cube ABCD EFGH with a rib length of 4 cm the surface area of the five side fields formed by ABF BCF CDF ADF and ABCD are?
Answers
Step-by-step explanation:
The five side surface area formed by ABF, BCF, CDF, ADF and ABCD is (32 + 16√2) cm² or 54.63 cm².
The cube is constructed of special three-dimensional space because it is composed of 6 square-shaped sides so that all the ribs have the same length.
In addition, the sides of the cube are facing each other so that the adjacent sides form a 90 ° angle which makes it easy for us to do any calculation with the phythagorean theorem and trigonometric comparisons.
Not only that, the calculation of diagonal sides and diagonal space of a cube is special. For cubes with a length of s, then the diagonal length of the sides is formulated with s√2 and the diagonal length of the space is formulated with s√3.
In the above problem, a combination of diagonal sides and diagonal cubes form a new space in which we are asked to calculate the area of the base.
Well, to be clearer, see the following discussion.
DISCUSSION:
Look again at the question above.
An ABCD cube is known. EFGH has 4 cm ribs. From the diagonal side of the AF and CF and the diagonal of the space DF form a square quadratic space F.ABCD with the side ABCD as the base and 4 right triangle ABF, BCF, CDF and ADF as the upright side.
Why is the field of CDF and ADF said right angles? Because the two fields are formed from a diagonal space and a diagonal side where the diagonals of each side - perpendicular to the base (cube ribs) caused by the sides of the ribs and diagonals form a 90 ° angle.
Back to the topic, the surface area of quadrilateral limas is the result of the total area of the base side with an area of 4 upright sides.
So, we must calculate one by one the area of the pyramid compiler.
Area ABCD (square) = s × s
Area ABCD = AB × BC
Area of ABCD = 4 cm × 4 cm
Area ABCD = 16 cm²
Area of ABF (right triangle) = ½ × a × t
Area ABF = 1/2 × AB × BF
The area of ABF = ½ × 4 cm × 4 cm
Area of ABF = 8 cm²
Area of BCF (right triangle) = ½ × a × t
Area of BCF = ½ × BC × BF
Area of BCF = ½ × 4 cm × 4 cm
Area of BCF = 8 cm²
Area of CDF (right triangle) = ½ × a × t
Area of CDF = ½ × CD × CF
The area of CDF = 1/2 × 4 cm × 4√2 cm
[remember, for cubes with lengths of s, then the diagonal length of the sides is formulated with s√2]
CDF area = 8√2 cm²
Area of ADF (right triangle) = ½ × a × t
Area of ADF = 1/2 × AD × AF
The area of ADF = 1/2 × 4 cm × 4√2 cm
[remember, for cubes with lengths of s, the diagonal length of the sides is formulated with s√2]
Area of ADF = 8√2 cm²
Thus, the surface area of the F.ABCD rectangular pyramid is:
L.ABCD + L.ABF + L.BCF + L.CDF + L.ADF
= 16 cm² + 8 cm² + 8 cm² + 8√2 cm² + 8√2 cm²
= (32 + 16√2) cm²
or
= 54.63 cm³
So, in the ABCD.EFGH cube, the five-sided surface area formed by ABF, BCF, CDF, ADF and ABCD is (32 + 16√2) cm² or 54.63 cm².
SOLUTION
Given,
the edge of the cube is 4cm.
When two cubes are joined end to end, then the resulting solid is a cuboid.
Length of the cuboid,l= 2× edge of cube= 2×4= 8cm
Breadth of the cuboid,b=Edge of the cube = 4cm
Height of the cuboid,h = Edge of the cube=4cm
Surface area of the cuboid
=) 2(lb+bh+lh)
=) 2(8cm×4cm+4cm×4cm+4cm×8cm)
=) 2(32cm^2 + 16cm^2 +32cm^2)
=) 2×80cm^2
=) 160cm^2
Thus, the surface area of the resulting cuboid is 160cm^2
hope it helps ☺️