Question

Triangle C F E is shown. Angle C F E is a right angle. An altitude is drawn from point F to point D on side C E, forming a right angle. The length of F D is 9, the length of C D is 5, and the length of D E is 2 x + 3.
What is the value of x and the length of segment DE?
StartFraction 5 Over 9 EndFraction = StartFraction 9 Over 2 x + 3 EndFraction
10x + 15 = 9(9)
x =
Length of = units

Answer 1

What is the value of x and the length of segment DE?

StartFraction 5 Over 9 EndFraction = StartFraction 9 Over 2 x + 3 EndFraction

10x + 15 = 9(9)

Answer: Keep it simple

x =  

6.6

Length of  =  

16.2

 units

Answer 2

The value of x and one of the sides of the triangle are needed.

$x=6.6\ \text{units}$

$DE=16.2\ \text{units}$

$CD=5\ \text{units}$

$FD=9\ \text{units}$

$DE=2x+3$

Finding CF

$CF^2=CD^2+FD^2\\\Rightarrow CF^2=106\ \text{units}$

In triangle DEF

$FE^2=DE^2+DF^2\\\Rightarrow FE^2=(2x+3)^2+81$

In triangle CFE

$CE^2=CF^2+FE^2\\\Rightarrow (5+2x+3)^2=106+(2x+3)^2+81\\\Rightarrow (8+2x)^2=187+(2x+3)^2\\\Rightarrow 64+4x^2+32x=187+4x^2+9+12x\\\Rightarrow x=\dfrac{187-64+9}{32-12}\\\Rightarrow x=6.6\ \text{units}$

$DE=2x+3=2\times 6.6+3\\\Rightarrow DE=16.2\ \text{units}$

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