The height of a triangle is increased by 10%. To retain the original area of the triangle, its corresponding base must be decreased by [SSC (CGL) Pre 2015] |
A) \[9\frac{1}{11}\]%
B) 10%
C) \[9\frac{1}{8}\]%
D) \[9\frac{1}{7}\]%
Correct Answer: A
Solution :
Let base be 10 cm and height be 10 cm. |
\[\therefore \] Area of triangle \[=\frac{1}{2}\times b\times h\] |
\[=\frac{1}{2}\times 10\times 10=50\,c{{m}^{2}}\] |
If height is increased by 10%, then new height = 11cm |
and \[\frac{1}{2}\times b\times 11=50\] |
\[\Rightarrow \] \[b=\frac{100}{11}cm\] |
\[\therefore \] Reduction in base |
\[=\frac{(10-\frac{100}{11})}{10}\times 100=\frac{10}{\frac{11}{10}}\times 100\] |
\[=\frac{1}{11}\times 100=9\frac{1}{11}\]% |
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