Right-angled 40961

A right-angled triangle ABC has sides a = 5 cm, b = 8 cm. The similar triangle A'B'C' is 2.5 times smaller. Calculate the percentage of the area of ​​triangle ABC that is the area of ​​triangle A'B'C'.

Correct answer:

p =  16 %

Step-by-step explanation:

$$\begin{gathered} a=5\text{cm} \\ b=8\text{cm} \\ k=1/2.5=\frac{2}{5}=0.4 \\ x=a\cdot k=5\cdot \frac{2}{5}=\frac{5\cdot 2}{5}=\frac{10}{5}=2\text{cm} \\ y=b\cdot k=8\cdot \frac{2}{5}=\frac{8\cdot 2}{5}=\frac{16}{5}=3.2\text{cm} \\ {S}_1=\frac{a\cdot b}{2}=\frac{5\cdot 8}{2}=20{\text{cm}}^2 \\ {S}_2=\frac{x\cdot y}{2}=\frac{2\cdot \frac{16}{5}}{2}=\frac{16}{5}=3.2{\text{cm}}^2 \\ p=100\cdot \frac{S_2}{S_1}=100\cdot \frac{\frac{16}{5}}{20}=16\% \end{gathered}$$

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