If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks.\[\frac{A\left( ∆ ABC \right)}{A\left( ∆ . . . . \right)} = \frac{80}{125} \therefore \frac{AB}{PQ} = \frac{......}{......}\]
Practice Set 1.4 | Q 3 | Page 25
If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks.\[\frac{A\left( ∆ ABC \right)}{A\left( ∆ . . . . \right)} = \frac{80}{125} \therefore \frac{AB}{PQ} = \frac{……}{……}\]
Given:
∆ABC ~ ∆PQR
A (∆ABC) = 80
A (∆PQR) = 125
According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
\[\therefore \frac{A\left( ∆ ABC \right)}{A\left( ∆ PQR \right)} = \frac{{AB}^2}{{PQ}^2}\]
\[ \Rightarrow \frac{80}{125} = \frac{{AB}^2}{{PQ}^2}\]
\[ \Rightarrow \frac{16}{25} = \frac{{AB}^2}{{PQ}^2}\]
\[\Rightarrow \frac{4^2}{5^2} = \frac{{AB}^2}{{PQ}^2}\]
\[ \Rightarrow \frac{AB}{PQ} = \frac{4}{5}\]
Therefore,
\[\frac{A\left( ∆ ABC \right)}{A\left( ∆ PQR \right)} = \frac{80}{125} and \frac{AB}{PQ} = \frac{4}{5}\]
Answer:-
Given:
- ∆ABC ~ ∆PQR
- A(∆ABC) = 80
- A(∆PQR) = 125
According to the theorem of areas of similar triangles, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Hence, A(∆ABC)/A(∆PQR) = AB^2/PQ^2
Substituting the given values, we get: 80/125 = AB^2/PQ^2
Simplifying the equation, we get: 16/25 = AB^2/PQ^2
Taking the square root of both sides, we get: AB/PQ = 4/5
Therefore, A(∆ABC)/A(∆PQR) = 80/125 and AB/PQ = 4/5.
Chapter 1. Similarity- Practice Set 1.4 – Page 25
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