Question

ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the area of ∆ANC and ∆ABC.

Answer 1

Given: In ΔABC,, , BC = 12cm and AC = 5cm.

TO FIND: Ratio of the triangles ∆ANC and ∆ABC.

In ∆ANC and ∆ABC,

$\angle ACN=\angle ACB$ (Common)

$\angle A=\angle ANC$ (90º each)

∴ ∆ANC $~$ ∆ABC (AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.



$$\begin{gathered} \therefore \frac{Ar\left(∆ANC\right)}{Ar\left(∆ABC\right)}={\left(\frac{AC}{BC}\right)}^2 \\ \Rightarrow \frac{Ar\left(∆ANC\right)}{Ar\left(∆ABC\right)}={\left(\frac{5\mathrm{cm}}{12\mathrm{cm}}\right)}^2 \end{gathered}$$

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