Chapter 9.1, Problem 27WE

a.

To determine

To find: The type of triangle and its area.

Answer to Problem 27WE

The given triangle is an isosceles triangle.

Explanation of Solution

Given information: The points are $A\left(5,-4\right),B\left(-3,4\right)\text{and C}\left(-2,-3\right)$ .

Formula used: $D=\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$

Calculation: Since a triangle is an isosceles, only if two of its sides are equal to each other. And if square of one side of the triangle is equal to the sum of the length of other two sides then, the triangle said to be right angle triangle.

  $\begin{array}{l}\Rightarrow AB=\sqrt{{\left(5+3\right)}^2+{\left(-4-4\right)}^2}\\ =\sqrt{64+64}\\ =8\sqrt{2}\\ \Rightarrow BC=\sqrt{{\left(-3+2\right)}^2+{\left(4+3\right)}^2}\\ =\sqrt{1+49}\\ =5\sqrt{2}\end{array}$

  $\begin{array}{l}\Rightarrow AC=\sqrt{{\left(5+2\right)}^2+{\left(-4+3\right)}^2}\\ =\sqrt{49+1}\\ =5\sqrt{2}\end{array}$

  $\begin{array}{l}\therefore BC=AC\\ \therefore 5\sqrt{2}=5\sqrt{2}\end{array}$

Hence, the given triangle is an isosceles triangle.

b.

To determine

To find: The type of triangle and its area.

Answer to Problem 27WE

The given triangle is not a right angle triangle.

Explanation of Solution

Given information: The points are $A\left(5,-4\right),B\left(-3,4\right)\text{and C}\left(-2,-3\right)$ .

Formula used: $D=\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$

Calculation: Since a triangle is an isosceles, only if two of its sides are equal to each other. And if square of one side of the triangle is equal to the sum of the length of other two sides then, the triangle said to be right angle triangle.

  $\begin{array}{l}\Rightarrow AB=\sqrt{{\left(5+3\right)}^2+{\left(-4-4\right)}^2}\\ =\sqrt{64+64}\\ =8\sqrt{2}\\ \Rightarrow BC=\sqrt{{\left(-3+2\right)}^2+{\left(4+3\right)}^2}\\ =\sqrt{1+49}\\ =5\sqrt{2}\end{array}$

  $\begin{array}{l}\Rightarrow AC=\sqrt{{\left(5+2\right)}^2+{\left(-4+3\right)}^2}\\ =\sqrt{49+1}\\ =5\sqrt{2}\end{array}$

  ${\left(5\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2\ne {\left(8\sqrt{2}\right)}^2$

Hence, the given triangle is not a right angle triangle.