Question

ABC~JKL with altitudes BX & KY. Find BXa. 19.2b. 21c. 24.6d. 28

Answer 1

SOLUTION:

Step 1:

In this question, we are given the following:

ABC~JKL with altitudes BX & KY. Find BX

a. 19.2

b. 21

c. 24.6

d. 28

Step 2:

From the question, we can see that using similar triangles:

ABC~JKL with altitudes BX & KY

$\begin{gathered} \frac{AB}{\text{JK}}\text{ }=\text{ }\frac{BX}{KY} \\ \text{where:} \\ AB\text{ = 32} \\ JK\text{ = 10} \\ KY\text{ = 6} \\ BX\text{ = ?} \end{gathered}$

Step 3:

Computing this, we have that:

$\begin{gathered} \frac{32}{10}=\frac{BX}{6} \\ \end{gathered}$

cross-multiply, we have that:

$\begin{gathered} 10BX\text{ = 32 x 6} \\ \text{Divide both sides by 10, we have that:} \\ BX\text{ =}\frac{32X\text{ 6}}{10} \\ BX\text{ = }\frac{192}{10} \\ BX\text{ = 19. 2 (OPTION A)} \end{gathered}$