-0.6875 is the answer to number 5
Answer:
I see two triangles
Step-by-step explanation:
Answer:
Step-by-step explanation:
To prove Δ ABC similar to ΔDBE we can consider
Segments AC and DE are parallel.
⇒ DE intersects AB and BC in same ratio.
AB is a transversal line passing AC and DE.
⇒∠BAC=∠BDE [corresponding angles]
Angle B is congruent to itself due to the reflexive property.
All of them are telling a relation of parts of ΔABC to ΔDBE.
The only option which is not used to prove that ΔABC is similar to ΔDBE is the first option ,"The sum of angles A and B are supplementary to angle C".
Let's begin by listing out the information given to us:
We start out by observing that Triangles MKR & ACD are similar or proportional
We will solve for the missing side by using the similar triangle theorem. This is shown below:~
There is a not so well-known theorem that solves this problem.
The theorem is stated as follows:
"Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides" (Coxeter & Greitzer)
This means that for a triangle ABC, where angle A has a bisector AD such that D is on the side BC, then
BD/DC=AB/AC
Here either
BD/DC=6/5=AB/AC, where AB=6.9,
then we solve for AC=AB*5/6=5.75,
or
BD/DC=6/5=AB/AC, where AC=6.9,
then we solve for AB=AC*6/5=8.28
Hence, the longest and shortest possible lengths of the third side are
8.28 and 5.75 units respectively.
