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Ratling [72]
3 years ago
7

The perimeter of a triangle is 44 cm.One of its sides is 4 cm smaller then the twice as big as the third side.Find the side of t

he triangle
Mathematics
1 answer:
lisabon 2012 [21]3 years ago
3 0

Answer:

The side of the triangle

a=8cm

b=20cm

c=16cm

Step-by-step explanation:

Perimeter = 44 cm

a= x

b=2x+4

c=2x

So the perimeter of a triangle is the sum of its sides

a+b+c=44cm\\x+(2x+4)+(2x)=44cm\\x+2x+2x-4=44cm\\5x=44-4\\x=\frac{40}{5} =8cm

a= 8cm

b=2*8+4=20 cm

c= 2*8= 16 cm

Check:

a+b+c=44cm

(8+20+16)cm=44cm

44cm=44cm

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Answer: (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

or  (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]

Taking square on both the sides , we get

(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0

Using quadratic formula : x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}

so, (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

since \sqrt{3}=1.732

so, (6+7(1.732),0)\text{ and }(6-7(1.732),0) are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

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a. The slope of the line passing through the points R(3, 5) and H(-1,2) is -3/4

b. The distance between two points (x₁, y₁) and (x₂,y₂) is 5 units

c.  The midpoint of the R(3, 5) and H(-1,2) is  (2, 4)

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a. The transformation rule is  (x,y) → (x + 1, y - 1)

b.

  • The x-coordinate is shifted 1 unit to the left and
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c. The image of B' the pre-image of B(5, 6) is (6, 5)

<h3 /><h3>7 a. How to find the slope of the line?</h3>

The slope of a line passing through the points (x₁, y₁) and (x₂,y₂) is m = (y₂ - y₁)/(x₂ - x₁)

Given that

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So, m = (y₂ - y₁)/(x₂ - x₁)

m = (2 - 5)/(-1 - 3)

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So, the slope of the line passing through the points R(3, 5) and H(-1,2) is -3/4

<h3>b. The distance between the points</h3>

The distance between two points (x₁, y₁) and (x₂,y₂) is d = √[(y₂ - y₁)² + (x₂ - x₁)²]

Given that

  • (x₁, y₁) = (3, 5) and
  • (x₂,y₂) = (-1, 2)

d = √[(y₂ - y₁)² + (x₂ - x₁)²]

d = √[(2 - 5)² + (-1 - 3)²]

d = √[(-3)² + (-4)²]

d = √[9 + 16]

d = √25

d = 5 units

So, the distance between two points (x₁, y₁) and (x₂,y₂) is 5 units

<h3>7 c How to find the midpoint of the R(3, 5) and H(-1,2) </h3>

The midpoint of the the points (x₁, y₁) and (x₂,y₂) is (x, y)  = [(x₁ + x₂)/2, (y₁ + y₂)/2]

Given that

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  • (x₂,y₂) = (-1, 2)

So, the midpoint (x, y)  = [(x₁ + x₂)/2, (y₁ + y₂)/2]

(x, y)  = [(3 + (-1))/2, (5 + 3)/2]

(x, y)  = [(3 - 1)/2, (5 + 3)/2]

(x, y)  = [4/2, 8/2]

(x, y)  = (2, 4)

So, the midpoint of the R(3, 5) and H(-1,2) is  (2, 4)

<h3>8. a The rule for the transformation of point A(1, 4) to point B(2, 3)</h3>

Given that point A(1, 4) and point B(2, 3) we see that point B(1 + 1, 4 - 1).

Let pont A be (x,y).

So, point B = (x + 1, y - 1)

So, the transformation rule is  (x,y) → (x + 1, y - 1)

<h3>b. Describe the transformation</h3>

Since the transformation rule is   (x,y) → (x + 1, y - 1), we see that

  • the x-coordinate is shifted 1 unit to the left and
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<h3>c. The image of B' the pre-image of B(5, 6)</h3>

Since the transformation rule is  (x,y) → (x + 1, y - 1) and point B is (5, 6), thus the image of B' is

(x,y) → (x + 1, y - 1)

(5,6) → (5 + 1, 6 - 1)

(5,6) → (6, 5)

So, the image of B' the pre-image of B(5, 6) is (6, 5)

Learn more about slope of a line here:

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Step-by-step explanation:

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