Question

In $△ABC$, $a+b=18$ units, $b+c=25$ units and $c+a=17$ units. What type of triangle is $ABC$? Give reason.

Answer 1

In $△ABC$, the followings are given:

$a+b=18........\left(1\right)$

$b+c=25........\left(2\right)$

$c+a=17........\left(3\right)$

Subtracting equation $\left(1\right)$ from equation $\left(2\right)$ :

$(b+c)-(a+b)=25-18$

$\Rightarrow (c-a)+(b-b)=25-18$

$\Rightarrow$$c-a=7.......\left(4\right)$

Now, adding equation $\left(3\right)$ and $\left(4\right)$:

$(c+a)+(c-a)=17+7$

$\Rightarrow$$c+a+c-a=24$

$\Rightarrow$$2c=24$

$\Rightarrow$$c=12$

Substituting $c=12$ in equation $\left(3\right)$:

$12+a=17$

$\Rightarrow$$a=17-12=5$

Now substitute $a=5$ in equation $\left(1\right)$:

$5+b=18$

$\Rightarrow$$b=18-5=13$

Finally consider $A{B}^2+B{C}^2$ as shown below:

$A{B}^2+B{C}^2$

$=c^2+a^2$

$={12}^2+5^2$

$=144+25$

$=169$

$=(13{)}^2$

$=b^2$

$=A{C}^2$

Since $A{B}^2+B{C}^2=A{C}^2$.

Hence, $△ABC$ is a right angled triangle.