Right triangle $ABC$ has legs of length $AB = 12$ and $BC = 6$. Point $D$ is randomly chosen within the triangle $ABC$. What is the probability that the triangle $ABD$ has area less than or equal to 12?

Guest Apr 12, 2017

#1**+3 **

**Right triangle ABC has legs of length AB = 12 and BC = 6. **

**Point D is randomly chosen within the triangle ABC. **

**What is the probability that the triangle ABD has area less than or equal to 12 ?**

\(\begin{array}{|rcll|} \hline ABC_{\text{Area } } &=& \frac{AB \cdot BC }{2} \quad & | \quad AB=12 \quad BC = 6 \\ &=& \frac{12 \cdot 6 }{2} \\ &=& 36 \\ \hline \end{array}\)

Let d = distance parallel from leg AB.

\(\begin{array}{rcll} ABD_{\text{Area } } = \frac{AB \cdot d }{2} &=& 12 \quad & | \quad AB=12 \\ \frac{12 \cdot d }{2} &=& 12 \\ \frac{d }{2} &=& 1 \\ d &=& 2 \\ \end{array} \)

\(\begin{array}{|rcll|} \hline \frac{BC-d}{x} &=& \frac{BC}{AB} \quad & | \quad AB=12 \quad BC = 6 \quad d = 2 \\ \frac{6-2}{x} &=& \frac{6}{12} \\ \frac{4}{x} &=& \frac{1}{2} \\ x &=& 8 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline ABD_{\text{Area } } = Area_{\text{trapezoid } } &=& (\frac{AB+x}{2}) \cdot d \quad & | \quad AB=12 \quad x = 8 \quad d = 2 \\ &=& (\frac{12+8}{2}) \cdot 2 \\ &=& 20 \\ \hline \end{array}\)

probability:

\(\begin{array}{|rcll|} \hline \frac{ ABD_{\text{Area } } } {ABC_{\text{Area } }} &=& \frac{20}{36} \\ &=& \frac{5}{9} \quad (55.\bar{5}\ \% ) \\ \hline \end{array}\)

heureka
Apr 12, 2017

#2**+1 **

Very nice, heureka.....!!!....while my approach isn't as "mathematical" as heureka's, it might be more intuitive.....

Look at the following pic :

We can let AB be the base of our triangle......and since the area of ABD cannot exceed 12, it must be that the height of ABD cannot exceed 2. And notice that GF will = this maximum height....

So......any point D chosen in area AGF [except on AG itself] will guarantee that triangle ABD will have an area <= to 12.

Also......any point D chosen in area BGFD [except on BG itself] will guarantee that triangle ABD will have an area <= 12.....because the base of such a triangle will= 12 and the height will be <= 2

So.....the probability that ABD has an area <=12 =

[ area AGF + area BGFD] / [area of triangle ABC] =

[4 + 16] / 36 =

20 / 36 =

5/9

CPhill
Apr 12, 2017