Question

Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length of one of its diagonals is 26 cm, find the length of the other diagonal.

Answer 1

First, we draw a figure of the parallelogram ABCD in which AB = 17 cm and BC = 11 cm, are the adjacent sides and AC and BD are the diagonals that bisect each other at O.


Thus, AO = OC = $\frac{AC}{2}$ and BO = OD = $\frac{\mathrm{BD}}{2}$ …(1)
In Δ ABC, BO is a median of the triangle. Thus, on using Appollonius' Theorem, we get:

$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{BC}}^2=2{\mathrm{BO}}^2+2{\mathrm{AO}}^2 \\ {\mathrm{AB}}^2+{\mathrm{BC}}^2=2{\left(\frac{\mathrm{BD}}{2}\right)}^2+2{\left(\frac{\mathrm{AC}}{2}\right)}^2 \\ {\mathrm{AB}}^2+{\mathrm{BC}}^2=2\times \frac{{\mathrm{BD}}^2}{4}+2\times \frac{{\mathrm{AC}}^2}{4} \\ {\mathrm{AB}}^2+{\mathrm{BC}}^2=\frac{{\mathrm{BD}}^2}{2}+\frac{{\mathrm{AC}}^2}{2} \\ \mathrm{AB}=17\mathrm{cm},\mathrm{BC}=11\mathrm{cm}\text{and}\mathrm{AC}=26\mathrm{cm} \\ \text{So},(17{)}^2+(11{)}^2=\frac{{\mathrm{BD}}^2}{2}+\frac{{\left(26\right)}^2}{2} \\ \Rightarrow 289+121=\frac{{\mathrm{BD}}^2}{2}+\frac{676}{2} \\ \Rightarrow 410=\frac{{\mathrm{BD}}^2}{2}+338 \\ \Rightarrow \frac{{\mathrm{BD}}^2}{2}=410-338 \\ \Rightarrow {\mathrm{BD}}^2=2\times 72 \\ \Rightarrow {\mathrm{BD}}^2=144 \\ \Rightarrow \mathrm{BD}=\sqrt{144}=12\mathrm{cm} \end{gathered}$$