Question

supposed a and b are the 2 sides of a right triangle such that satisfying value of :

Answer 1

\[\sqrt{a^{2}-4a \sqrt{3}+21} + \sqrt{2b^{2} - 4b \sqrt{2}+20} = 7\]

Answer 2

the question is determine the perimeter of triangle

Answer 3

here the choices :
A. sqrt(23) + 2 + sqrt(3)
B. 2sqrt(3) + sqrt(2) + 3sqrt(3)
C. 2sqrt(3) + 2 + sqrt(14)
D. 2sqrt(3) + sqrt(2) + sqrt(14)
E. 4sqrt(3) + 2sqrt(2) + sqrt(14)

Answer 4

i dont know the start to do this

Answer 5

if you hit the alt key and then the letter v it makes the sqrt sign √

Answer 6

simply complete the square

Answer 7

\[\sqrt{a^{2}-4a \sqrt{3}+21} + \sqrt{2b^{2} - 4b \sqrt{2}+20} = 7\]

\[\sqrt{(a-2\sqrt{3})^2+ 9} + \sqrt{2(b-\sqrt{2})^2+16} = 7\]

\[\implies a = 2\sqrt{3}~~, b=\sqrt{2}~~\color{gray}{\text{(why?)}}\]

Answer 8

nice, i got the hypotenuse of the triangle :
c = sqrt ((2sqrt(3))^2 + (sqrt(2))^2) = sqrt(12+2) = sqrt(14)
so, the perimeter a + b + c = 2sqrt(3) + sqrt(2) + sqrt(14)

thank you very much, @rational :)

Answer 9

yw:)