Chapter 11.7, Problem 11PSB

To determine

To calculate:Theratio of area of triangle ABCandarea of triangle LMN .

Answer to Problem 11PSB

Theratiois $5:4$ .

Explanation of Solution

Given information:

SideAB = 5 ,

Side AC = 13 ,

Side LN = 10 ,

Side MN = 8 .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of triangle: $A=\frac{1}{2}\times b\times h$

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Calculation:

  

Sides BC and LM can be calculated by Pythagoras theorem.

In right angled triangle ABC , we get

  ${\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2$

  $\begin{array}{l}{\left(BC\right)}^2={\left(AC\right)}^2-{\left(AB\right)}^2\\ {\left(BC\right)}^2={\left(13\right)}^2-{\left(5\right)}^2\\ {\left(BC\right)}^2=169-25\\ {\left(BC\right)}^2=144\\ BC=\sqrt{144}=12\end{array}$

In right angled triangle LMN , we get

  ${\left(LN\right)}^2={\left(LM\right)}^2+{\left(MN\right)}^2$

  $\begin{array}{l}{\left(LM\right)}^2={\left(LN\right)}^2-{\left(MN\right)}^2\\ {\left(LM\right)}^2={\left(10\right)}^2-{\left(8\right)}^2\\ {\left(LM\right)}^2=100-64\\ {\left(LM\right)}^2=36\\ LM=\sqrt{36}=6\end{array}$

  $\begin{array}{l}\frac{A(△ABC)}{A(△LMN)}=\frac{\frac{1}{2}\times BC\times AB}{\frac{1}{2}\times MN\times LM}\\ \frac{A(△ABC)}{A(△LMN)}=\frac{\frac{1}{2}\times 12\times 5}{\frac{1}{2}\times 6\times 8}\\ \frac{A(△ABC)}{A(△LMN)}=\frac{30}{24}\\ \frac{A(△ABC)}{A(△LMN)}=\frac{5}{4}\end{array}$