Asked by kaylen
a rectangle has length (5-squareroot12)m and breadth (4+6/squareroot3)m. express in the form a+bsquareroot3, where a and b are integers,
i)the area of the rectangle
ii)the square of the length of the diagonal of the rectangle
i)the area of the rectangle
ii)the square of the length of the diagonal of the rectangle
Answers
Answered by
collins
area=2(L+B)
A=2[(5-root(12)+10/root(3)
A=2[(5-root(12)]+10/root(3)
A=2[(5-root(12)root(3)+10/root(3)]
A=2[(5root(3)-root(36)+10/root(3)
A=2[5root(3))-6+10)/root(3)
A=(10root(3)+8root(3)
A=18root(3)/3
A=6root(3)........
Diagonal=[5-root(3)]^2+10/root(3)^2
dere go just play with that
A=2[(5-root(12)+10/root(3)
A=2[(5-root(12)]+10/root(3)
A=2[(5-root(12)root(3)+10/root(3)]
A=2[(5root(3)-root(36)+10/root(3)
A=2[5root(3))-6+10)/root(3)
A=(10root(3)+8root(3)
A=18root(3)/3
A=6root(3)........
Diagonal=[5-root(3)]^2+10/root(3)^2
dere go just play with that
Answered by
collins
diagonal=[5-root(12)^2]+(10/root(3)^2
corrected......Now solve
corrected......Now solve
Answered by
Reiny
area = LW , not 2(L+W)
that would be perimeter
area = (5 - √12)(4 + 6/√3)
= 20 + 30/√3 - 4√12 - 6√4
= 20 + 30/√3 *√3/√3 - 8√3 - 12
= 23 + 10√3 - 8√3
= 8 + 2√3
D^2 = (5-√12)^2 + (4+6/√3)^2
= (5 - 2√3)^2 + (4 + 2√3)^2
= 25 - 20√3 + 12 + 16 + 16√3 + 12
= 65 - 4√3
that would be perimeter
area = (5 - √12)(4 + 6/√3)
= 20 + 30/√3 - 4√12 - 6√4
= 20 + 30/√3 *√3/√3 - 8√3 - 12
= 23 + 10√3 - 8√3
= 8 + 2√3
D^2 = (5-√12)^2 + (4+6/√3)^2
= (5 - 2√3)^2 + (4 + 2√3)^2
= 25 - 20√3 + 12 + 16 + 16√3 + 12
= 65 - 4√3
Answered by
collins
yeah i read perimeter.....you right
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.