1) A dilation maps triangle HIJ onto triangle H'I'J. If HI= 7 cm, IJ= 7 cm, I’J’= 5.25 cm, and H’J= 9 cm, the find HJ.
a. 4.08 cm^^^
b. 6.75 cm
c. 9 cm
d. 12 cm
2) if A' is the image of a(3, 4) after a dilation with scale factor 7 centered at the origin, what is the distance between A and A’!
a. 7 units
b. 28 units
c. 30 units^^^
d. 35 units
HJ/H'J' = IJ/I'J'
so,
HJ/9 = 7/5.25 = 12
A=(3,4)
A'=(21,28)
√((21-3)^2+(28-4)^2 = √(18^2+24^2) = 30
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1) Well, it seems like triangle HIJ is getting a bit of a makeover! Let's see if we can figure out the final look of HJ. We know that HI = 7 cm and IJ = 7 cm. Now, since we're dealing with a dilation, we need to consider the scale factor. The ratio of corresponding sides in a dilation is constant, so we can set up a proportion:
HI / IJ = H'J / I'J'
Substituting the given values, we have:
7 / 7 = 9 / 5.25
Now, let's cross multiply and solve for the missing side, H'J:
(7)(5.25) = 7(H'J)
36.75 = 7(H'J)
H'J = 36.75 / 7
H'J ≈ 5.25 cm
Ah, but we're interested in finding HJ, not H'J. Since HJ is a corresponding side to H'J in the original triangle, their lengths are proportional too! So, we can set up another proportion:
HI / IJ = HJ / H'J
Substituting the known values, we get:
7 / 7 = HJ / 5.25
Simplifying further:
1 = HJ / 5.25
So, multiplying both sides by 5.25, we find:
HJ ≈ 5.25 cm
Looks like option (a) 4.08 cm was just clowning around. The correct answer is approximately 5.25 cm.
2) Ah, A and A' are going through a little transformation of their own! We have A at coordinates (3, 4) and A' after a dilation with a scale factor of 7 centered at the origin. To find the distance between the two, we can use the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Substituting the given coordinates, we get:
Distance = √((0 - 3)^2 + (0 - 4)^2)
Distance = √(9 + 16)
Distance = √25
Distance = 5
Wait a minute, that's not any of the options! Maybe I should stop clowning around. My apologies for the diversion. The correct answer is approximately 5 units, which isn't listed as an option. So, I guess we're still left hanging without a correct choice. Blame it on the clowning coordinates!
1) To solve this question, we need to use the concept of similarity of triangles. In a dilation, corresponding sides of similar triangles are proportional.
Let's compare the sides of triangle HIJ and triangle H'I'J:
HI / H'I' = IJ / I'J' = HJ / H'J
Plugging in the given values, we have:
7 / H'I' = 7 / 5.25 = HJ / 9
To find HJ, we can cross-multiply and solve for HJ:
7 * 9 = H'I' * 7
HJ = (7 * 9) / 7 = 9 cm
Therefore, the answer is c. 9 cm.
2) To solve this question, we will find the distance between A and A' using the distance formula.
The distance formula is:
Distance = √((x₂ - x₁)² + (y₂ - y₁)²)
Given that A is (3, 4) and A' is the image of A after a dilation of scale factor 7 centered at the origin, we can find A' by multiplying the coordinates of A by the scale factor:
A' = (7 * 3, 7 * 4) = (21, 28)
Now we can substitute the values into the distance formula:
Distance = √((21 - 3)² + (28 - 4)²)
Distance = √(18² + 24²)
Distance = √(324 + 576)
Distance = √900
Distance = 30 units
Therefore, the answer is c. 30 units.
A dilation maps \bigtriangleup \mathrm{H}\mathrm{I}\mathrm{J}△HIJ onto \bigtriangleup \mathrm{H}'\mathrm{I}'\mathrm{J}'△H
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I
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. If \mathrm{H}\mathrm{I}=8\ \mathrm{i}\mathrm{n}.HI=8 in., \mathrm{H}'\mathrm{I}'=16\ \mathrm{i}\mathrm{n}.H
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=16 in. and \mathrm{I}\mathrm{J}=5\ \mathrm{i}\mathrm{n}.IJ=5 in., then \mathrm{I}'\mathrm{J}'I
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