$If the sides of a triangle are a^{2} - 3 a b + 8 units, 4 a^{2} + 5 a b - 3 units and 6 - 3 a^{2} + 4 a b units, then its perimeter is ____ units.$

$2 a^{2} + 6 a b + 11$

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$2 a^{2} + 5 a b + 11$

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$a^{2} + 4 a b + 11$

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$2 a^{2} + 6 a b + 12$

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Solution

The correct option is A
$2 a^{2} + 6 a b + 11$

$Given sides of the triangle are (a^{2} - 3 a b + 8) units, (4 a^{2} + 5 a b - 3) units and (6 - 3 a^{2} + 4 a b) units.$

$Perimeter of a triangle = Sum of all three sides$
$= (a^{2} - 3 a b + 8) + (4 a^{2} + 5 a b - 3) + (6 - 3 a^{2} + 4 a b)$

(Grouping the like terms)
$= (a^{2} + 4 a^{2} - 3 a^{2}) + (- 3 a b + 5 a b + 4 a b) + (8 - 3 + 6)$
$= 2 a^{2} + 6 a b + 11$ units

$Hence, the perimeter of the triangle is 2 a^{2} + 6 a b + 11 units.$

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