| In the figure given below, \[\angle ABC=90{}^\circ ,\] \[AD=15cm\]and \[DC=20cm\]. If BD is the bisector of \[\angle ABC,\] What is the perimeter of the triangle ABC? |
 |
A) \[74\,cm\]
B) \[84\,cm\]
C) \[91\,cm\]
D) \[105\,cm\]
Correct Answer: B
Solution :
| [b] Since BD is the angle bisector of\[\angle B\], therefore by angle bisector theorem, we get \[\Delta ABD\tilde{\ }\Delta CBD\] |
| \[\frac{AB}{BC}=\frac{AD}{DC}=\frac{15}{20}\Rightarrow \frac{AB}{BC}=\frac{3}{4}\] (1)Now, by pythagoras theorem, we get, |
| \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] |
| \[{{(AD+DC)}^{2}}={{\left( \frac{3}{4}BC \right)}^{2}}+B{{C}^{2}}\] |
| \[{{(35)}^{2}}=\frac{25}{16}B{{C}^{2}}\Rightarrow 1225\times \frac{16}{25}=B{{C}^{2}}\] |
| \[B{{C}^{2}}=49\times 16\Rightarrow BC=7\times 4=28\,cm\] From eq. (1), We get, |
| \[AB=\frac{3}{4}\times BC=\frac{3}{4}\times 28=21\,cm\] |
| Thus, the perimeter of |
| \[\Delta ABC=(28+21+35)\,cm=84\,cm\] |
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