If a- b and c are the side lengths of the triangle such that a-b-c-a-b-c-5-4 - The perimeter of the triangle will be-A- 5 a - c B- 3b-cC- 10 a -9D- 10 a - c

The correct option is C a+b+c/a b c=5/4 Using componendo and dividendo (a+b+c)+(a b c)/(a+b+c) (a b c)=9/1 a=9(b+c) Perimeter of the triangle = a+b+c nbsp; ...

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Standard X

Mathematics

Area of a Triangle Given Its Vertices

If a, b and c...

Question

If a, b and c are the side lengths of the triangle such that $\frac{a + b + c}{a - b - c} = \frac{5}{4}$ .The perimeter of the triangle will be:

3(b+c)

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5(a+c)

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10(a+c)

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Solution

The correct option is C

$\frac{a + b + c}{a - b - c} = \frac{5}{4}$
Using componendo and dividendo
$\frac{(a + b + c) + (a - b - c)}{(a + b + c) - (a - b - c)} = \frac{9}{1}$
$a = 9 (b + c)$
Perimeter of the triangle = a+b+c
= $a + \frac{1}{9} a$
= $\frac{10}{9} a$ .

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If a, b and c are the side lengths of the triangle such that a+b+ca−b−c=54.The perimeter of the triangle will be:

The correct option is C a+b+ca−b−c=54 Using componendo and dividendo (a+b+c)+(a−b−c)(a+b+c)−(a−b−c)=91 a=9(b+c) Perimeter of the triangle = a+b+c = a+19a = 109a.

If a, b and c are the side lengths of the triangle such that $\frac{a + b + c}{a - b - c} = \frac{5}{4}$ .The perimeter of the triangle will be:

The correct option is C

$\frac{a + b + c}{a - b - c} = \frac{5}{4}$
Using componendo and dividendo
$\frac{(a + b + c) + (a - b - c)}{(a + b + c) - (a - b - c)} = \frac{9}{1}$
$a = 9 (b + c)$
Perimeter of the triangle = a+b+c
= $a + \frac{1}{9} a$
= $\frac{10}{9} a$ .