We are given that triangle ABC is similar to triangle EDF, denoted as \(\Delta ABC \sim \Delta EDF\). This similarity means that the ratio of their corresponding sides is constant.
The proportion of corresponding sides is:
\( \frac{AB}{ED} = \frac{AC}{EF} = \frac{BC}{DF} \)
First, determine the ratio of similarity using the known corresponding sides, AB and ED:
Ratio \(k = \frac{AB}{ED} = \frac{5}{12.5} = 0.4\)
This ratio \(k=0.4\) applies to all corresponding sides.
Use the similarity proportion \(\frac{AB}{ED} = \frac{BC}{DF}\):
\( \frac{5}{12.5} = \frac{BC}{15} \)
Solve for BC:
\( BC = 15 \times \frac{5}{12.5} = 15 \times 0.4 \)
\( BC = 6 \text{ cm} \)
Use the similarity proportion \(\frac{AB}{ED} = \frac{AC}{EF}\):
\( \frac{5}{12.5} = \frac{7.5}{EF} \)
Solve for EF:
\( EF = 7.5 \times \frac{12.5}{5} = 7.5 \times 2.5 \)
\( EF = 18.75 \text{ cm} \)
The remaining sides of the two triangles are BC (from \(\Delta ABC\)) and EF (from \(\Delta EDF\)).
Calculate their sum:
Sum \(= BC + EF = 6 \text{ cm} + 18.75 \text{ cm}\)
Sum \(= 24.75 \text{ cm}\)
In the given figure, if \(RS = 3\sqrt{3} \text{ cm}\) and \(RPS\) is an equilateral triangle, then find the value of \(QR\)
For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?

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What is the angle at $B$ in the triangle $ABC$, if the angle made at $K$ is $30^\circ$ and the triangles shown in the figure are similar?

What is \(AB + BC\) equal to?
In the given figure, if \(RS = 3\sqrt{3} \text{ cm}\) and \(RPS\) is an equilateral triangle, then find the value of \(QR\)
For a pair of similar triangles shown in the figure, the angles made at $B$ and $R$ are same in both the triangles $BAC$ and $RAQ$. If the ratio of the areas of the two triangles is 2:1, and if the area of $BAC$ is $100\text{ cm}^2$ then what is the area of $RAQ$ in sq.cm.?
