Here, side of the square = 6 cm
∴ Area of the square = Side x Side
= 6 cm x 6 cm
= 36 cm2
Since, [Area of the rhombus] = [Area of the square]
∴ Area of the rhombus =
∵ Side of the square = 15 cm
∴ Perimeter of the square = 4 x 15 cm = 60 cm
Since, [Perimeter of the rectangle] = [Perimeter of the square]
∴ [Perimeter of the rectangle] = 60 cm
Let length of the rectangle = x cm
∴ 2[x + 10] = 60 [∵ Breadth of the rectangle = 10 cm]
2x + 20 = 60
2x = 60 - 20 = 40
Length of the rectangle = 20 cm
Now, Area of the rectangle = Length x Breadth
= 20 x 10 cm2
= 200 cm2
Area of the square = Side x Side
= 15 x 15 cm2
= 225 cm2
∴ Difference in areas = 225 cm2 - 200 cm2 = 25 cm2
The edge (side) of the given cube = 2 cm
Since, there such cubes are joined, then
Total length (l) = (2+2+2) cm
= 6 cm
Breadth (b) = 2 cm
Height (h) = 2 cm
∴ Total surface area of the resultant cuboid
= 2[lb + bh + hl]
= 2[6 x 2 + 2 x 2 + 2 x 6] cm2
= 2[28] cm2
= 56 cm2
Thus, the required total surface area of the cuboid = 56 cm2.
Let the length of the diagonal BD = x cm
Area of the quadrilateral ABCD = 32 cm2
∵ Area of the quadrilateral = x (A diagonal) x (Sum of the length of the perpendicular on the diagonal from the opposite vertices)
∴ Area of the quadrilateral ABCD =
=