Ratio of corresponding sides of similar triangles, given the areas. The area of two similar triangles are 72 and 162. what is the ratio of their corresponding sides?
 A: When linear dimensions a scaled by the factor $\lambda$, area is scaled by the factor $\lambda^2$. Here, we have $\lambda^2=\dfrac{162}{72}=\dfrac{9}{4}$. So $\lambda=\dfrac{3}{2}$. 
A: Hint:  the units of area are length$^2$
A: 
if $\Delta ABC $ and $\Delta MNO$ are similar triangle and AD and MP are perpendicular on BC and ON then
$$\dfrac {AB^2}{MN^2}=\dfrac {AC^2}{MO^2}=\dfrac {BC^2}{NO^2}=\dfrac {AD^2}{MP^2} =\dfrac {area \Delta ABC}{area\Delta MNO}$$ (This is a theorem)
so in question : $$\dfrac {area \Delta ABC}{area\Delta MNO}=\dfrac{72}{162}$$
so ratio of corresponding sides are $$ \dfrac {AB^2}{MN^2}=\dfrac {72}{162}$$
$$ \dfrac {AB}{MN}=\sqrt {\dfrac {72}{162}}$$
$$\dfrac {AB}{MN}=\dfrac23$$
so ratio will be $$2:3$$
A: If the side lengths of the first triangle are $a,b,c$, then the side lengths of the second triangle are $ka,kb,kc$ for some $k>0$. An area of the first triangle is $72=\frac{1}{2}ab\sin \alpha$ and an area of the second is $162=\frac{1}{2}kakb\sin\alpha$ so $\frac{72}{162}$ is a square of the ratio of their corresponding sides.
A: Since the triangles are similar and no other constraint is available, you can take them to be equilateral with sides $ a $ and $ b $ units.
$$ A_1 = 72x = \frac{\sqrt{3}}{4} a^2 $$
$$ A_2 = 162x = \frac{\sqrt{3}}{4} b^2 $$
Therefore,
$$ \left( \frac{a}{b} \right)^2 = \frac{72}{162} = \frac{4}{9}$$
Thus, sides are in the ratio $ 2 : 3 $
