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worty [1.4K]
3 years ago
9

HELP

Mathematics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

=18.38cm

Step-by-step explanation:

The longest line segment in a right rectangular prism is the diagonal that connects two opposite vertices. On the first diagram attached, the green line segment connecting A and G is one such diagonals. The goal is to find the length of segment .

Look at the attached picture. The longest segment we can draw is AG (or any of its equivalent: BH, CE, DF). Let's focus on AG for example.

We can think of AG as the hypotenuse of triangle ACG. So, we need AC first. AC is itself the hypotenuse of triangle ABC,

so we have,

AC^2 = AB^2+BC^2=8^2+7^2=113

Now we can resume where we stopped with triangle AGC, and we have

AG=\sqrt{AC^2+CG^2}=\sqrt{113+225}=\sqrt{338}

=18.38cm

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3 years ago
A triangle has a 60 degree angle, and the two adjacent sides are 12 and 12 times the square root of 3. Find the radius of a circ
Natasha_Volkova [10]

The radius of the circle is a line from the center of the circle to its circumference

The radius of the circle with the same vertex as a center is 10.16 units

<h3>How to determine the radius</h3>

From the question, we understand that:

  • The adjacent sides of the triangle are 12 and 12\sqrt 3
  • The triangle has an angle of 60 degrees

Start by calculating the area of the triangle using:

Area = 0.5ab\sin C

So, we have:

Area = 0.5 * 12 * 12\sqrt 3 *\sin (60)

This gives

Area = 0.5 * 12 * 12\sqrt 3 * \frac{\sqrt 3}{2}

Area = 108

The arc divides the triangle into two equal regions.

So, the area of the arc is:

Arc = 0.5 * Area = 0.5 *  108

Arc = 54

Also, the area of the arc is:

Arc = \frac{\theta}{360} * \pi r^2

So, we have:

Arc = \frac{60}{360} *3.14 * r^2

Simplify

Arc = 0.523 * r^2

This gives

0.523 * r^2 = 54

Divide both sides by 0.523

r^2 = 103.25

Take the square roots of  both sides

r = 10.16

Hence, the radius of the circle with the same vertex as a center is 10.16 units

Read more about circumcised triangles at:

brainly.com/question/4268382

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