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3 years ago

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The ratio of the lengths of the sides of a triangle ABC is 3:4:6. M, N, and K are the midpoints of the sides. Perimeter of the △

MNK equals 5.2 in. Find the length of the sides of the △ABC. No picture.

1 answer:

juin [17]3 years ago

7 0

Given that in a trianlgle the sides AB, BC, CA are in the ratio 3:4:6.

Let AB = 3k, BC = 4k and CA = 6k.

Then perimeter =3k+4k+6k = 13k

M, N, K are mid points of the sides.

By mid point theorem MN = 3k/2, NK = 4k/2 and KM = 6k/2

Hence perimeter of MNK = 13k/2 =5.2 (given)

Solve for k

k=2(5.2)/13 = 0.8

Hence sides are

AB = 3k = 3(0.8) = 2.4 in

BC = 4k= 3.2 in

CA = 4.8 in

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Let us take a decimal number $2.14$ and a fraction $\frac{5}{20}$

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$2.14+\frac{5}{20} = \frac{214}{100} +\frac{5}{20}$

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$\frac{5}{20} + 2.14 = \frac{5}{20} + \frac{214}{100}$

$= \frac{5 \times 5}{20 \times 5} +\frac{214}{100} \\ \\= \frac{25}{100} + \frac{214}{100} \\ \\= 0.25 + 2.14 \\ \\ = 2.39$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

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For Associated Property of Addition

Let us take two same decimal numbers $2.14$ , $7.25$ and a fraction $\frac{5}{20}$

Now, according to the Associated property of Addition:

For any three numbers $a$ , $b$ and $c$

$a + (b+c) =(a+b) + c$

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$a + (b+c)$

$2.14 + (7.25 + \frac{5}{20}) = 2.14 + (\frac{725}{100} + \frac{5 \times 5}{20\times 5})$

$= 2.14 + (\frac{725}{100} + \frac{25 }{100})$

$= 2.14 + (\frac{750}{100} )$

$= \frac{214}{100} + \frac{750}{100}$

$= \frac{964}{100}$

$=9.64$

The Right hand side:

$(a + b )+c$

$(2.14 + 7.25 )+ \frac{5}{20}= ( 9.39 ) + \frac{5 }{20}$

$= 9.39 + \frac{5 \times 5 }{20 \times 5}$

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$= 9.39 + 0.25$

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Thus,

$(2.14 + (7.25 + \frac{5}{20} )= (2.14 + 7.25 )+ \frac{5}{20}$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Associative Property of Addition is satisfied.

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