Question

# Prove that the ratio of the area of two similar triangles is equal to the ratio of squares on the corresponding sides.

Solution

## $$Given:\triangle ABC\sim\triangle PQR$$To prove :$$\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}={\left(\dfrac{AB}{PQ}\right)}^{2}={\left(\dfrac{BC}{QR}\right)}^{2}={\left(\dfrac{AC}{PR}\right)}^{2}$$Proof:-$$arABC=\dfrac { 1 }{ 2 } \times base\times height=\dfrac { 1 }{ 2 } \times BC\times AM---(i)$$$$ar\left( PQR \right) =\dfrac { 1 }{ 2 } \times base\times hight=\dfrac { 1 }{ 2 } \times QP\times PN----(ii)$$$$\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}=\dfrac {BC \times AM}{QR\times PN}----(A)$$In $$\triangle ABM$$ and $$\triangle PQN, <B=<Q, <M=<N$$$$\triangle ABM\sim \triangle PQN$$$$\therefore \dfrac{AB}{PQ}=\dfrac{AM}{PN}---(B)$$From $$\left(A\right)$$$$\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}= \dfrac{BC\times AM}{QR\times PN}= \dfrac{BC}{QR}\times \dfrac{AB}{PQ}---(C)$$Now, given $$\triangle ABC \sim \triangle PQR \Rightarrow \dfrac{AB}{PQ}=\dfrac{BC}{QP}= \dfrac{AC}{PR}$$Putting in $$\left(C\right) \quad \dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}= {\left(\dfrac{AB}{PQ}\right)}^{2}$$Similarly$$\Rightarrow\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}= {\left(\dfrac{AB}{PQ}\right)}^{2}{\left(\dfrac{BC}{QR}\right)}^{2}={\left(\dfrac{AC}{PR}\right)}^{2}$$Hence provedMathematics

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