SOLUTION TO PROBLEM 7

Let T denote the area of the triangle $\triangle ABC$, and denote by T₁, T₂ and T₃ the areas of t₁,t₂ and t₃, respectively. Moreover, let c be the length of AB, and let c₁,c₂, and c₃ be the lengths of the bases parallel to AB of t₁,t₂ and t₃ respectively. Then, in view of the similarity of the four triangles, one has

$$\frac{\sqrt{T_1}}{\sqrt{T}}=\frac{c_1}{c},\quad \frac{\sqrt{... ...uad \mbox{and}\quad \frac{\sqrt{T_3}}{\sqrt{T}}=\frac{c_3}{c}.$$

Since c₁+c₂+c₃=c, it follows that

$$\frac{\sqrt{T_1}}{\sqrt{T}}+\frac{\sqrt{T_2}}{\sqrt{T}}+ \frac{\sqrt{T_3}}{\sqrt{T}}=\frac{c_1+c_2+c_3}{c}=1,$$

and therefore

$$T=(\sqrt{T_1}+\sqrt{T_2}+\sqrt{T_3})^2=(\sqrt{4}+\sqrt{9}+\sqrt{49})^2 =144.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder
 

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Last modified:   Thu Feb 22 20:28:27 CST 2001