The (Example) and ①, ② and ③ in Fig. P are the figures which cut off a 3cm cube as shown in a lower figure at one plane and looked at the solid below a cut surface from the direction of A (right above) and B (front).
The portion into which the cut surface is in sight is shaded by slant line.
As for ①，② and ③, as shown in Fig. Q, write a cut surface to the sketch and shade by slant line.
Moreover, find the volume of the solid below a cut surface.
Noted that the volume of a pyramid is calculated by (base area) × (height) × 1/3.
The portion into which the cut surface is in sight is shaded by slant line.
As for ①，② and ③, as shown in Fig. Q, write a cut surface to the sketch and shade by slant line.
Moreover, find the volume of the solid below a cut surface.
Noted that the volume of a pyramid is calculated by (base area) × (height) × 1/3.
Answer
In the figure seen from the direction of B, it turns out that the face of the original cube remains in the trapezoid form. A cut surface becomes as it is shown in Fig. 1.
It can be found the volume of the solid below the cut surface by (base area) × (average of height).
It is 3 × 3 × {(3 + 2 + 2 + 1) /4} = 18cm3.
② The top and bottom surface on the cube remains after cutting are shown in Fig. 2.
The surface where the original cube remains in the figure seen from the direction of B has a right triangle of 3cm in length, 2cm in width.
A cut surface becomes a trapezoid as shown in Fig. 3.
The volume of the solid below the cut surface can be found by subtracting the volume of the solid above the cut surface from the volume of the original cube.
The volume of the solid above cut surface can be found by subtracting the volume of triangular pyramid A-EFG from the volume of the triangular pyramid A-BCD as shown in Fig. 3.
The homothetic ratio of △ABC and △AEF is BC : EF = 3cm : 1cm = 3 : 1.
Thus, AC : AF is also 3 : 1.
The length of FC is equivalent to 3 - 1 = 2 which is corresponding value of the ratio.
The length of AF which is equivalent to 1 is 3cm / 2 × 1 = 1.5cm.
The volume of big triangular pyramid is 3 × 3 / 2 × (3 + 1.5) × 1/3 = 6.75cm3.
The volume of small triangular pyramid is 1 × 1 / 2 × 1.5 × 1/3 = 0.25cm3.
The volume of the solid above the cut surface is 6.75 - 0.25 = 6.5cm3.
Therefore, the volume of a lower solid is 3 × 3 × 3 - 6.5 = 20.5cm3.
③ Fig. 4 expresses the bottom surface of the original cube which remains after cutting.
The surface of the original cube which remains in the seen figure from the direction of B is a right triangle 2cm in width.
A cut surface becomes a pentagon as shown in Fig. 5.
The volume of the solid below the cut surface is found by subtracting the volume both of triangular pyramid I-BFC and J-DEH from the volume of triangular pyramid A-KIJ as shown in Fig. 5.
I-BFC and J-DEH are congruent.
The homothetic ratio of △IFC and △DGC is FC : GC = 2cm : 1cm = 2 : 1.
Then IF : DG is also 2 : 1 and since DG = 1cm, IF=2cm.
The homothetic ratio of △ABL and △IBF is AL : IF = 3cm : 2cm = 3 : 2.
Thus, since LB : BF is also 3 : 2, BF = 3cm × 2/(3+2) = 1.2cm.
The volume of triangular pyramid A-KIJ is (3+2) × (3+2) / 2 × 3 × 1/3 = 12.5cm3.
The volume of triangular pyramid I-BFC is 2 × 1.2 / 2 × 2 × 1/3 = 0.8cm3.
The volume of triangular pyramid J-DEH is also 0.8 cm3.
Therefore, the volume of the solid below the cut surface is 12.5 - 0.8 × 2 = 10.9cm3.
① Cut surface : Fig.1 Volume : 18cm3
② Cut surface : Fig.3 Volume : 20.5cm3
③ Cut surface : Fig.5 Volume : 10.9cm3
② Cut surface : Fig.3 Volume : 20.5cm3
③ Cut surface : Fig.5 Volume : 10.9cm3
Solution
① As for the figure seen from the direction of A, only the cut surface is visible.
In the figure seen from the direction of B, it turns out that the face of the original cube remains in the trapezoid form. A cut surface becomes as it is shown in Fig. 1.
It can be found the volume of the solid below the cut surface by (base area) × (average of height).
It is 3 × 3 × {(3 + 2 + 2 + 1) /4} = 18cm3.
② The top and bottom surface on the cube remains after cutting are shown in Fig. 2.
The surface where the original cube remains in the figure seen from the direction of B has a right triangle of 3cm in length, 2cm in width.
A cut surface becomes a trapezoid as shown in Fig. 3.
The volume of the solid below the cut surface can be found by subtracting the volume of the solid above the cut surface from the volume of the original cube.
The volume of the solid above cut surface can be found by subtracting the volume of triangular pyramid A-EFG from the volume of the triangular pyramid A-BCD as shown in Fig. 3.
The homothetic ratio of △ABC and △AEF is BC : EF = 3cm : 1cm = 3 : 1.
Thus, AC : AF is also 3 : 1.
The length of FC is equivalent to 3 - 1 = 2 which is corresponding value of the ratio.
The length of AF which is equivalent to 1 is 3cm / 2 × 1 = 1.5cm.
The volume of big triangular pyramid is 3 × 3 / 2 × (3 + 1.5) × 1/3 = 6.75cm3.
The volume of small triangular pyramid is 1 × 1 / 2 × 1.5 × 1/3 = 0.25cm3.
The volume of the solid above the cut surface is 6.75 - 0.25 = 6.5cm3.
Therefore, the volume of a lower solid is 3 × 3 × 3 - 6.5 = 20.5cm3.
③ Fig. 4 expresses the bottom surface of the original cube which remains after cutting.
The surface of the original cube which remains in the seen figure from the direction of B is a right triangle 2cm in width.
A cut surface becomes a pentagon as shown in Fig. 5.
The volume of the solid below the cut surface is found by subtracting the volume both of triangular pyramid I-BFC and J-DEH from the volume of triangular pyramid A-KIJ as shown in Fig. 5.
I-BFC and J-DEH are congruent.
The homothetic ratio of △IFC and △DGC is FC : GC = 2cm : 1cm = 2 : 1.
Then IF : DG is also 2 : 1 and since DG = 1cm, IF=2cm.
The homothetic ratio of △ABL and △IBF is AL : IF = 3cm : 2cm = 3 : 2.
Thus, since LB : BF is also 3 : 2, BF = 3cm × 2/(3+2) = 1.2cm.
The volume of triangular pyramid A-KIJ is (3+2) × (3+2) / 2 × 3 × 1/3 = 12.5cm3.
The volume of triangular pyramid I-BFC is 2 × 1.2 / 2 × 2 × 1/3 = 0.8cm3.
The volume of triangular pyramid J-DEH is also 0.8 cm3.
Therefore, the volume of the solid below the cut surface is 12.5 - 0.8 × 2 = 10.9cm3.
