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Given:
Length of each side of cube = 5 cm
Formula used:
Surface area of the cuboid = 2(LB + BH + HL)
Where,
L = length of the cuboid
B = breadth of the cuboid
H = height of the cuboid
Calculation:
Now, 3 cubes are joined end to end
The length of the resulting cuboid will be = 5 + 5 + 5 = 15 cm
But, the breadth and height will remain the same.
Breadth = 5 cm and Height = 5 cm
Surface area of the cuboid = 2(LB + BH + HL)
⇒ Surface area of the cuboid = 2(15 × 5 + 5 × 5 + 5 × 15) = 2(75 + 25 + 75)
⇒ Surface area of the cuboid = 350 cm2
∴ The surface area of the resulting solid is 350 cm2
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A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter
In fig. cone ABC is cut out by a plane parallel to the base FG. DEFG is the frustum so obtained. Let O be the centre of the base of the cone and O’ the centre of the base of the frustum.
It is given that ∠BAC = 60° ∠OAC = 30°
In right triangle AOC, tan
And, C =
Height of the frustum = P'O =
volume of the frustum =
Radius of the wire =
Volume of wire =