Question

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Select the correct answer.

LMN is an isosceles triangle. What is the approximate length of side LM, and what is the approximate perimeter of triangle LMN?

Round your answer to the nearest tenth.


A. The approximate length of side LM is 6 units, and the approximate perimeter of triangle LMN is 22.8 units.

B. The approximate length of side LM is 6 units, and the approximate perimeter of triangle LMN is 22 units.

C. The approximate length of side LM is 6.4 units, and the approximate perimeter of triangle LMN is 22.8 units.

D. The approximate length of side LM is 6.4 units, and the approximate perimeter of triangle LMN is 22 units.

Answer 1

The points are L(1,1), M(6,-3) and N(-4,-3)

Now we will use distance formula.
$Distance = \sqrt{ {( x_{1} - x_{2)}}^{2} +{( y_{1} - y_{2)}}^{2} }$

$LM = \sqrt{ {(1 - 6)}^{2} + {(1 + 3)}^{2} } = \sqrt{25 + 16} = \sqrt{41} = 6.4 \: units$

$MN = \sqrt{ {(6 + 4)}^{2} + {( - 3 + 3)}^{2} } = \sqrt{100 + 0} = \sqrt{100} = 10 \: units$

$LN= \sqrt{ {(1 + 4)}^{2} + {( 1 + 3)}^{2} } = \sqrt{25 + 16} = \sqrt{41} = 6.4 \: units$

Perimeter=LM+MN+LN=6.4+10+6.4 = 22.8 units.

So, option (C) is correct.✔✔✔

C. The approximate length of side LM is 6.4 units, and the approximate perimeter of ∆LMN is 22.8 units.

Answer 2

Answer:

The approximate length of side LM is 6.4 units, and the approximate perimeter of ∆LMN is 22.8 units.

Step-by-step explanation:

The points are L(1,1), M(6,-3), and N(-4,-3)

Now we will use the distance formula.

Step 1

Distance $=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

$L M=\sqrt{(1-6)^{2}+(1+3)^{2}}=\sqrt{25+16}=\sqrt{41}=6.4$units

$M N=\sqrt{(6+4)^{2}+(-3+3)^{2}}=\sqrt{100+0}=\sqrt{100}=10$ units

Step 2

$L N=\sqrt{(1+4)^{2}+(1+3)^{2}}=\sqrt{25+16}=\sqrt{41}=6.4$units

Therefore,

Perimeter=LM+MN+LN=6.4+10+6.4 = 22.8 units.

Hence, The approximate length of side LM is 6.4 units, and the approximate perimeter of ∆LMN is 22.8 units.

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