Chapter 11.8, Problem 4PSB

a.

To determine

To find: The area of a triangle with sides 2, 5 and 7.

Answer to Problem 4PSB

The area of a triangle with sides 2, 5 and 7 is 0.

Explanation of Solution

Given Information:

Sides of the triangle are:

  $a=2,b=5\text{and}c=7$

Formula used:

Using Heron’s formula,

Area of a triangle $\left(A\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Here, $s=\frac{a+b+c}{2}$ is the semi-perimeter of a triangle

  $a,b\text{and}c$ are the sides of a triangle

Calculation:

Let $a,b\text{and}c$ be the sides of the triangle.

  $\therefore a=2,b=5\text{and}c=7$

We know that, Area of an inscribed quadrilateral $\left(A\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

  $"s"$ is the semi-perimeter of the quadrilateral

Now, $s=\frac{a+b+c}{2}$

  $\begin{array}{l}s=\frac{2+5+7}{2}\\ s=\frac{14}{2}\\ s=7\end{array}$

So, $A=\sqrt{\left(7-2\right)\left(7-5\right)\left(7-7\right)}$

  $\begin{array}{l}A=\sqrt{5\times 2\times 0}\\ A=0\end{array}$

So, the area of a triangle is 0.

Hence, there is no triangle with the sides 2, 5 and 7.

b.

To determine

To calculate: The area of a triangle with sides 4, 6 and 12.

Answer to Problem 4PSB

There does not exist any triangle with sides 4, 6 and 12.

Explanation of Solution

Given Information:

Sides of the triangle are:

  $a=4,b=6\text{and}c=12$

Formula used:

Using Heron’s formula,

Area of a triangle $\left(A\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Here, $s=\frac{a+b+c}{2}$ is the semi-perimeter of a triangle

  $a,b\text{and}c$ are the sides of a triangle

Calculation:

Let $a,b\text{and}c$ be the sides of the triangle.

  $\therefore a=4,b=6\text{and}c=12$

We know that, Area of an inscribed quadrilateral $\left(A\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

  $"s"$ is the semi-perimeter of the quadrilateral

Now, $s=\frac{a+b+c}{2}$

  $\begin{array}{l}s=\frac{4+6+12}{2}\\ s=\frac{22}{2}\\ s=11\end{array}$

So, $A=\sqrt{\left(11-4\right)\left(11-6\right)\left(11-12\right)}$

  $A=\sqrt{7\times 5\times \left(-1\right)}$

This is not defined as it is a square root of a negative number..

Hence, there is no triangle with the sides 4, 6 and 12.

c.

To determine

To explain: the results obtained in parts a and b.

Answer to Problem 4PSB

No such triangles exist.

Explanation of Solution

Given Information:

1^st triangle has sides:

  $a_1=2,b_1=5\text{and}{c}_1=9$

2^nd triangle has sides:

  $a_2=4,b_2=6\text{and}{c}_2=12$

Formula used:

Sum of any two sides of a triangle is always greater than the third side

1^st triangle has sides:

  $a_1=2,b_1=5\text{and}{c}_1=9$

2^nd triangle has sides:

  $a_2=4,b_2=6\text{and}{c}_2=12$

In a triangle, sum of any two sides is always greater than the third side.

But $a_1+b_1=7<9$

  $\therefore a_1+b_1<c_1$

So, the sides $a_1,b_1\text{and}{c}_1$ does not satisfy the criteria for being a triangle.

Also, $a_2+b_2=10<12$

  $\therefore a_2+b_2<c_2$

So, the sides $a_2,b_2\text{and}{c}_2$ does not satisfy the criteria for being a triangle.

Hence, no such triangles exist.