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# Another one of my Triangle Questions

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In $$\Delta ABC$$, line segments are drawn parallel to each of the sides dividing the triangle into six regions. The areas of three regions are shown in the figure. What is the area of $$\Delta ABC$$?

Thank You :P

Nov 15, 2018

#1
+22343
+15

Another one of my Triangle Questions

In $$\Delta ABC$$, line segments are drawn parallel to each of the sides dividing the triangle into six regions.
The areas of three regions are shown in the figure. What is the area of $$\Delta ABC$$?

$$\begin{array}{l} \text{Let Area A =[ABC] } \\\\ \text{Let Area A_4=[E'EP] =a^2 = 4, ~a=2 }\\ \text{Let Area A_5=[D'DP] =b^2 = 9, ~b=3 }\\ \text{Let Area A_6=[F'FP] =c^2 = 16 ~c=4} \\\\ \text{Let Area A_1=[E'AD] }\\ \text{Let Area A_2=[D'BF] }\\ \text{Let Area A_3=[F'CE] } \end{array}$$

$$\text{The triangles are similar \\( \triangle ABC \sim \triangle E'EP \sim \triangle D'DP \sim \triangle F'FP \sim \triangle E'AD \sim \triangle D'BF \sim \triangle F'CE ).}$$

The key theorem we apply here is that the ratio of the areas of 2 similar triangles is
the ratio of a pair of corresponding sides squared.

$$\begin{array}{|rcll|} \hline \dfrac{A_4}{A} &=& \left(\dfrac{E'P}{CB}\right)^2 \\ \dfrac{A_5}{A} &=& \left(\dfrac{PD}{CB}\right)^2 \\ \hline \dfrac{A_4}{A}\cdot \dfrac{A}{A_5} &=& \dfrac{E'P^2}{CB^2} \cdot \dfrac{CB^2}{PD^2} \\ \dfrac{A_4}{A_5} &=& \dfrac{E'P^2}{PD^2} \\ \dfrac{a^2}{b^2} &=& \dfrac{E'P^2}{PD^2} \\ \dfrac{a}{b} &=& \dfrac{E'P}{PD} \\ \boxed{\dfrac{E'P}{PD}=\dfrac{a}{b} } \\ \hline \end{array}$$

$$\begin{array}{|rclcrl|} \hline \dfrac{E'P}{PD} &=& \dfrac{a}{b} && E'P &=& \dfrac{a}{a+b}E'D \\ && && E'D &=& \dfrac{a+b}{a} E'P \\ \dfrac{E'D}{PD} &=& \dfrac{a+b}{a} \cdot \dfrac{E'P}{PD} \\ \dfrac{E'D}{PD} &=& \dfrac{a+b}{a} \cdot \dfrac{a}{b} \\ \boxed{\dfrac{E'D}{PD} = \dfrac{a+b}{b} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{A_1}{A_5} &=& \left( \dfrac{E'D}{PD} \right)^2 \\ A_1 &=& A_5\left( \dfrac{E'D}{PD} \right)^2 \\ A_1 &=& b^2\left( \dfrac{a+b}{b} \right)^2 \\ \mathbf{A_1} & \mathbf{=} & \mathbf{(a+b)^2} \quad & | \quad A_1 = (2+3)^2=5^2=25 \\ \hline \end{array}$$

analogous

$$\begin{array}{|rclcrl|} \hline \dfrac{D'P}{PF} &=& \dfrac{b}{c} && D'P &=& \dfrac{b}{b+c}D'F \\ && && D'F &=& \dfrac{b+c}{b}D'P \\ \dfrac{D'F}{PF} &=& \dfrac{b+c}{b} \cdot \dfrac{D'P}{PF} \\ \dfrac{D'F}{PF} &=& \dfrac{b+c}{b} \cdot \dfrac{b}{c} \\ \boxed{\dfrac{D'F}{PF} = \dfrac{b+c}{c} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{A_2}{A_6} &=& \left( \dfrac{D'F}{PF} \right)^2 \\ A_2 &=& A_6\left( \dfrac{D'F}{PF} \right)^2 \\ A_2 &=& c^2\left( \dfrac{b+c}{c} \right)^2 \\ \mathbf{A_2} & \mathbf{=} & \mathbf{(b+c)^2} \quad & | \quad A_2 = (3+4)^2=7^2=49 \\ \hline \end{array}$$

analogous

$$\begin{array}{|rclcrl|} \hline \dfrac{F'P}{PE} &=& \dfrac{c}{a} && F'P &=& \dfrac{c}{a+c}F'E \\ && && F'E &=& \dfrac{a+c}{c}F'P \\ \dfrac{F'E}{PE} &=& \dfrac{a+c}{c} \cdot \dfrac{F'P}{PE} \\ \dfrac{F'E}{PE} &=& \dfrac{a+c}{c} \cdot \dfrac{c}{a} \\ \boxed{\dfrac{F'E}{PE} = \dfrac{a+c}{a} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{A_3}{A_4} &=& \left( \dfrac{F'E}{PE} \right)^2 \\ A_3 &=& A_4\left( \dfrac{F'E}{PE} \right)^2 \\ A_3 &=& a^2\left( \dfrac{a+c}{a} \right)^2 \\ \mathbf{A_3} & \mathbf{=} & \mathbf{(a+c)^2} \quad & | \quad A_3 = (2+4)^2=6^2=36 \\ \hline \end{array}$$

$$\mathbf{A=\ ?}$$

$$\begin{array}{|rcll|} \hline A_1+A_2+A_3 &=& \mathbf{A}+ A_4+A_5 + A_6 \\ 25+49+36 &=& A+ 4+9+16 \\ 110 &=& A + 29 \\ A &=& 110 - 29 \\ \mathbf{A} & \mathbf{=} & \mathbf{81} \\ \hline \end{array}$$

Nov 16, 2018
#2
+101322
+2

Thanks, heureka!!!!

That was a tough one....!!!

CPhill  Nov 16, 2018
#3
+702
+3

Awesome, thanks heureka!

CoolStuffYT  Nov 17, 2018
#4
+3

Here is another way:

Triangle PD'D:PF'F and 9:16 and sqrt(9):sqrt(16)=3:4
Line PD':FP =3:4 and FD' =3 + 4=7. So, triangle PFF':FD'B.
FD'B =7^2 - 16 - 9 =24 area of PDF'B. By similar reasoning:
EF' =2 + 4 =6 and the area of EF'C=6^2 - 4 - 16=16 area of E'PFC. And the same applies to DE' =2 + 3 =5 and the area of ADE' =5^2 - 4 - 9=12 area of AEPD'.
So, the area of ABC triangle= 4 +9+16+24+16+12=81 units^2

Nov 17, 2018
#5
+101322
+1

Thanks, Guest!!!

THAT is a REALLY nice method  !!!!

CPhill  Nov 17, 2018
#6
+101768
+4

Thanks Heureka and guest,

I think my logic is similar to guests but I did not really understand guest's that well.

Here is my diagram.

1) On my diagram I have labeled the angles of the triangle ABC

2) Then using only the fact that corresponding angles on parallel lines are congruent, I have labelled all the relevant angles in black.

3) Now, using the fact that the angle sum of a triangle is always the same (180 degrees but that does not matter) I have filled in the 3rd angle on all of the triangles - they are in pink.

4) Now it can be seen that there are 4 similar triangles. (3 equal angles test)

Also there are 3 parallelograms.

5) Now the ratio of the areas of the 3 small similar triangles is                    4:9:16

6) so the ratio of corresponding sides of the 3 small similar triangles is       2:3:4

7) On the next diagram I will label 3 coresponding sides with their lengths   (in black)

8) Then I will use the fact that parallelograms have opposite sides of equal length to fill in the missint length secions on AC (in red)

9)   3+2+4=9

10)   So now I know that the ratio of the coresponding sides of the 4 similar triangles is    2:3:4:9

11) So the ratio of the areas of the 4 similar triangles is                                                      4:9:16:81

Since the areas of the little ones are exactly 4,9 and 16, the area of the big one, i.e. triangle ABC is 81 units squared.

Nov 18, 2018
#7
+702
+4

Wow, now I get it, thanks Melody (and everyone else too)!

CoolStuffYT  Nov 19, 2018