Question
The base and altitude of a right-angled triangle are 12 cm and 5 cm respectively. The perpendicular distance of its hypotenuse from the opposite vertex is :
Answer: Option B
Area of the triangle :
$$\eqalign{
& = \left( {\frac{1}{2} \times 12 \times 5} \right)c{m^2} \cr
& = 30\,c{m^2} \cr} $$
Hypotenuse :
$$\eqalign{
& = \sqrt {{{12}^2} + {5^2}} \,cm \cr
& = \sqrt {169} \,cm \cr
& = 13\,cm \cr} $$
Let the perpendicular distance of the hypotenuse from the opposite vertex be x cm
Then,
$$\eqalign{
& \Rightarrow \frac{1}{2} \times 13 \times x = 30 \cr
& \Rightarrow x = \frac{{60}}{{13}} \cr
& \Rightarrow x = 4\frac{8}{{13}}\,cm \cr} $$
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Area of the triangle :
$$\eqalign{
& = \left( {\frac{1}{2} \times 12 \times 5} \right)c{m^2} \cr
& = 30\,c{m^2} \cr} $$
Hypotenuse :
$$\eqalign{
& = \sqrt {{{12}^2} + {5^2}} \,cm \cr
& = \sqrt {169} \,cm \cr
& = 13\,cm \cr} $$
Let the perpendicular distance of the hypotenuse from the opposite vertex be x cm
Then,
$$\eqalign{
& \Rightarrow \frac{1}{2} \times 13 \times x = 30 \cr
& \Rightarrow x = \frac{{60}}{{13}} \cr
& \Rightarrow x = 4\frac{8}{{13}}\,cm \cr} $$
Was this answer helpful ?
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