Question
ABC is a triangle in which $\angle\text{A}=90^\circ,\ \text{AN}\perp\text{BC}$ BC = 12cm and AC = 5cm. Find the ratio of the area of $\triangle\text{ANC}$ and $\triangle\text{ABC}.$
✓
Answer
Given: In $\triangle\text{ABC},\angle\text{A}=90^\circ,\text{AN}\perp\text{BC},$ BC = 12cm and AC = 5cm.
To find: Retio of the triangle $\triangle\text{ANC}$ and $\triangle\text{ABC}.$
In $\triangle\text{ANC}$ and $\triangle\text{ABC}$
$\angle\text{ACN}=\angle\text{ACB}$ (Common)
$\angle\text{A}=\angle\text{ANC}$ (90° each)
$\therefore\triangle\text{ANC}\sim\triangle\text{ABC}$ (AA Similarity)
We know that the ratio of areas of two similar triangle isequal to the ratio of squares of their corresponding sides.
$\therefore\frac{\text{Ar}(\triangle\text{ANC})}{\text{Ar}(\triangle\text{ABC})}=\Big(\frac{\text{AC}}{\text{BC}}\Big)^2$
$\Rightarrow\frac{\text{Ar}(\triangle\text{ANC})}{\text{Ar}(\triangle\text{ABC})}=\Big(\frac{5\text{cm}}{12\text{cm}}\Big)^2$
$\Rightarrow\frac{\text{Ar}(\triangle\text{ANC})}{\text{Ar}(\triangle\text{ABC})}=\frac{25}{144}$
To find: Retio of the triangle $\triangle\text{ANC}$ and $\triangle\text{ABC}.$
In $\triangle\text{ANC}$ and $\triangle\text{ABC}$
$\angle\text{ACN}=\angle\text{ACB}$ (Common)
$\angle\text{A}=\angle\text{ANC}$ (90° each)
$\therefore\triangle\text{ANC}\sim\triangle\text{ABC}$ (AA Similarity)
We know that the ratio of areas of two similar triangle isequal to the ratio of squares of their corresponding sides.
$\therefore\frac{\text{Ar}(\triangle\text{ANC})}{\text{Ar}(\triangle\text{ABC})}=\Big(\frac{\text{AC}}{\text{BC}}\Big)^2$
$\Rightarrow\frac{\text{Ar}(\triangle\text{ANC})}{\text{Ar}(\triangle\text{ABC})}=\Big(\frac{5\text{cm}}{12\text{cm}}\Big)^2$
$\Rightarrow\frac{\text{Ar}(\triangle\text{ANC})}{\text{Ar}(\triangle\text{ABC})}=\frac{25}{144}$
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