Answer:
5.9 inches
Step-by-step explanation:
Triangle BCD is a special 30-60-90 triangle, so it has special properties. If we call side BD "x," then side BC will be x![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
![14 = x\sqrt{3} \\x = \frac{14}{\sqrt{3} }](https://tex.z-dn.net/?f=14%20%3D%20x%5Csqrt%7B3%7D%20%5C%5Cx%20%3D%20%5Cfrac%7B14%7D%7B%5Csqrt%7B3%7D%20%7D)
Now, we know that side BD is equal to ![\frac{14}{\sqrt{3} }](https://tex.z-dn.net/?f=%5Cfrac%7B14%7D%7B%5Csqrt%7B3%7D%20%7D)
We can also see that triangle ABC is an isoceles triangle, since one angle is 45 degrees, and the other is 90, leaving the last angle to also be 45 degrees.
This means that side AB = side BC --> side AB = 14
All we need to do is subtract
from 14, and we are left with something around 5.9 inches
Answer fast? What?
x + 2y = 17...Equation A
8x + 3y = 45...Equation B
Solve A for x.
x = 17 - 2y
Plug this into B to solve for y.
8(17 - 2y) + 3y = 45
136 - 16y + 3y = 45
136 - 13y = 45
-13y = 45 - 136
-13y = -91
y = -91/-13
y = 91/13
y = 7
To find y, plug x = 7 into either equation to find x.
I will use Equation A.
x + 2y = 17
x + 2(7) = 17
x + 14 = 17
x = 17 - 14
x = 3
Answer: x = 3, y = 7
Step-by-step explanation:
QR is parallel to TU | Given
S is the midpoint of QT | Given
QS = ST | Definition of midpoint
<QSR = <TSU | definition of vertical angles
<QRS = <STU | Definition of Alternate Interior Angles
QSR = TSU | ASA
Pretty sure that's right :D
Also, wherever it says "=", I mean congruency symbol