Question

Triangles ABC and PQR are similar to each other, then ______.

Options

• \[\frac{\operatorname{Ar} \cdot(\triangle \mathrm{ABC})}{\operatorname{Ar} \cdot(\triangle \mathrm{PQR})}=\frac{\mathrm{BC}^{2}}{\mathrm{PQ}^{2}}\]

• \[\frac{\mathrm{AB}^{2}}{\mathrm{PQ}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{PR}^{2}}\]

• \[\frac{\operatorname{Ar} \cdot(\triangle \mathrm{BAC})}{\operatorname{Ar} \cdot(\triangle \mathrm{QPR})} \neq \frac{\mathrm{AB}^{2}}{\mathrm{QP}^{2}}\]

• \[\frac{\mathrm{AC}^{2}}{\mathrm{PR}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{PQ}^{2}}\]

Answer 1

Triangles ABC and PQR are similar to each other, then \[\underline{\mathbf{\frac{AB^{2}}{PQ^{2}} = \frac{AC^{2}}{PR^{2}}}}\]

Explanation:

We know that,

The areas of two similar triangles are proportional to the squares on their corresponding sides.

⇒ `(Ar.(△ABC)​)/(Ar.(△PQR)) = (AB^2)/(PQ^2​)`   .......(1)

⇒ `(Ar.(△ABC)​)/(Ar.(△PQR)) = (AC^2)/(PR^2​)`   .......(2)

From equation (1) and (2), we get:

⇒ `(AB^2​)/(PQ^2) = (AC^2)/(PR^2)​`