| The area of an isosceles \[\Delta ABC\] with AB = AC and altitude AD = 3 cm is 12 sq cm. What is its perimeter? |
A) 18 cm
B) 16 cm
C) 14 cm
D) 12 cm
Correct Answer: A
Solution :
| Let \[AB=CA=a\,\,cm\]and base = b cm |
| Now, area of the |
| \[\Delta ABC=\frac{1}{2}\times b\times h\] |
| \[\Rightarrow \] \[12=\frac{1}{2}\times b\times 3\] |
| \[\therefore \] \[b=\frac{12\times 2}{3}=8\,\,cm\] |
| Here, \[BD=CD=\frac{b}{2}\] |
| \[=\frac{8}{2}=4\,\,cm\] |
| In right angled \[\Delta ABD,\] by Pythagoras theorem, |
| \[AB=\sqrt{B{{D}^{2}}+A{{D}^{2}}}\] |
| \[\Rightarrow \] \[a=\sqrt{{{4}^{2}}+{{3}^{2}}}\] |
| \[\Rightarrow \]\[\sqrt{16+9}=\sqrt{25}=5\,\,cm\] |
| Now, perimeter of an isosceles triangle |
| \[=2a+b=2\times 5+8\] |
| \[=10+8=18\,\,cm\] |
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