Chapter 11.7, Problem 14PSB

To determine

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

Answer to Problem 14PSB

Theratiois $5:8$ .

Explanation of Solution

Given information:

In triangle ABC,

SideAB = 13 ,

Side AC = 13 ,

Side BC = 10 .

In triangle PQR,

Side PR = 20,

Side PQ = 12,

Side QR = 16.

Formula 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: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

SideAD can be calculated by Pythagoras theorem.

In right angled triangle ADC , we get

  ${\left(AC\right)}^2={\left(AD\right)}^2+{\left(DC\right)}^2$

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

The ratio is as follows:

  $\begin{array}{l}\frac{A(△ABC)}{A(△PQR)}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PQ}\\ \frac{A(△ABC)}{A(△PQR)}=\frac{\frac{1}{2}\times 10\times 12}{\frac{1}{2}\times 16\times 12}\\ \frac{A(△ABC)}{A(△PQR)}=\frac{60}{96}\\ \frac{A(△ABC)}{A(△PQR)}=\frac{5}{8}\end{array}$