Question

Given triangle IMJ with altitude JL, JL = 32, and IL =24, find IJ, JM, LM, and IM.

Answer 1

Okay, so I really hope that you can read all my work. I just spent the last 45 mins doing this problem as neatly as possible with as much detail as possible. So, I really hope my work doesn't confuse you. 

My Final Answers were:
JI= 40 units (found using the Pythagorean theorem)
IM= 66.66 units = 66 and 2/3rds (found by dividing the length of JL by the cosine of ∠JIM )
LM=42.66 units = 42 and 2/3rds (found by IM= 24+LM; solve for LM since we know IM=66.66)
JM= 53.33 units= 53 and (1/3rd) (found by using the Pythagorean theorem; this time using JI as "a" and IM as c)

Hope this helped and all made sense!

(Pythagorean theorem is a²+b²=c²)

Answer 2

Part A:

From the given figure, IJ represents the hypothenuse of the right triangle IJL.

By the Pythagoras theorem,

$IJ^2=24^2+32^2 \\ \\ =576+1024=1600 \\ \\ \Rightarrow IJ= \sqrt{1600} =40$



Part B:

From the given figure, angle J is obtained as follows:

$\tan I= \frac{32}{24} = \frac{4}{3} \\ \\ \Rightarrow I^o=\tan^{-1}\left( \frac{4}{3} \right)$

Line JM can be obtained as follows:

$\tan I= \frac{JM}{40} \\ \\ \frac{4}{3}= \frac{JM}{40} \\ \\ \Rightarrow JM= \frac{40\times4}{3} = \frac{160}{3}$



Part C:

From the given triangle, LM is one of the legs of the right triangle JLM with the other leg, JL = 32 and the hypothenuse, JM = 160/3.

By the Pythagoras theorem,

$LM^2=JM^2-JL^2 \\ \\ =\left( \frac{160}{3} \right)^2-32^2= \frac{25,600}{9} -1,024 \\ \\ = \frac{16,384}{9} \\ \\ \Rightarrow LM= \sqrt{\frac{16,384}{9}} = \frac{128}{3}$



Part D:

From the figure, IM = IL + LM

$24+ \frac{128}{3} = \frac{200}{3}$