Ratio of Corresponding Sides of Similar Triangles
For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Let the areas of the two similar triangles be $A_1$ and $A_2$, and let their corresponding sides be $s_1$ and $s_2$. The relationship is given by the formula:
$ \frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2 $
Given areas are $A_1 = 16\text{ m}^2$ and $A_2 = 36\text{ m}^2$. Substitute these values into the formula:
$ \frac{16\text{ m}^2}{36\text{ m}^2} = \left(\frac{s_1}{s_2}\right)^2 $
Simplify the ratio of the areas:
$ \frac{16}{36} = \frac{4 \times 4}{9 \times 4} = \frac{4}{9} $
So, we have:
$ \left(\frac{s_1}{s_2}\right)^2 = \frac{4}{9} $
To find the ratio of the corresponding sides ($s_1 : s_2$), take the square root of both sides:
$ \frac{s_1}{s_2} = \sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3} $
The ratio of the corresponding sides is $2:3$. Comparing this ratio to the given options, option C ($4:6$) simplifies to $2:3$.
Similar Questions
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Important Questions from Triangles
What is the circumcenter of the triangle ABC?
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
