If areas of two similar triangles …
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Sia ? 6 years, 6 months ago
Given : {tex}\triangle \mathrm{ABC}{\sim} \triangle \mathrm{PQR}{/tex} &
ar {tex}\triangle \mathrm{ABC}{/tex} = ar {tex}\Delta \mathrm{PQR}{/tex}
{tex}{/tex}To prove: {tex}\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}{/tex}
Since, {tex}\triangle \mathrm{ABC}{\sim} \Delta \mathrm{PQR}{/tex}
ar {tex}\triangle A B C=\text { ar } \triangle P Q R{/tex} (given)
{tex}\frac{{\Delta ABC}}{{ar\Delta PQR}} = 1{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{{A{B^2}}}{{P{Q^2}}} = \frac{{B{C^2}}}{{Q{R^2}}} = \frac{{C{A^2}}}{{P{R^2}}} = 1{/tex}
[Using Theorem of area of similar Triangles]
{tex}\Rightarrow{/tex} AB = PQ, BC = QR & CA = PR
Thus, {tex}\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}{/tex}
0Thank You