Sides of two similar triangles are in the ratio 4 ∶ 9. Area of these triangles are in the ratio:
16 ∶ 81
When two triangles are similar, their corresponding sides are in proportion, and their corresponding angles are equal. There is a specific relationship between the ratio of their sides and the ratio of their areas.
The theorem relating the sides and areas of similar triangles states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Let $\triangle ABC$ be similar to $\triangle PQR$. If the ratio of their corresponding sides is $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = k$, then the ratio of their areas is:
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = k^2 = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$.
In this problem, the sides of two similar triangles are in the ratio 4 ∶ 9.
Let the ratio of the sides be $s_1 : s_2 = 4 : 9$. This can be written as $\frac{s_1}{s_2} = \frac{4}{9}$.
According to the theorem, the ratio of their areas ($A_1 : A_2$) is the square of the ratio of their sides:
$\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2$
Substitute the given side ratio:
$\frac{A_1}{A_2} = \left(\frac{4}{9}\right)^2$
Calculate the square:
$\frac{A_1}{A_2} = \frac{4^2}{9^2} = \frac{16}{81}$
Thus, the ratio of the areas of the two similar triangles is 16 ∶ 81.
Therefore, the area of these triangles are in the ratio 16 ∶ 81.
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