Question
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
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
collins
diagonal=[5-root(12)^2]+(10/root(3)^2
corrected......Now solve
corrected......Now solve
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
collins
yeah i read perimeter.....you right