Question

Right triangle ABC, $\overline{BC}$ = 5, $\overline{AC}$ = 12, and $\overline{AM}$ = x; MN is perpendicular to AC, NP is perpendicular BC; N is on AB. If y= $\overline{MN}+\overline{NP}$, one-half the perimeter of rectangle MCPN, then:

• A. y= $\frac{1}{2}$(5+12)
• B. y= $\frac{5x}{12}+\frac{12}{5}$
• C. y= $\frac{144-7x}{12}$
• D. y= 12
• E. y= $\frac{5x}{12}$ + 6

Answer 1

The correct option is C y= $\frac{144-7x}{12}$

Since triangles $\Delta ABC$ and $\Delta NBP$ are similar triangles(by

$AAA$ criteria), so:

$⟹\frac{AC}{NP}=\frac{CB}{PB}$

$⟹\frac{12}{12-x}=\frac{5}{PB}$

$⟹PB=5-\frac{5x}{12}$

Now, $MC=NP=12-x$ and

$MN=CP=5-PB=5-5+\frac{5x}{12}=\frac{5x}{12}$.

Hence, $y=MN+NP=12-x+\frac{5x}{12}$

$⟹y=\frac{144-7x}{12}.$