ABCD-EFGH in a figure is a box of rectangular prism 60cm in length, 80cm in width and 40cm in height and there is no lid.
There is a 40cm high pillar is straightly erected on the intersection M of a diagonal line of the bottom.
The vertex O of the pillar and chalk are tied up with string 50cm in length.
The length of MG is 50cm.
Pi is assumed to be 3.14.
(1) Find the area of the part painted with chalk in the outside of the box.
(2) Find the area of the part painted with chalk in the inside of the box.
There is a 40cm high pillar is straightly erected on the intersection M of a diagonal line of the bottom.
The vertex O of the pillar and chalk are tied up with string 50cm in length.
The length of MG is 50cm.
Pi is assumed to be 3.14.
(1) Find the area of the part painted with chalk in the outside of the box.
(2) Find the area of the part painted with chalk in the inside of the box.
Answer
Both the length of the string and the length of the diagonal line at the bottom is 50 cm.
The most length that chalk reaches on the outside of the box is 50 cm.
The part painted with chalk is an inner side of the circle centering on O in Fig. 1 and it is turns into the shadow area except the inside of rectangle ABCD.
The area is 50 × 50 × 3.14 - 60 × 80 = 3050cm2.
The length of the pillar is 40cm and the length of the string is 50cm.
Thus, when you look at the box from the direction of ABFE, the most distant place that chalk reaches on the bottom of the box is the place of 30cm from O, as shown in Fig. 2.
Next, the range side face BCGF can be painted is checked.
The chalk passes along the middle point P of the side FG.
Furthermore, since OB = OC = 50cm, it passes along the points B and C.
Thus, the inner side of the semicircle BCP with BC a diameter can be painted as shown in Fig. 4.
Next, the painted range is checked in side face ABFE.
The most distant place from O in side face ABFE is the point Q in Fig. 5 which looked at the box from the direction of BCGF.
As shown in Fig. 6, the inner side of the semicircle ABQ with radius of AR = BR = RQ = 30cm can be painted.
That is, on all the sides of the box, the area of the range painted is the area of every two semicircles with BC and AB a diameter.
Therefore, the area of the part painted is (area of circle centering on M with a radius of 30 cm) +(area of semicircle with BC a diameter) × 2 + (area of semicircle with AB a diameter) × 2.
It becomes 30 × 30 × 3.14 + 40 × 40 × 3.14 / 2 × 2 + 30 × 30 × 3.14 / 2 × 2 = 10676cm2.
(1) 3050 cm2
(2) 10676 cm2
(2) 10676 cm2
Solution
(1) The development view of the box as shown in Fig. 1 is drawn.
Both the length of the string and the length of the diagonal line at the bottom is 50 cm.
The most length that chalk reaches on the outside of the box is 50 cm.
The part painted with chalk is an inner side of the circle centering on O in Fig. 1 and it is turns into the shadow area except the inside of rectangle ABCD.
The area is 50 × 50 × 3.14 - 60 × 80 = 3050cm2.
(2) The part painted in the inner side of the box is considered by dividing a box into the bottom surface and the side surface.
The length of the pillar is 40cm and the length of the string is 50cm.
Thus, when you look at the box from the direction of ABFE, the most distant place that chalk reaches on the bottom of the box is the place of 30cm from O, as shown in Fig. 2.
As shown in Fig. 3, the part when the box is looked at from the top is the inside of a circle centering M with a radius of 30cm.
Next, the range side face BCGF can be painted is checked.
The chalk passes along the middle point P of the side FG.
Furthermore, since OB = OC = 50cm, it passes along the points B and C.
Thus, the inner side of the semicircle BCP with BC a diameter can be painted as shown in Fig. 4.
Next, the painted range is checked in side face ABFE.
The most distant place from O in side face ABFE is the point Q in Fig. 5 which looked at the box from the direction of BCGF.
As shown in Fig. 6, the inner side of the semicircle ABQ with radius of AR = BR = RQ = 30cm can be painted.
That is, on all the sides of the box, the area of the range painted is the area of every two semicircles with BC and AB a diameter.
Therefore, the area of the part painted is (area of circle centering on M with a radius of 30 cm) +(area of semicircle with BC a diameter) × 2 + (area of semicircle with AB a diameter) × 2.
It becomes 30 × 30 × 3.14 + 40 × 40 × 3.14 / 2 × 2 + 30 × 30 × 3.14 / 2 × 2 = 10676cm2.
