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3 years ago

What is the perimeter of the quadrilateral if the two triangles are congruent using SSS

Mathematics

1 answer:

Westkost [7]3 years ago

6 0

Answer:

20 units (I used units because that's what i was taught when there isn't a specific unit)

Step-by-step explanation:

If the triangles are congruent, that means that everything is congruent/the same. So what we do is, we set both equations equal to each other.

6x - 3 = 4x + 9

And now we simplify.

6x - 3 = 4x + 9

-4x -4x

--------------------

2x - 3 = 9

+3 +3

-------------------

2x = 12

÷2 ÷2

---------------

x = 6

To find the perimeter, just multiply 6 by 4 since there's four sides. 6 x 4 = 20

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Answer:

point c is the intersection of line AB and ED

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3 years ago

5. Each polygon circumscribes a circle. Find the perimeter.

Oksana_A [137]

See the diagram of the circumscribed polygon attached below:

Answer:

A. 14.2 in.

Step-by-step explanation:

Perimeter = 3.7 + 3.6 + 3.4 + 1.9 + g

Let's find g.

a = 1.9 in. (Tangents drawn from an external point of a circle are congruent)

b = 3.4 - 1.9 = 1.5 in.

b = c = 1.5 in. (Tangents drawn from an external point)

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Step-by-step explanation:

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3 years ago

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10.5 is greater than 10 but also less than 11.

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3 years ago

<img src="https://tex.z-dn.net/?f=%28%20%5Csin%5E%7B2%7D%20%28%20%5Cfrac%7B%5Cpi%7D%7B%204%20%7D%20%20-%20%20%5Calpha%20%29%20%2

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Step-by-step explanation:

$\sin^2 (\frac{\pi}{4} - \alpha) = \frac{1}{2}(1 - \sin 2\alpha)$

Use the identity

$\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$

on the left side.

$\dfrac{1 - \cos [2(\frac{\pi}{4} - \alpha)]}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

$\dfrac{1 - \cos (\frac{\pi}{2} - 2\alpha)}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

Now use the identity

$\sin \theta = \cos(\frac{\pi}{2} - \theta)$

on the left side.

$\dfrac{1 - \sin 2\alpha}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

$\frac{1}{2}(1 - \sin 2\alpha) = \frac{1}{2}(1 - \sin 2\alpha)$

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2 years ago