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wolverine [178]
4 years ago
11

Prove or disprove the figure defined by points A, B, C, d is a rhombus

Mathematics
1 answer:
just olya [345]4 years ago
4 0

Answer:

It is not a rhombus, because all the internal angles are equal.

Step-by-step explanation:

A rhombus is a figure that has four sides of equal length, and 2 internal angles that are smaller than the other two internal angles.

First, let's look at figure ABCD.

We can see that all the sides have equal length by just counting the number of squares that each side takes (each side is 12 units long)

So now we need to see the angles.

Remember that if two lines are perpendicular, then the angle formed by those two lines is exactly 90°.

We can see that the segment BC is horizontal  (the y-value of B is the same than the y-value of C)

While the segment CD is vertical (the x-value of C is the same than the x-vale of D)

Then one segment is horizontal, and the other is vertical, which means that the segments are perpendicular, thus the angle C is 90°.

Now we can do the same for segments CD and DA

We already know that CD is a vertical segment.

For segment DA we can see that the y-value of D is the same as the y-value of A, then this segment is horizontal, this means that segments CD and DA are perpendiculars, which means that D = 90°.

Because DA is parallel to BC, and CD is parallel to BA, we can conclude that:

A = 90°

B = 90°

Then all the angles are equal, so this is not a rhombus. (it is a square)

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Step-by-step explanation:

The Three expressions that have a sum or difference of 3 /4 .

are

B . 11/12 − 1/6

\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{11}{12}-\dfrac{1\times 2}{6\times 2}\\\\\dfrac{11}{12}-\dfrac{1}{6}=\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}

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\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{3}{4}

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\dfrac{2}{3}+\dfrac{1}{12}=\dfrac{2\times 4}{3\times 4}+\dfrac{1}{12}\\\\\dfrac{3}{5}+\dfrac{3}{20}=\dfrac{8+1}{12}=\dfrac{9}{12}=\dfrac{3}{4}

Therefore,

\dfrac{2}{3}+\dfrac{1}{12}=\dfrac{3}{4}

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