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= √(1600) =40`



Part B:

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


`\tan I= (32)/(24) = (4)/(3) \\ \\ \Rightarrow I^o=\tan^(-1)\left( (4)/(3) \right)`

Line JM can be obtained as follows:


`\tan I= (JM)/(40) \\ \\ (4)/(3)= (JM)/(40) \\ \\ \Rightarrow JM= (40*4)/(3) = (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( (160)/(3) \right)^2-32^2= (25,600)/(9) -1,024 \\ \\ = (16,384)/(9) \\ \\ \Rightarrow LM= \sqrt{(16,384)/(9)} = (128)/(3)`



Part D:

From the figure, IM = IL + LM


`24+ (128)/(3) = (200)/(3)`