# What identity allows finding the area of a triangle given the area of adjacent triangles?

In my daughter's math book, there is a question that asks what the area of the larger of the two bottom triangles is given the total area of the figure (112) and the areas of two other triangles (17 and 11). The solution given in the book is to find the combined area of the two unknown triangles:

$$112-17-11=84$$

Then use the following equation to find the area of each of the other triangles:

$$84*\frac{11}{11+17}=33$$

$$84*\frac{17}{11+17}=51$$

My question is what is the mathematical identity in the above equation?

• The ratio of the areas of two triangles with the same height is equal to the ratio of their bases. – franz lemmermeyer Mar 1 '17 at 15:18

## 3 Answers Area of $\triangle\rm AZY = AY\times B/2 = (11+17)$

Area of $\triangle\rm AXY = AY\times R/2 = 112-(11+17)=84$

Thus, $\,\rm R/B=84/(11+17)$

$$\rm Area\ of\ \triangle\rm OXY = OY\times R\times\frac12 = OY\times\left(\frac{84}{11+17}\right)\times B\times\frac12$$ $$=\left(\frac{84}{11+17}\right)\times \rm Area\ of\ \triangle OYZ$$ $$=84\times\left(\frac{11}{11+17}\right)=33$$

Similarly,

$$\rm Area\ of\ \triangle\rm OXA = OA\times R\times\frac12 = OA\times\left(\frac{84}{11+17}\right)\times B\times\frac12$$ $$=\left(\frac{84}{11+17}\right)\times \rm Area\ of\ \triangle OAZ$$ $$=84\times\left(\frac{17}{11+17}\right)=51$$

Hint: Call $DOC$ the triangle of area $17$, $COB$ the triangle of area $11$, so the other triangles are $DOA$ with area $x$ and $AOB$ with area $y$. Consider $DO$ as the common basis of the triangles $DOC$ and $DOA$, then the quotient of their areas is the quotient of their heights: $$\frac{17}{x}=k$$ Now do the same for the trangles $COB$ and$AOB$ and note that the two heights ( with respect to the common basis $OB$) are the same so also their quotient is $k$ and we have: $$\frac{11}{y}=\frac{17}{x}=k$$

and you can complete • Red lines are parallel
• Green lines are parallel

So:

• $|BQ|=|OC|=|DR|$
• $BQCO$ and $DRCO$ are parallelograms, so $p_3=p_6$ and $p_4=p_7$

Then (according to the similarity of triangles): $$\frac{p_5}{p_1}=\left(\frac{|BQ|}{|AO|}\right)^2$$ By denoting $\frac{|BQ|}{|AO|}=k$ we obtain: $$p_5=k^2 p_1$$ $$\frac{p_1+p_3+p_5+p_6}{p_1}=\left(\frac{|AC|}{|AO|}\right)^2=\left(\frac{|AO|+|OC|}{|AO|} \right)^2=\left(\frac{|AO|+|BQ|}{|AO|} \right)^2 =(k+1)^2$$ so: $$p_1+p_3+p_5+p_6 = (k+1)^2 p_1$$

See, that $p_1+p_3+p_5+p_6 = (k^2+1)p_1+2 p_3$, so $$(k+1)^2 p_1= (k^2+1)p_1+2 p_3$$ $$k p_1= p_3$$ $$k = \frac{p_3}{p_1}$$

On the other hand: $$\frac{p_8}{p_2}=\left(\frac{|DR|}{|AO|}\right)^2=\left(\frac{|BQ|}{|AO|}\right)^2=k^2$$ By continuing this reasoning in analogous way, we obtain $$k = \frac{p_4}{p_2}$$

So: $$\frac{p_4}{p_2} = \frac{p_3}{p_1}$$ $$\frac{p_4}{p_3} = \frac{p_2}{p_1}$$ $$p_4 = p_3\frac{p_2}{p_1}$$

Now we have to solve the following system of equations: $$\begin{cases}p_1=17 \\ p_2=11 \\ p_1+p_2+p_3+p_4=112 \\ p_4 = p_3\frac{p_2}{p_1}\end{cases}$$ and we obtain the solution: $$(p_1,p_2,p_3,p_4)=(17,11,51, 33)$$