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Answer :
To solve this problem, we need to set up an equation for the perimeter of the isosceles triangle. We know that an isosceles triangle has two sides of equal length. Let's call the length of these equal sides [tex]\( x \)[/tex].
We are given:
- The perimeter of the triangle is 7.5 meters.
- The shortest side [tex]\( y \)[/tex] is 2.1 meters.
For an isosceles triangle, the perimeter is the sum of all its sides, which is:
[tex]\[ 2x + y = 7.5 \][/tex]
Since we know [tex]\( y = 2.1 \)[/tex], we can substitute this value into the equation:
[tex]\[ 2x + 2.1 = 7.5 \][/tex]
This equation needs to be solved to find [tex]\( x \)[/tex].
From the options provided in the question, the correct equation that matches this setup is:
[tex]\[ 2.1 + 2x = 7.5 \][/tex]
To solve for [tex]\( x \)[/tex], we subtract 2.1 from both sides of the equation:
[tex]\[ 2x = 7.5 - 2.1 \][/tex]
[tex]\[ 2x = 5.4 \][/tex]
Now, divide both sides by 2 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{5.4}{2} \][/tex]
[tex]\[ x = 2.7 \][/tex]
So, the correct equation to use is [tex]\( 2.1 + 2x = 7.5 \)[/tex], which is also the correct option provided.
We are given:
- The perimeter of the triangle is 7.5 meters.
- The shortest side [tex]\( y \)[/tex] is 2.1 meters.
For an isosceles triangle, the perimeter is the sum of all its sides, which is:
[tex]\[ 2x + y = 7.5 \][/tex]
Since we know [tex]\( y = 2.1 \)[/tex], we can substitute this value into the equation:
[tex]\[ 2x + 2.1 = 7.5 \][/tex]
This equation needs to be solved to find [tex]\( x \)[/tex].
From the options provided in the question, the correct equation that matches this setup is:
[tex]\[ 2.1 + 2x = 7.5 \][/tex]
To solve for [tex]\( x \)[/tex], we subtract 2.1 from both sides of the equation:
[tex]\[ 2x = 7.5 - 2.1 \][/tex]
[tex]\[ 2x = 5.4 \][/tex]
Now, divide both sides by 2 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{5.4}{2} \][/tex]
[tex]\[ x = 2.7 \][/tex]
So, the correct equation to use is [tex]\( 2.1 + 2x = 7.5 \)[/tex], which is also the correct option provided.
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Rewritten by : Jeany