ BCD =$\frac{1}{2}\left((-4)\left(-6-5\right)+(-1)\left(5+5\right)+4\left(-5+6\right)\right]$
=$\frac{1}{2}(44-10+4)=19$ square units

Therefore area of the quadrilateral ABCD = 53+19 = 72 square units

Note: To find the area of the polygon, divide it into triangles regions, which have no common area and add the areas of these regions


Area of the given above $\Delta$ ABC = $\frac{1}{2}\left(2\left(0+4\right)+(-1)\left(-4-3\right)+2\left(3-0\right)\right]$
= $\frac{1}{2}(8+7+6)$
= 10.5 square units