In the quadrilateral ABCD, ab ∥ CD, AC bisects ⊥ DAB, ad = BC, AC ⊥ BC, ⊥ DAB = ⊥ B, ab = 4cm

In the quadrilateral ABCD, ab ∥ CD, AC bisects ⊥ DAB, ad = BC, AC ⊥ BC, ⊥ DAB = ⊥ B, ab = 4cm


It is an isosceles trapezoid with other three sides of 2cm and perimeter of 10cm



Given that the volume of an inscribed cube of a sphere is 8, what is the volume of the sphere


The volume of inscribed cube is 8
So the side length of the cube is 2
The opposite vertex angle is the diameter of the circle
The diagonal of each square of the cube is 2
The diameter of the circle is 2 and 3
The radius is root 3
Sphere volume v = 4 π R & # 179 / 3
V = 4 radical 3 π



If the length of a cuboid is increased by 5cm, the surface area of the cuboid will be increased by 80cm 2. Calculate the surface area of the original cuboid
Why are my formulas different+_+


The increased surface area is the area of four rectangles whose length is 5cm and width is the side length of the square cross section of the original cuboid. The area of such a rectangle is 80 / 4 = 20 square centimeter. The length of the rectangle is 5cm and the width is 20 / 5 = 4cm. Thus, the square cross section area of the original cuboid is 4 × 4



It is known that the length of a cuboid is 4cm, the width is 3cm, and the height is 5cm


The length, width and height of the cuboid are 4cm, 3cm and 5cm respectively. (1) the sum of the edge lengths of the cuboid is 4 × (4 + 3 + 5) = 48CM, so the sum of all the edge lengths of the cuboid is 48CM. (2) the surface area is 2 × (4 × 3 + 4 × 5 + 3 × 5) = 2 × 47 = 94cm2, so the surface area of the cuboid is 94cm2