Question

question 42, 2.1.35
part 4 of 5
plot each point and form the right triangle abc. verify that the triangle is a right triangle. find its area.
a = (-5,3); b = (4,0); c = (1,6)
to show that the triangle is a right triangle, show that the sum of the squares of the lengths of two of the sides (the legs) equals the square of the length of the third side (the hypotenuse).
find the length of each line segment.
d(a,b)=3\sqrt{10}
d(a,c)=3\sqrt{5}
d(b,c)=3\sqrt{5}
(simplify your answers. type exact answers, using radicals as needed.)
find the squared distance of each segment.
d(a,b)^2 = 90
d(a,c)^2 = 45
d(b,c)^2 = 45
(simplify your answers. type exact answers, using radicals as needed.)
the sum of the squares of the lengths of the legs of the triangle is
the square of the length of the hypotenuse of the triangle is
(simplify your answers.)

Explanation:

Step1: Identify legs and hypotenuse

In a right - triangle, the hypotenuse is the longest side. Here, since $d(A,B)=3\sqrt{10}$, $d(A,C) = 3\sqrt{5}$, and $d(B,C)=3\sqrt{5}$, the legs are $d(A,C)$ and $d(B,C)$ and the hypotenuse is $d(A,B)$.

Step2: Calculate sum of squares of legs

$[d(A,C)]^{2}+[d(B,C)]^{2}=45 + 45=90$.

Step3: Calculate square of hypotenuse

$[d(A,B)]^{2}=90$.

Step4: Calculate area of right - triangle

The area of a right - triangle is $A=\frac{1}{2}\times\text{base}\times\text{height}$. Here, base and height are the lengths of the legs. So $A=\frac{1}{2}\times3\sqrt{5}\times3\sqrt{5}=\frac{1}{2}\times45 = 22.5$.

Answer:

The sum of the squares of the lengths of the legs of the triangle is $90$. The square of the length of the hypotenuse of the triangle is $90$. The area of the right - triangle is $22.5$.