The ratio of the areas of two similar figures is the square of the ratio of their corresponding leng

Question

The ratio of the areas of two similar figures is the square of the ratio of their corresponding lengths, i.e. $\frac{A_1}{A_2} = \left( \frac{l_1}{l_2} \right)^2$ where $A_1$ and $l_1$ are the area and the length of the first figure respectively, and $A_2$ and $l_2$ are the area and the length of the second figure respectively.
If the triangles are not similar, then check if they have a common height.
For triangles with common height, the ratio of the areas of the two triangles is equal to the ratio of the lengths of the bases of the two triangles.

$\frac{A_1}{A_2} = \frac{\frac{1}{2} \times b_1 \times h}{\frac{1}{2} \times b_2 \times h} = \frac{b_1}{b_2}$

Triangles with a common height need not be similar.

where $b_1$ and $A_1$ is the base and area of $\Delta ABD$ respectively,

and $b_2$ and $A_2$ is the base and area of $\Delta BCD$ respectively.

The ratio of the volumes of two similar solids is the cube of the ratio of their corresponding lengths, i.e. $\frac{V_1}{V_2} = \left( \frac{l_1}{l_2} \right)^3$ where $V_1$ and $l_1$ are the volume and the length of the first solid respectively, and $V_2$ and $l_2$ are the volume and the length of the second solid respectively.
For solids of equal density, masses are proportional to volumes, i.e. $\frac{m_1}{m_2} = \frac{V_1}{V_2}$ and $\frac{m_1}{m_2} = \left( \frac{l_1}{l_2} \right)^3$ where $m_1$ is the mass of the first solid, and $m_2$ is the mass of the second solid.

Accepted Answer

$\dfrac{A_1}{A_2}=\left(\dfrac{l_1}{l_2}\right)^2$, $\dfrac{V_1}{V_2}=\left(\dfrac{l_1}{l_2}\right)^3$, and for equal density, $\dfrac{m_1}{m_2}=\dfrac{V_1}{V_2}$.

Area and Volume of Similar Figures and Solids

Area and Volume of Similar Figures and Solids

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