Answer:
Area = 18 square units
Step-by-step explanation:
To find the area of the triangle, let's go through the following steps:
(i) Let the vertices be;
A = (7, -3)
B = (4, -3)
C = (4, 9)
(ii) The sides of the triangle are therefore,
AB, BC and CA
(iii) Using the distance formula, calculate the lengths of AB, BC and CA
[tex]AB = \sqrt{(7-4)^2 + ( -3 - (-3))^2}[/tex]
[tex]AB = \sqrt{3^2 + (0)^2}\\[/tex]
[tex]AB = \sqrt{9}[/tex]
[tex]AB = 3[/tex]
[tex]BC = \sqrt{(4-4)^2 + ( -3 - 9)^2}[/tex]
[tex]BC = \sqrt{0^2 + (-12)^2}[/tex]
[tex]BC = \sqrt{144}[/tex]
[tex]BC = 12[/tex]
[tex]CA = \sqrt{(4-7)^2 + ( 9 - (-3))^2}[/tex]
[tex]CA = \sqrt{(-3)^2 + (12)^2}[/tex]
[tex]CA = \sqrt{9 + 144}[/tex]
[tex]CA = \sqrt{153}[/tex]
[tex]CA = 12.4[/tex]
(iv) Now that we have all the sides, let's calculate the area of the triangle using the Heron's formula.
Area = [tex]\sqrt{p(p-a)(p-b)(p-c)}[/tex]
Where;
p = [tex]\frac{a + b + c}{2}[/tex]
a, b and c are the sides of the triangle.
In our case,
let
a = AB = 3
b = BC = 12
c = CA = 12.4
∴ p = [tex]\frac{3 + 12 + 12.4}{2}[/tex]
p = 13.7
Area = [tex]\sqrt{p(p-a)(p-b)(p-c)}[/tex]
Area = [tex]\sqrt{13.7(13.7-3)(13.7-12)(13.7-12.4)}[/tex]
Area = [tex]\sqrt{13.7(10.7)(1.7)(1.3)}[/tex]
Area = [tex]\sqrt{323.9639}[/tex]
Area = 17.999
Area = 18 square units
OR
To get the area of the triangle, we can use a much simpler approach.
Since the triangle is a right triangle,
(i) the hypotenuse, which is the longest side is CA = 12.4
(ii) the other two sides are AB and BC. These two sides will form the right angle.
Therefore, we can use the relation:
Area = [tex]\frac{1}{2}[/tex] x base x height
Where;
the base or height can either be AB or BC
Area = [tex]\frac{1}{2}[/tex] x 3 x 12
Area = 18 square units
PS: In a right triangle, the other two sides apart from the hypotenuse form the right angle.
