Question

In right angled triangle ABC, BC = 5, AC = 12, and AM = X ; $MN\perp AC,NP\perp BC;$ N is on AB. if y = MN + NP, then

• A. $y=\frac{1}{2}(5x+12)$
• B. $y=\frac{5x}{12}+\frac{12}{5}$
• C. $y=\frac{144-7x}{12}$
• D. y = 12x

Answer 1

Answer: (C)

$y=\frac{144-7x}{12}$

In right angled triangle ABC, BC = 5, AC = 12, and AM = X ; N is on AB.
Then $(AB{)}^2=(12{)}^2+(5{)}^2=144+25=169\Rightarrow AB=13$
NP $\perp$BC&MN$\perp$AC and AM=x
Then NP=MC=12-x and
$\frac{BC}{AC}=\frac{MN}{AM}\Rightarrow \frac{5}{12}=\frac{MN}{x}\Rightarrow MN=\frac{5X}{12}$
Then y=MN+NP=$(12-x)+\frac{5x}{12}=\frac{144-12x+5x}{12}=\frac{144-7x}{12}$