Question

In the given figure, $\frac{DE}{BC}$, AE = 15 cm, EC = 9 cm, NC = 6 cm, and BN = 24 cm. Find the value of ME and DM.

Answer 1

In the given figure,

$DE\parallel BC$

AE = 15 cm, EC = 9 cm

NC = 6 cm and BN = 24 cm

In $\Delta ABC$

$DE\parallel BC$

$\therefore$ Pairs of similar triangles are

(a) $\Delta ADE\sim \Delta ABC$

(b) $\Delta ADM\sim \Delta ABN$

(c) $\Delta AME\sim \Delta ANC$

In $\Delta AME\sim \Delta ANC$ (corresponding sides are proportional)

$\Rightarrow \frac{15}{15+9}=\frac{ME}{6}=\frac{DM}{24}$

$\Rightarrow ME=\frac{15\times 6}{24}=\frac{15}{4}=3.75\text{cm}$

Similarly

In $\Delta ADE\sim \Delta ABC$ (corresponding sides are proportional)

$\Rightarrow \frac{15}{15+9}=\frac{DM+ME}{24+6}$

$\Rightarrow \frac{15}{24}=\frac{DM+3.75}{30}$

$DM+3.75=18.75$

$DM=15cm$