Q:

# What is the area of a triangle whose vertices are J(−2, 1) , K(4, 3) , and L(−2, −5) ?Enter your answer in the box.

Accepted Solution

A:
we know thatThe Heron's Formula is a method for calculating the area of a triangle when you know the lengths of all three sides. The formula is equal to$$Area=\sqrt{p(p-a)(p-b)(p-c)}$$where a,b,c -----> are the lengths of the sides of a trianglep ----> is half the perimeterwe have$$J(-2,1)\ K(4,3)\ L(-2,-5)$$The formula to calculate the distance between two points is equal to$$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$$Step 1Find the distance JK$$J(-2,1)\ K(4,3)$$Substitute the values in the formula of distance$$d=\sqrt{(3-1)^{2}+(4+2)^{2}}$$$$d=\sqrt{(2)^{2}+(6)^{2}}$$$$dJK=\sqrt{40}\ units$$Step 2Find the distance KL$$K(4,3)\ L(-2,-5)$$Substitute the values in the formula of distance$$d=\sqrt{(-5-3)^{2}+(-2-4)^{2}}$$$$d=\sqrt{(-8)^{2}+(-6)^{2}}$$$$dKL=10\ units$$Step 3Find the distance JL$$J(-2,1)\ L(-2,-5)$$Substitute the values in the formula of distance$$d=\sqrt{(-5-1)^{2}+(-2+2)^{2}}$$$$d=\sqrt{(-6)^{2}+(0)^{2}}$$$$dJL=6\ units$$Step 4Find the perimeter of the trianglewe know thatthe perimeter of a triangle is the sum of the length sides of the triangleso$$P=dJK+dKL+dJL$$substitute the values$$P=\sqrt{40}\ units+10\ units+6\ units=22.32\ units$$Find half the perimeter$$p=22.32/2=11.16\ units$$Step 5Find the area of the triangleApplying the Heron's Formula$$Area=\sqrt{p(p-a)(p-b)(p-c)}$$we have$$p=11.16\ units$$$$a=dJK=\sqrt{40}\ units=6.32\ units$$$$b=dKL=10\ units$$$$c=dJL=6\ units$$substitute the values$$Area=\sqrt{11.16(11.16-6.32)(11.16-10)(11.16-6)}$$$$Area=\sqrt{11.16(4.84)(1.16)(5.16)}$$$$Area=17.98\ units^{2}$$$$Area=18\ units^{2}$$thereforethe answer isThe area of the triangle is $$18\ units^{2}$$