If DE is the midsegment of ABC what is the value of x AND y?
1 answer:
Answer:
3.) x = 4.5
4.) y = 7
Step-by-step explanation:
3 is an easier one than 4. By the triangle midsegment theorem, the midsegment of a triangle will be half of the base if it is parallel to it. AB is our base, so the equation is like this!
Now for 4, it actually is pretty easy. By the same triangle midsegment theorem, we automatically know that EC and EA are congruent, and CD and DB are also congruent. Meaning y equals 7.
But another way we can determine this (the long, kinda unecessary, but useful way) is through examining the relationship that triangles ABC and EDC have! We know that they are similar triangles by SAS or AA. Meaning, their equation looks something like this...
Like I said earlier, 7 is equal to y, but I figured I'd prove it so we are on the same page. Nothing wrong with double-checking ;)
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Let x = negative real number ⇒x<0
from the statement above, we can generate an equation:
(x + 5)² = 48
= 
⇒ eliminate the square by getting the square root on both sides
⇒ the perfect square of a real number has one positive real number and a negative real number
transposing 5 to other side, we will arrive at two (2) values for x:
= -5 - 4√3 = -11.928
= -5 + 4√3 = 1.928
Since we are only looking at the negative real number, our answer will be -11.928, also equal to -5 - 4√3.
from the statement above, we can generate an equation:
(x + 5)² = 48
⇒ eliminate the square by getting the square root on both sides
⇒ the perfect square of a real number has one positive real number and a negative real number
transposing 5 to other side, we will arrive at two (2) values for x:
Since we are only looking at the negative real number, our answer will be -11.928, also equal to -5 - 4√3.
Making the usual substitutions (x=r·cos(θ), y=r·sin(θ)), you have
(r·cos(θ))² +4r·cos(θ) +4 + (r·sin(θ))² -4r·sin(θ) +4 = 8
r(r +4·cos(θ) -4·sin(θ)) = 0
Dividing by r then subtracting the non-r terms gives
r = 4(sin(θ) -cos(θ))
(r·cos(θ))² +4r·cos(θ) +4 + (r·sin(θ))² -4r·sin(θ) +4 = 8
r(r +4·cos(θ) -4·sin(θ)) = 0
Dividing by r then subtracting the non-r terms gives
r = 4(sin(θ) -cos(θ))