Chapter 9, Problem 32RP

To determine

To calculate:The unknown segment of the given triangle i.e. AB .

Answer to Problem 32RP

The measure of required segment = $2\sqrt{29}$ units

Explanation of Solution

Given information:

Angle C in triangle ABC is right angle

E and F are midpoints of AC and BC

AF= $\sqrt{41}$ , BE= $2\sqrt{26}$

Formula used:

Midpoint Theorem

Pythagoras Theorem

Calculation :

On using Pythagoras theorem, we have

In triangle AFC, we have

  ${\text{AF}}^2{\text{=AC}}^2{\text{ + CF}}^2\mathrm{.......}\left(i\right)$

In triangle BEC., we have

  ${\text{BE}}^2{\text{=EC}}^2{\text{ + BC}}^2\mathrm{.......}\left(ii\right)$

On adding (i) and (ii), we get

  $\begin{array}{l}{\text{AF}}^2{\text{+BE}}^2{\text{=EC}}^2{\text{+ BC}}^2{\text{+AC}}^2{\text{ + CF}}^2\\ \Rightarrow {\text{AF}}^2{\text{+BE}}^2\text{=}\left({\text{EC}}^2{\text{+ CF}}^2\right)+\left({\text{BC}}^2{\text{+AC}}^2\right)\\ \Rightarrow {\text{AF}}^2{\text{+BE}}^2={\text{EF}}^2{\text{+AB}}^2\mathrm{........}\left(iii\right)\end{array}$

Because, in triangle ECF, we have

  ${\text{EC}}^2{\text{+ CF}}^2={\text{EF}}^2$

And in triangle ACB, we have

  ${\text{BC}}^2{\text{+AC}}^2={\text{AB}}^2$

Now, given E and F are the midpoints of AC and BC

On using midpoint theorem, we have

  $\text{EF=}\frac{1}{2}\text{AB}$

From eq 9iii), we have

  $\begin{array}{l}{\text{AF}}^2{\text{+BE}}^2={\text{EF}}^2{\text{+AB}}^2\\ \Rightarrow 41+104={\left(\frac{1}{2}\text{AB}\right)}^2+{\text{AB}}^2=\frac{1}{4}{\text{AB}}^2+{\text{AB}}^2=\frac{5}{4}{\text{AB}}^2\\ \Rightarrow \frac{5}{4}{\text{AB}}^2=145\\ \Rightarrow {\text{AB}}^2=\frac{145\times 4}{5}=4\times 29\\ \Rightarrow \text{AB}=\text{2}\sqrt{29}\text{ units}\end{array}$