Question

The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.

Answer 1

For the first triangle, we have,
Base = 15 cm and altitude = 7 cm
Thus, area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Altitude}$

= $\frac{1}{2}\times 15\times 7=52.5{\mathrm{cm}}^2$
It is given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 cm²
One side of the second triangle = 10.5 cm
Therefore,
The other side of the second triangle = $\frac{2\times \mathrm{Area}}{\mathrm{One}\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{triangle}}$

= $\frac{2\times 52.5}{10.5}=10\mathrm{cm}$
Hence, the other side of the second triangle will be 10 cm.