The areas of two similar triangles are $81 \mathrm{~cm}^{2}$ and $49 \mathrm{~cm}^{2}$ respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?
Given: The area of two similar triangles is 81cm2 and 49cm2 respectively.
To find:
(1) Ratio of their corresponding heights.
(2) Ratio of their corresponding medians.
(1) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
$\frac{\operatorname{ar}(\text { triangle } 1)}{\operatorname{ar}(\text { triangle } 2)}=\left(\frac{\text { altitudel }}{\text { altitude } 2}\right)^{2}$
$\frac{81}{49}=\left(\frac{\text { altitudel }}{\text { altitude2 }}\right)^{2}$
Taking square root on both sides, we get
$\frac{9}{7}=\frac{\text { altitude } 1}{\text { altitude } 2}$
altitudel :altitude $2=9: 7$
(2) We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.
$\frac{\operatorname{ar}(\text { trianglel })}{\operatorname{ar}(\text { triangle } 2)}=\left(\frac{\text { median1 }}{\text { median } 2}\right)^{2}$
$\frac{81}{49}=\left(\frac{\text { median } 1}{\text { median } 2}\right)^{2}$
Taking square root on both sides, we get
$\frac{9}{7}=\frac{\text { median } 1}{\text { median } 2}$
median $1:$ median $2=9: 7$