Question

finding a segment length
f e
what is the value of x and the length of segment de?

1. $\frac{5}{9} = \frac{9}{2x + 3}$
2. $10x + 15 = 9(9)$

$x = \square$
length of $overline{de} = \square$ units
c d
5 9 $2x + 3$

Explanation:

Step1: Cross - multiply the proportion

Given the proportion $\frac{5}{9}=\frac{9}{2x + 3}$, cross - multiplying (which means multiplying the numerator of the first fraction by the denominator of the second fraction and setting it equal to the numerator of the second fraction times the denominator of the first fraction) gives us $5\times(2x + 3)=9\times9$.
Expanding the left - hand side: $5\times2x+5\times3 = 10x + 15$, and the right - hand side is $81$. So we have the equation $10x+15 = 81$.

Step2: Solve for x

Subtract 15 from both sides of the equation $10x + 15=81$.
$10x+15 - 15=81 - 15$
$10x=66$
Divide both sides by 10: $x=\frac{66}{10}=6.6$? Wait, no, wait. Wait, the second step in the problem says $10x + 15=9(9)$. Wait, $9\times9 = 81$, so $10x+15 = 81$. Then $10x=81 - 15=66$, so $x=\frac{66}{10}=6.6$? Wait, no, maybe I made a mistake in cross - multiplication. Wait, the original proportion: if we have $\frac{a}{b}=\frac{c}{d}$, then $a\times d=b\times c$. So $\frac{5}{9}=\frac{9}{2x + 3}$, so $5\times(2x + 3)=9\times9$. So $10x+15 = 81$. Then $10x=81 - 15 = 66$, so $x=\frac{66}{10}=6.6$? But let's check the second step given in the problem: it says $10x + 15=9(9)$. Wait, maybe there was a typo in the problem's second step? Wait, no, maybe I misread the proportion. Wait, the segment lengths: CD is 5, FD is 9, DE is $2x + 3$, and maybe there is a geometric mean theorem (altitude - on - hypotenuse theorem) here. The altitude - on - hypotenuse theorem states that in a right triangle, the length of the altitude drawn to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse. Also, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg.
Wait, maybe the proportion is $\frac{CD}{FD}=\frac{FD}{DE}$, so $\frac{5}{9}=\frac{9}{DE}$, and $DE = 2x+3$. So cross - multiplying: $5\times DE=9\times9$, so $5\times(2x + 3)=81$. Then $10x+15 = 81$. Then $10x=81 - 15=66$, so $x=\frac{66}{10}=6.6$? But 66 divided by 10 is 6.6, but maybe we can simplify. Wait, 66 and 10 have a common factor of 2, so $x=\frac{33}{5}=6.6$. Then $DE=2x + 3=2\times\frac{33}{5}+3=\frac{66}{5}+ \frac{15}{5}=\frac{81}{5}=16.2$. Wait, but let's do the steps as per the problem's given second step: $10x + 15=9(9)$. So $10x+15 = 81$. Subtract 15: $10x=66$. Divide by 10: $x = 6.6$ or $\frac{33}{5}$. Then $DE=2x + 3=2\times\frac{33}{5}+3=\frac{66 + 15}{5}=\frac{81}{5}=16.2$.

Wait, maybe the problem's second step has a typo, but following the given step: $10x + 15=9(9)$ (i.e., $10x+15 = 81$)

Step1 (re - doing with the given second step):

We have the equation $10x+15 = 81$ (since $9\times9 = 81$)

Step2: Subtract 15 from both sides

$10x+15-15 = 81 - 15$
$10x=66$

Step3: Divide both sides by 10

$x=\frac{66}{10}=\frac{33}{5}=6.6$

Step4: Find the length of DE

Since $DE = 2x+3$, substitute $x=\frac{33}{5}$ into it.
$DE=2\times\frac{33}{5}+3=\frac{66}{5}+\frac{15}{5}=\frac{81}{5}=16.2$ or $DE = 2\times6.6+3=13.2 + 3=16.2$

Answer:

$x=\frac{33}{5}$ (or 6.6), Length of $\overline{DE}=\frac{81}{5}$ (or 16.2) units