The areas

The areas of two similar triangles are 12 sq cm and 48 sq cm. If the height of the smaller one is 2.1 cm, then find the height of the bigger one.

Correct answer:

h₂ =  4.2 cm

Step-by-step explanation:

$$\begin{gathered} A_1=12{\text{cm}}^2 \\ {A}_2=48{\text{cm}}^2 \\ {A}_1<A_2 \\ {h}_1<h_2 \\ {h}_1=2.1\text{cm} \\ {S}_1=\frac{a_1\cdot h_1}{2} \\ {a}_1=\frac{2\cdot A_1}{h_1}=\frac{2\cdot 12}{2.1}=\frac{80}{7}\doteq 11.4286\text{cm} \\ {A}_2=\frac{a_2\cdot h_2}{2} \\ {A}_2=\frac{(k\cdot a_1)\cdot (k\cdot h_1)}{2} \\ {A}_2=k^2\cdot A_1 \\ k=\sqrt{A_2/A_1}=\sqrt{48/12}=2 \\ {a}_2=k\cdot a_1=2\cdot 11.4286=\frac{160}{7}\doteq 22.8571 \\ {h}_2=k\cdot h_1=2\cdot 2.1=\frac{21}{5}=4.2=4.2\text{cm} \\ \text{ Verifying Solution: } \\ {S}_{22}=\frac{a_2\cdot a_2}{2}=\frac{22.8571\cdot 22.8571}{2}=\frac{12800}{49}\doteq 261.2245{\text{cm}}^2 \end{gathered}$$

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