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# coordinates

0
40
1

FInd the area of the triangle with vertices (1,6), (1,11), and (7,43).

Aug 3, 2022

#1
+113
+2

This is probably pre vector algebra or something.

Let's call these vertices A(1,6), B(1,11) and C(7,43).

We can find the length of each of the sides using the pythagorean theorem.

AB = (0,5), length = 5

AC = (6,37), length = $$\sqrt{6^2 + 37^2} = \sqrt{1,405} \approx 37.48$$

BC = (6,32), length = $$\sqrt{6^2 + 32^2} = \sqrt{1,405} \approx 32.56$$

Use herons formula, see https://www.mathsisfun.com/geometry/herons-formula.html

s = 1/2 ( a + b + c ) = 1/2 ( 5 + 37.48 + 32.56 ) = 1/2 (75.04) = 37.52

The area is A = $$\sqrt{s\times(s-a)\times(s-b)\times(s-c)} = \sqrt{37.52\times32.52\times 0.04\times 4.96} = \sqrt{242.08} \approx15.56$$

Aug 3, 2022

#1
+113
+2

This is probably pre vector algebra or something.

Let's call these vertices A(1,6), B(1,11) and C(7,43).

We can find the length of each of the sides using the pythagorean theorem.

AB = (0,5), length = 5

AC = (6,37), length = $$\sqrt{6^2 + 37^2} = \sqrt{1,405} \approx 37.48$$

BC = (6,32), length = $$\sqrt{6^2 + 32^2} = \sqrt{1,405} \approx 32.56$$

Use herons formula, see https://www.mathsisfun.com/geometry/herons-formula.html

s = 1/2 ( a + b + c ) = 1/2 ( 5 + 37.48 + 32.56 ) = 1/2 (75.04) = 37.52

The area is A = $$\sqrt{s\times(s-a)\times(s-b)\times(s-c)} = \sqrt{37.52\times32.52\times 0.04\times 4.96} = \sqrt{242.08} \approx15.56$$

tuffla2022 Aug 3, 2022