The point of intersection of two straight line graphs is the point with coordinates that satisfies both lines equations
The length of segment EF is presented as follows;
[tex]EF =\underline{\mathbf{ 17\dfrac{1}{7} \ ft.}}[/tex]
The reason why the above value is correct is as follows;
The given parameters are;
Height of right triangle ACD = 40 ft.
Height of right triangle BCD = 30 ft.
Required:
To find the length of EF
Solution:
The equation of the lines AC and BD are found and equated to find the height of EF as follows;
The slope of AC = [tex]\dfrac{-40}{CD}[/tex]
The equation of AC is presented as follows;
[tex]y - 40 = \dfrac{-40}{CD} \times (x - 0)[/tex]
[tex]y = \dfrac{-40}{CD} \times x + 40[/tex]
The slope of BD = [tex]\dfrac{30}{CD}[/tex]
The equation of BD is given as follows;
[tex]y - 30 = \dfrac{30}{CD} \times (x - CD)[/tex]
[tex]y = \dfrac{30}{CD} \times (x - CD)+ 30[/tex]
Equating both values of y to find the value of y at the intersecting point E, gives;
[tex]\dfrac{-40}{CD} \times x + 40 = \dfrac{30}{CD} \times (x - CD)+ 30[/tex]
Which gives;
[tex]\dfrac{40 \cdot CD - 40 \cdot x}{CD} = \dfrac{30 \cdot x}{CD}[/tex]
Therefore;
40·CD - 40·x = 30·x
[tex]CD = \dfrac{70\cdot x}{40} = \dfrac{7\cdot x}{4}[/tex]
[tex]CD = \dfrac{7\cdot x}{4}[/tex]
Therefore, at E, we have;
[tex]y = EF= \dfrac{40 \cdot CD - 40 \cdot x}{CD} = \dfrac{30 \cdot x}{CD}[/tex]
[tex]y = EF = \dfrac{40 \times \dfrac{7}{4}\cdot x - 40 \cdot x}{\dfrac{7}{4}\cdot x} = \dfrac{30 \cdot x}{\dfrac{7}{4}\cdot x} = \dfrac{120}{7} = 17\dfrac{1}{7}[/tex]
[tex]\underline {EF = 17\dfrac{1}{7} \ ft.}[/tex]
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