In the figure,ABC is congruent to ADC . If the square ABCD is dilated by a factor of 2 to form A'B'C'D', the ratio of the area of A'B'C'D' to the area of ABCD is 1/2. Since the dilation is by a factor of 2, so the area of the A'B'C'D' should be smaller than the area of ABCD.
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Answer:The claim is correct
Explanation:Assume the given triangle ABCperimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> IIFrom I and II, we can prove that:perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)
Answer:
The Answer: x^2 + x / 15 - 2 / 15