In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm. (i) Write all possible pairs of similar triangles. (ii) Find the lengths of ME and DM.
In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm. (i) Write all possible pairs of similar triangles. (ii) Find the lengths of ME and DM.

Selina Solutions Concise Class 10 Maths Chapter 15 ex. 15(A) - 9

Answer:

(I) In\[\Delta \text{ }AME\text{ }and\text{ }\Delta \text{ }ANC\],

\[\angle AME\text{ }=\angle ANC\] \[\left[ Since\text{ }DE\text{ }\left| \left| \text{ }BC\text{ }in\text{ }this\text{ }way,\text{ }ME\text{ } \right| \right|\text{ }NC \right]\]

\[\angle MAE\text{ }=\angle NAC\text{ }\left[ Common\text{ }angle \right]\]

Thus, \[\vartriangle AME\text{ }\sim\text{ }\vartriangle ANC\]by \[AA\]rule for similitude

In\[\Delta \text{ }ADM\text{ }and\text{ }\Delta \text{ }ABN\],

\[\angle ADM\text{ }=\angle ABN\text{ }\left[ Since\text{ }DE\text{ }\left| \left| \text{ }BC\text{ }in\text{ }this\text{ }way,\text{ }DM\text{ } \right| \right|\text{ }BN \right]\]

\[\angle DAM\text{ }=\angle BAN\text{ }\left[ Common\text{ }angle \right]\]

Thus, \[\vartriangle ADM\text{ }\sim\text{ }\vartriangle ABN\]by \[AA\]rule for similitude

In\[\Delta \text{ }ADE\text{ }and\text{ }\Delta \text{ }ABC\],

\[\angle ADE\text{ }=\angle ABC\text{ }\left[ Since\text{ }DE\text{ }\left| \left| \text{ }BC\text{ }in\text{ }this\text{ }way,\text{ }ME\text{ } \right| \right|\text{ }NC \right]\]

\[\angle AED\text{ }=\angle ACB\text{ }\left[ Since\text{ }DE\text{ }||\text{ }BC \right]\]

Thus, \[\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\]by \[AA\]rule for similitude

(ii) Proved over that, \[\vartriangle AME\text{ }\sim\text{ }\vartriangle ANC\]

So as relating sides of comparative triangles are corresponding, we have

\[ME/NC\text{ }=\text{ }AE/AC\]

\[ME/6\text{ }=\text{ }15/24\]

\[ME\text{ }=\text{ }3.75\text{ }cm\]

Furthermore, \[\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\][Proved above]

So as relating sides of comparative triangles are corresponding, we have

\[AD/AB\text{ }=\text{ }AE/AC\text{ }=\text{ }15/24\text{ }\ldots .\text{ }\left( 1 \right)\]

Likewise, \[\vartriangle ADM\text{ }\sim\text{ }\vartriangle ABN\][Proved above]

So as relating sides of comparative triangles are corresponding, we have

\[DM/BN\text{ }=\text{ }AD/AB\text{ }=\text{ }15/24\text{ }\ldots \text{ }.\text{ }From\text{ }\left( 1 \right)\]

\[DM/24\text{ }=\text{ }15/24\]

\[DM\text{ }=\text{ }15\text{ }cm\]