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

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Solution

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

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

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

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

The sides of a triangle are $a^{2} - 3 a b + 8$ , $4 a^{2} + 5 a b - 3$ and $6 - 3 a^{2} + 4 a b$ . Then its perimeter is

^{2}

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4 a^{2} + 5 b^{2} - 6 a b, 3 a b, 6 a^{2} - 2 b^{2}

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