| The area of two similar triangles are \[\text{25 c}{{\text{m}}^{\text{2}}}\] and\[\text{121 c}{{\text{m}}^{\text{2}}}\]. |
| The ratio of their sides is: (CBSE 2019) |
A) \[\text{25}:\text{121}\]
B) \[\text{5}:\text{11}\]
C) \[121:25\]
D) \[11:5\]
Correct Answer: B
Solution :
| [b] By using the areas of similar triangles theorem, |
| \[\frac{ar(\Delta ABC)}{ar(\Delta PQR)}={{\left( \frac{AB}{PQ} \right)}^{2}}={{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\frac{25}{121}={{\left( \frac{AB}{PQ} \right)}^{2}}={{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{5}{11}\] |
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