# Thread: geometry problem

1. ## geometry problem

1. pic 7

A. 7
B. 21
C. 26
D. 52

2. Pic 8

A. -1
B. 2
C. -2
D. 1/2

2. Originally Posted by dgenerationx2
1. pic 7

A. 7
B. 21
C. 26
D. 52
A median of a triangle is a segment drawn a vertex to the midpoint of the opposite side.

AD = BD

5x - 2 = 2x + 7
3x = 9
x=3

AD + BD = AB
5(3) - 2 + 2(3) + 7=26

Originally Posted by dgenerationx2
2. Pic 8

A. -1
B. 2
C. -2
D. 1/2
Did you mean to say the slope of the altitude =4?

The product of the two slopes must be -1 since the lines are perpendicular.

$\displaystyle 4\cdot\frac{-x}{2}=-1$

$\displaystyle \frac{-4x}{2}=-1$

$\displaystyle -2x=-1$

$\displaystyle x={\color{red}\frac{1}{2}}$

3. 1) $\displaystyle \overline{AD}=\overline{BD}\rightarrow 5x-2=2x+7\leftrightarrow 3x=9\leftrightarrow x=3\rightarrow\overline{AB}=2\cdot(5\cdot 3-2)=\boxed{26}$.
2) From Pythagoras' theorem: $\displaystyle 4^2+\overline{DC}^2=\overline{AC}^2$ and $\displaystyle 4^2+\overline{BD}^2=\overline{AB}^2$. From the slope: $\displaystyle \tan\alpha=\frac {\overline{AD}}{\overline{BC}}=\frac {4}{\overline{BC}}=\frac {-x}2\rightarrow \overline{BC}=\frac 8{-x}$. Can you continue?
EDIT: I think masters is right as doing this way is hard and stupid.:P