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