The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is
A$30 \sqrt{7} cm$
B$\frac{15 \sqrt{7}}{2} cm$
C$\frac{15 \sqrt{7}}{4} cm$
D
30 cm
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C$\frac{15 \sqrt{7}}{4} cm$
C
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In the given figure, the ratio $AD$ to $DC$ is $3$ to $2$. If the area of $\triangle\text{ABC}$ is $40 \mathrm{~cm}^2$, what is the area of $\triangle\text{BDC}?$
The area of a right triangle is $28 cm^2$. If one of its perpendicular sides exceeds the other by 10 cm . then the length of the longest of the perpendicular is