The area of a triangle is found using the formula: A=½bh
1 answer:
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Answer: The triangles are congruent by the HL theorem
HL = hypotenuse leg
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Explanation:
Through the segment addition postulate, we can say:
WZ = WT+TZ
YT = TZ+ZY
Notice how both WZ and YT have TZ as part of their segments.
Since WZ and YT are the same length, we can then say,
WZ = YT
WT+TZ = TZ+ZY
WT = ZY
I used substitution for the second step.
The last step is where I subtracted TZ from both sides, and they canceled out.
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We have the following info:
- WT = ZY (just found)
- AT = UZ (given)
- angle AWT = angle UYZ (given)
Normally, the conclusion from here would be "not enough info" because this is a SSA case. Note how the angles are not between the congruent sides. So we wouldn't use SAS.
However, despite us having SSA, it doesn't mean we can't keep going. The reason why is because we have two right triangles. So we can use the HL theorem instead. Statement 1 above refers to the L of HL, while statement 2 is the H.
HL = hypotenuse leg
Again, HL only works for right triangles. It's the only time we can use SSA to prove right triangles are congruent; otherwise, SSA is not a valid congruence theorem.
Answer:
See attached
Step-by-step explanation:
For some point X, center C, and dilation factor k, the image point X' will be ...
... X' = C + k(X -C)
Assuming you want k=12 in both cases, for the written problem, the first point is ...
... X' = (-3, 0) + 12((-5, 4) -(-3, 0)) = (-3, 0) +12(-2, 4)
... = (-3, 0) +(-24, 48)
... X' = (-27, 48)
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In the problem in the image, the lower right point will be ...
... Q' = C + k(Q -C)
... Q' = (-3, 1) +12((2, 2) -(-3, 1)) = (-3, 1) + 12(5, 1)
... = (-3, 1) + (60, 12)
... Q' = (57, 13)
