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lyudmila [28]
3 years ago
5

A square has the sides length 2k-1 units. An equilateral triangle has the sides of the length k+2 units. the square and the tria

ngle have the same perimeter. what is the value of k?
I got k=3 I think I got wrong can someone please correct me???
Mathematics
1 answer:
AlladinOne [14]3 years ago
8 0
I think that you assumed that:
2k - 1 = k + 2

However we can't do this, because a square has 4 sides whilst a triangle only has 3 sides. 
This means that we can say
4(2k - 1) = 3(k + 2) 

We multiply by 4 due to 4 sides of the square, and by 3 due to the 4 sides of the triangle.

Lets expand the brackets, and solve it:

4(2k - 1) = 3(k + 2)
8k -4 = 3k + 6
5k -4 = 6    ( subtract both sides by 3x to collect the x values)
5k = 10      (add both sides by 4 to get the x's alone)
k = 2           (divide both sides by 5 to get what just x is)

So k = 2 units
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An object is launched from a platform. Its height (in meters), xxx seconds after the launch, is modeled by: h(x)=-5x^2+20x+60h(x
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Base = 24 cm or 10cm

Step-by-step explanation:

REMEMBER:
An isosceles triangle ABC with base BC = ‘b' & height AD = ‘h' & its equal sides =13 cm & area = 60 cm²

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A=\frac{bh_b}{2}

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There are 2 solutions for b

b=2\sqrt{2}   \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2+\sqrt{13^4-4*60^2} } } ≈ 10cmb=2\sqrt{2}   \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2-\sqrt{13^4-4*60^2} } }=24cm

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Area of a triangle = 1/2 * b * h = 60

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=> b^4 - 676 b² + 57600 = 0

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=> b²= 676 +- √(226576) /2

=> b² = (676 +- 476 )/2

=> b² = 1152/2 , 200 /2

=> b² = 576 , 100

=> b = 24, 10

So, Base = 24 cm or 10cm

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