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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

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

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

(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

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

Suggest Corrections

Similar questions

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

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

State True or False:

On subtracting $a^{2} + a b + b^{2}$ from $4 a^{2} - 3 a b + 2 b^{2}$ , the answer is $3 a^{2} - 4 a b + b^{2}$ .

Add:
$8 a b, - 5 a b, 3 a b, - a b$

\frac{4 a^{2} - 8 a + 5 a b - 10 b}{4 a + 5 b} = . . . . . . . . . . . . . .

Join BYJU'S Learning Program