Given: △ABC, m∠B=90° AB=12, BC=16, BK ⊥ AC . Find: AC and BK.

Question

Given: △ABC, m∠B=90°

AB=12, BC=16,

BK

 ⊥

AC

 . 

    Find: AC and BK.

Answer 1

Short leg 90 degrees Long leg   Hypotenuse

  AB=12      90             BC=16         AC= ?

  AK = ?      90             BK = ?         AB=12

AC = SQRT (AB*AB + BC*BC) = 20 [right triangle; Pythagorean Theorem]

Similar triangles:[Note: In diagram, share two angles. Therefore share three angles]

BK / 16 = AB / AC

BK / 16 = 12 / 20

BK = (3/5)16

BK = 48/5

Answer 2

After you draw the figure, look for the easy way out. ABC is a right triangle, because it has a 90-degree angle at B. Factor the given two side lengths and notice a 3-4-5 special-right-triangle situation. So AC is 5x4 or 20. BK is a common side to triangles AKB and CKB. Define AK as X and CK as 20-X. Using the Pythagorean Theorem for both BK triangles and rearranging to get BK equals, we can set the equals sides equal to each other and solve for X, which is 7.2. Putting that value into BK^2 = 12^2 - X^2 gives us the value of BK, which is 9.6.