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aliya0001 [1]
3 years ago
5

How many unique triangles can be drawn with side lengths 8 in., 12 in., and 24 in.? Explain. A. Any three sides form a unique tr

iangle, so these lengths can be used to draw exactly 1 triangle. B. If you draw two of the sides together, the third side could angle in or out from the existing angle, so these lengths can be used to draw exactly 2 triangles. C. Two of the side lengths add to a sum that is less than the third side length, so these lengths cannot be used to draw any triangles. D. Any three sides form many triangles with different angles, so these lengths can be used to draw infinitely many triangles. i need help quick please
Mathematics
1 answer:
netineya [11]3 years ago
6 0

Answer: C. Two of the side lengths add to a sum that is less than the third side length, so these lengths cannot be used to draw any triangles.

Explanation:

Those two sides in question are 8 and 12. They add to 8+12 = 20, but this sum is less than the third side 24. A triangle cannot be formed.

Try it out yourself. Cut out slips of paper that are 8 units, 12 units, and 24 units respectively. The units could be in inches or cm or mm based on your preference.

Then try to form a triangle with those side lengths. You'll find that it's not possible. If we had the 24 unit side be the horizontal base, so to speak, then we could attach the 8 and 12 unit lengths on either side of this horizontal piece. But then no matter how we rotate those smaller sides, they won't meet up to form the third point for the triangle. The sides are simply too short. Other possible configurations won't work either.

As a rule, the sum of any two sides of a triangle must be larger than the third side. This is the triangle inequality theorem.

That theorem says that the following three inequalities must all be true for a triangle to be possible.

  • x+y > z
  • x+z > y
  • y+z > x

where x,y,z are the sides of the triangle. They are placeholders for positive real numbers.

So because 8+12 > 24 is a false statement, this means that a triangle is not possible for these given side lengths. Therefore, 0 triangles can be formed.

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The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f
WITCHER [35]

Answer:

Part 1) A=36(\pi-2)\ cm^2

Part 2) P=6(\pi+2\sqrt{2})\ cm

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)

we have that the base and the height of triangle is equal to the radius of the circle

r=12\ cm\\b=12\ cm\\h=12\ cm

substitute

A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2

simplify

Factor 36

A=36(\pi-2)\ cm^2

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r

substitute the given values

C=\frac{1}{2}\pi (12)\\C=6\pi\ cm

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm

simplify

AC=12\sqrt{2}\ cm

Find the perimeter

P=(6\pi+12\sqrt{2})\ cm

simplify

Factor 6

P=6(\pi+2\sqrt{2})\ cm

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The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

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and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
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