The corresponding altitudes of two similar triangles are 6cm and 9cm respectively, Find the ratio of their areas.
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Let the two triangle be ABC and DEF with altitudes AP and DQ, respectively.
It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}.$
We know that the ratio of areas of two similar triangle is equia to the ratio of squares of their corresponding altitudes.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{(\text{AP})^2}{(\text{DQ})^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{6}^2}{\text{9}^2}$
$=\frac{36}{81}$
$=\frac{4}{9}$
Hence, the ratio of their areas is 4 : 9
It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}.$
We know that the ratio of areas of two similar triangle is equia to the ratio of squares of their corresponding altitudes.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{(\text{AP})^2}{(\text{DQ})^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{6}^2}{\text{9}^2}$
$=\frac{36}{81}$
$=\frac{4}{9}$
Hence, the ratio of their areas is 4 : 9
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