For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?
To solve for the area of triangle \(RAQ\), we need to use the property of similar triangles. Here are the steps to find the solution:
- Since the triangles \(BAC\) and \(RAQ\) are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. This can be expressed as: \(\text{Area of } BAC : \text{Area of } RAQ = \left(\frac{\text{Side of } BAC}{\text{Side of } RAQ}\right)^2\)
- Given that the ratio of the areas of the triangles is 2:1, we have: \(\frac{\text{Area of } BAC}{\text{Area of } RAQ} = 2:1\)
- The area of triangle \(BAC\) is given as \(100 \ \text{cm}^2\). Let the area of triangle \(RAQ\) be \(x \ \text{cm}^2\). \(\frac{100}{x} = 2\)
- Solve for \(x\): \(x = \frac{100}{2} = 50\)
Therefore, the area of triangle \(RAQ\) is 50 cm2.
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