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MrMuchimi
3 years ago
5

Hi I need help on my homework plz help ​

Mathematics
1 answer:
nordsb [41]3 years ago
8 0
Okay I can help you !
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A square has a side length 15.8cm.If 3 such squares are linked side to side in a straight line, what is the perimeter of the who
Novay_Z [31]

Answer:

126.4 cm

Step-by-step explanation:

1. 15.8 * 3 = 47.4 (This is the length)

2. Find half of the rectangle

47.4 + 15.8 (width) = 63.2

3. Find the whole perimeter

63.2 * 2 = 126.4

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Which pair is a base and corresponding height for the triangle?
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<span> B. b = 20 units and h = 16 units </span>
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Mark is making 10 kites.he uses 5yards of ribbon for each kite.he has already made 2 of the kites.how many yards of ribbon will
Andreyy89

Answer:

40 yards

Step-by-step explanation:

8×5≈40

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88 in²

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If three tangents to a circle form an equilateral triangle, prove that the tangent points form an equilateral triangle inscribed
weqwewe [10]
I added a figure so you can guide yourself throughout the proof I'm about to write, so I recommend that you download the picture beforehand and have this window and the picture's window open. Alright, let's get started!
Assuming that FED is an equilateral triangle according to the wording of the problem, we have that the angles \widehat{AFB}=\widehat{BEC}=\widehat{CDA}=60.
We also know that the circle in green is Inscribed in FED.

The following applies to every inscribed circle inside a triangle:
The center of the inscribed circle of a triangle is the intercept of all three angle bisectors of the triangle.

The above theorem implies that the line (FO) is an angle bisector because it goes through the vertex F and the center of the inscribed circle.

The previous statement implies that the angle \widehat{OFB}=30.

Now let's work on the OFB triangle.
Knowing that \widehat{OBF}=90 (because (EF) is tangent to the circle at B and OB is a radius of the circle. If you're lost here, remember that a tangent to a circle is always perpendicular to the radius of the circle.) we can then derive that \widehat{FOB}=180-90-30=60 (because the sum of the measures of all angles in a triangle is always equal to 180 degrees).

In the same way, we can prove also that:
\widehat{BOE}=\widehat{EOC}=\widehat{COD}=\widehat{DOA}=\widehat{AOF}=60 degrees.
Knowing the above we notice that \widehat{BOC}=\widehat{COA}=\widehat{AOB}=120degrees.

We're at the last part of our proof here:
Now notice that \widehat{BOA} subtends the same arc on the circle  that \widehat{BCA}.
According to the inscribed angle theorem, an angle \theta inscribed in a circle is half of the central angle 2\theta that subtends the same arc on the circle. 
Therefore \widehat{BCA}=\frac{\widehat{BOA}}{2}=60degrees

We can prove in a similar fashion that:  \widehat{CAB}=\widehat{ABC}=60degrees 
Therefore all the angles of the ABC triangle have a measure of 60 degrees, we conclude then that  ABC is equilateral.


5 0
3 years ago
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