When Perimeters are Equal!

Geometry Level 3

Among the given shapes, choose a rectangle namely R \color{#3D99F6}R that has the same perimeter of a square S \color{#20A900}S .

If the sides of R \color{#3D99F6}R are in the ratio: { 2 3 } \big\{\frac{2}{3}\big\} , then the ratio of the area of R \color{#3D99F6}R to the ratio of the area of S \color{#20A900}S is: { A B } \big\{\frac{A}{B}\big\} .

What is A + B = ? \color{#D61F06}A+B=?


The answer is 49.

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Hana Wehbi
Jun 12, 2018

Relevant wiki: Similar Polygons - Area and Perimeter Relations

Let s s be the side of the square ( S ) S) , and a and b a \text{ and } b the sides of the rectangle ( R ) (R) .

Given a b = 2 3 ⟹ b = 3 a 2 ⟹ ( a + b ) = 5 a 2 \frac{a}{b}=\frac{2}{3}\implies b=\frac{3a}{2}\implies (a+b)=\frac{5a}{2} .

Also, 4 s = 2 ( a + b ) (both perimeters of S and R are equal.) 4s=2(a+b) \color{#3D99F6} \text { (both perimeters of S and R are equal.)}

⟹ 2 s = a + b ⟹ s = 5 a 4 \implies 2s=a+b\implies s=\frac{5a}{4} .

Question is: A S A R = a b s 2 = 3 a 2 2 × 16 25 a 2 = 24 25 ⟹ A = 24 and B = 25 ⟹ A + B = 49 \color{#D61F06}\frac{A_S}{A_R}= \frac{ab}{s^2}=\color{#333333}\frac{3a^2}{2}\times \frac{16}{25a^2}=\frac{24}{25}\implies A=24 \text{ and } B=25 \implies A+B=\boxed{49}

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