A rectangle departs from the position in a figure and is moving in the direction of an arrow at 2 cm/s.
(1) Find the time when the rectangle is completely inside of the triangle.
(2) Find the area of the portion of the rectangle out of a triangle 12 seconds after leaving.
(3) Find the time when the area of the portion of the rectangle which is contained in the triangle is 30 cm2 or more.
Solution
(1) While the rectangle is moving the length of the arrow in Fig.1, it is completely inside of the triangle.
In order to find this length, the length of DE is to be found.
△ABC and △ADE are homothetic and the ratio of height is 24 : (24 - 6) = 24 : 18 = 4 : 3.
Since BC : DE = 4 : 3, the length of DE is 20 cm × 3/4= 15 cm.
Thus, the length of the arrow is 15 - 12 =3 cm.
The time to be found is 3 / 2 = 1.5 seconds since a rectangle moves at 2 cm/s.
(3) When a part of area of the rectangle in a triangle is 30 cm2, the sum of the length of a upper base and a lower base is 30 × 2 / 6 = 10cm.
According to (2), the difference of the length of DG and BF is 4 - 1.5 = 2.5 cm.
This difference of length does not change while the form where a rectangle and a triangle overlap is a trapezoid.
A part of area of the rectangle in a triangle becomes 30 cm2 for the first time is at the time of BF= (10 + 2 - 5) / 2 = 6.25cm.
It is again becomes 30 cm2 is at the time of QC = 6.25cm as shown in Fig. 3.
That is, the time to find is the time that the rectangle moved the length of the arrow in Fig. 3.
The length is 12 + (20 - 6.25 × 2) = 19.5 cm.
Time to move this length is for 19.5 / 2 = 9.75 seconds.
(1) Find the time when the rectangle is completely inside of the triangle.
(2) Find the area of the portion of the rectangle out of a triangle 12 seconds after leaving.
(3) Find the time when the area of the portion of the rectangle which is contained in the triangle is 30 cm2 or more.
Answer
(1) 1.5 seconds
(2) 55.5 cm2
(3) 9.75 seconds
(2) 55.5 cm2
(3) 9.75 seconds
(1) While the rectangle is moving the length of the arrow in Fig.1, it is completely inside of the triangle.
In order to find this length, the length of DE is to be found.
△ABC and △ADE are homothetic and the ratio of height is 24 : (24 - 6) = 24 : 18 = 4 : 3.
Since BC : DE = 4 : 3, the length of DE is 20 cm × 3/4= 15 cm.
Thus, the length of the arrow is 15 - 12 =3 cm.
The time to be found is 3 / 2 = 1.5 seconds since a rectangle moves at 2 cm/s.
(2) In 12 seconds, a rectangle moves 2 cm × 12= 24 cm.
Fig. 2 is a figure of 12 seconds after.
BF = 24 - 20 = 4 cm.
The area to find is the area which trapezoid DBFG area which is the inside of triangle is subtracted from the rectangular area.
In Fig. 2, △ABI and △HBF are homothetic and as BI : AI =10: 24 = 5 : 12, BF : HF = 5 : 12.
HF = 4 cm × 12/5= 48/5 cm. HG=48/5 - 6 = 18/5 cm.
Since △HBF and △HDG are also homothetic, DG = HG × 5/12 = 18/5 × 5/12 = 1.5cm.
Therefore, since the area of trapezoid DBFG is (1.5 + 4) × 6 / 2 = 16.5 cm2, the area to be found is 6 × 12 - 16.5=55.5cm2.
Fig. 2 is a figure of 12 seconds after.
BF = 24 - 20 = 4 cm.
The area to find is the area which trapezoid DBFG area which is the inside of triangle is subtracted from the rectangular area.
In Fig. 2, △ABI and △HBF are homothetic and as BI : AI =10: 24 = 5 : 12, BF : HF = 5 : 12.
HF = 4 cm × 12/5= 48/5 cm. HG=48/5 - 6 = 18/5 cm.
Since △HBF and △HDG are also homothetic, DG = HG × 5/12 = 18/5 × 5/12 = 1.5cm.
Therefore, since the area of trapezoid DBFG is (1.5 + 4) × 6 / 2 = 16.5 cm2, the area to be found is 6 × 12 - 16.5=55.5cm2.
(3) When a part of area of the rectangle in a triangle is 30 cm2, the sum of the length of a upper base and a lower base is 30 × 2 / 6 = 10cm.
According to (2), the difference of the length of DG and BF is 4 - 1.5 = 2.5 cm.
This difference of length does not change while the form where a rectangle and a triangle overlap is a trapezoid.
A part of area of the rectangle in a triangle becomes 30 cm2 for the first time is at the time of BF= (10 + 2 - 5) / 2 = 6.25cm.
It is again becomes 30 cm2 is at the time of QC = 6.25cm as shown in Fig. 3.
That is, the time to find is the time that the rectangle moved the length of the arrow in Fig. 3.
The length is 12 + (20 - 6.25 × 2) = 19.5 cm.
Time to move this length is for 19.5 / 2 = 9.75 seconds.
