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3 years ago

The perimeter of a triangle is 32 feet. One side of the triangle is 3 times the second side. The third side is 12 feet longer

Mathematics

1 answer:

Ivenika [448]3 years ago

4 0

We have a triangle whose perimeter is 32 feet. Some information regarding the sides of the triangle is given:

GiveN:

One side is 3 times the second side.
Third side is 12 feet longer than the second side.

As we can see that, two sides of the triangle are defined the second sides. It means First and third side can be expressed in the form of second side.

So let the second side be x

Then,

First side = 3x
And, third side = x + 12

We know that, perimeters is the sum of the lengths of all sides of the triangle, So it can be written as:

$3x + x + x + 12 = 32$

Solving it further,

$5x + 12 = 32$

Subtracting 12 from both sides,

$5x + 12 - 12 = 32 - 12$

$5x = 20$

Dividing 5 from both sides,

$\frac{5x}{5} = \frac{20}{5}$

$x = 4$

Then,

First side = 3x = 12 feet
Second side = 4 feet
Third side = x + 12 = 16 feet

And we are done with the answer !!

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PLEASE HELP! Explain your answer, THANK YOU!

Misha Larkins [42]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.