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12. In an isosceles triangle ABC, if AB = AC =

    \[\mathbf{13cm}\]

and the altitude from A on BC is
CBSE | Class 10 | Maths | RD Sharma | Triangles
12. In an isosceles triangle ABC, if AB = AC =

    \[\mathbf{13cm}\]

and the altitude from A on BC is

    \[\mathbf{5cm},\]

find BC.

Solution:

Given,

An isosceles triangle ABC, AB = AC =

    \[13cm,\text{ }AD\text{ }=\text{ }5cm\]

Required to find: BC

In ∆ ADB, by using Pythagoras theorem, we have

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> A{{D}^{2}}~+\text{ }B{{D}^{2}}~=\text{ }{{13}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> {{5}^{2}}~+\text{ }B{{D}^{2}}~=\text{ }169 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> B{{D}^{2}}~=\text{ }169\text{ }\text{ }25\text{ }=\text{ }144 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \Rightarrow BD\text{ }=\text{ }\surd 144\text{ }=\text{ }12\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

                                                       

    \[\]

Similarly, applying Pythagoras theorem is ∆ ADC we can have,

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> A{{C}^{2}}~=\text{ }A{{D}^{2}}~+\text{ }D{{C}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> {{13}^{2}}~=\text{ }{{5}^{2}}~+\text{ }D{{C}^{2}} \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \Rightarrow DC\text{ }=\text{ }\surd 144\text{ }=\text{ }12\text{ }cm \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

                                                       

    \[\]

Thus, BC =

    \[~BD\text{ }+\text{ }DC\text{ }=\text{ }12\text{ }+\text{ }12\text{ }=\text{ }24\text{ }cm\]

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