1.
A) A square piece of paper, PQRS, has side length 40 cm. The page is grey on one side and white on the other side. Point M is the midpoint of side PQ and point N is the midpoint of side PS. The paper is folded along MN so that P touches the paper at the point P'
Point T lies on QR and point U lies on SR such that TU is parallel to MN, and
when the paper is folded along TU, the point R touches the paper at the point R' on MN.
What is the area of hexagon NMQTUS?
Here are some known properties of the diagonals of a square that may be useful:
• the diagonals are equal in length; and
• the diagonals right bisect each other; and
• the diagonals bisect the corner angles.
b) A rectangular piece of paper, PQRS, has PQ = 30 cm and P S = 40 cm. The paper has grey lines on one side and is plain white on the other. The paper is folded so that the two diagonally opposite corners P and R coincide.
This creates a crease along line segment AC, with A on PS and C on QR.
Determine the length of AC.
This question can be found at Waterloo problem of the week A = Grade 9/10
B = Grade 11/12
a) To find the area of hexagon NMQTUS, we need to break it down into smaller shapes that we can calculate.
First, let's look at triangle NMU. Since MN is the folded edge, it is also the height of the triangle. We know that the side length of the square is 40 cm, so MN is half of that, which is 20 cm. The base of the triangle is MU, which is the same as the side length of the square, 40 cm.
The formula for the area of a triangle is A = 0.5 * base * height. Plugging in the values, we get A = 0.5 * 40 cm * 20 cm = 400 cm^2.
Next, let's find the area of trapezoid QTUS. The formula for the area of a trapezoid is A = 0.5 * (sum of parallel sides) * height. In this case, the parallel sides are QT and US, both of which are the same as the side length of the square, 40 cm. The height of the trapezoid is TU. Since TU is parallel to MN, it is also 20 cm.
Plugging in the values, we get A = 0.5 * (40 cm + 40 cm) * 20 cm = 800 cm^2.
Finally, to find the area of hexagon NMQTUS, we add the areas of the triangle and trapezoid: 400 cm^2 + 800 cm^2 = 1200 cm^2.
b) To determine the length of AC, we need to use the properties of diagonals in a rectangle. We know that the diagonals of a rectangle are equal in length and bisect each other.
Since PR is a diagonal of the rectangle, it is equal in length to PQ or PS. Therefore, PR = PQ = 30 cm.
When folding the paper along PR, the crease line AC coincides with the diagonal PR. Thus, AC must also be equal to PR.
Therefore, the length of AC is 30 cm.
a) To find the area of hexagon NMQTUS, we need to break it down into smaller shapes.
1. Start by finding the area of square PQRS. Since the side length is 40 cm, the area is equal to (side length)^2 = 40^2 = 1600 cm^2.
2. The folded paper creates a right triangle NMQ. The base of this triangle is NM, half the side length of the square, which is 40/2 = 20 cm. The height of the triangle is also 20 cm since M is the midpoint of PQ. Therefore, the area of triangle NMQ is (base x height)/2 = (20 x 20)/2 = 200 cm^2.
3. The paper is folded along TU, creating a trapezoid QTUS. The two parallel sides, QT and US, are equal in length and have a length of 20 cm each, as TU is parallel to MN. The height of the trapezoid is equal to NM, which is 20 cm. Therefore, the area of trapezoid QTUS is (sum of parallel sides x height)/2 = ((20 + 20) x 20)/2 = 400 cm^2.
4. The hexagon NMQTUS can be divided into triangle NMQ and trapezoid QTUS. Therefore, the area of hexagon NMQTUS is equal to the area of triangle NMQ plus the area of trapezoid QTUS = 200 cm^2 + 400 cm^2 = 600 cm^2.
So, the area of hexagon NMQTUS is 600 cm^2.
b) To determine the length of AC, we can use the properties of diagonals of a rectangle.
1. The diagonals of a rectangle are equal in length. In this case, the diagonal PR would be equal to the length of AC.
2. The diagonals of a rectangle bisect each other. Since AC is a fold that coincides the diagonally opposite corners P and R, AC will bisect the diagonal PR into two equal parts.
3. The length of diagonal PR can be found using the Pythagorean theorem. The length of PQ is given as 30 cm and the length of PS is given as 40 cm.
Applying the Pythagorean theorem, we have:
PR^2 = PQ^2 + PS^2
PR^2 = 30^2 + 40^2
PR^2 = 900 + 1600
PR^2 = 2500
PR = √2500
PR = 50 cm
Since AC bisects PR, the length of AC is half of the length of PR.
AC = PR/2
AC = 50/2
AC = 25 cm
So, the length of AC is 25 cm.
To find the area of hexagon NMQTUS, we first need to understand the properties of the shape and use the given information to determine its dimensions.
1. Start by drawing a square with side length 40 cm and label its vertices as P, Q, R, and S.
2. Draw the line segment MN through the midpoint of side PQ and label the midpoint as M.
3. Fold the square along the line MN, making P touch the paper at P'.
4. Now, locate the points T and U on the folded side QR such that TU is parallel to MN.
5. Note that since MN is the median of the right triangle PQM, it is also the perpendicular bisector of side QR. This means that QR is split into two equal segments, each measuring 40 cm/2 = 20 cm.
6. Extend the points T and U until they meet the other side of the folded paper at A and C, respectively. These points will form a hexagon, NMQTUS.
7. Since AC is a vertical line after folding, it will also be the height of the hexagon.
To find the area of the hexagon:
1. Calculate the length of AC. Since AC is a vertical line, it is equal to the distance between the folded sides. In this case, PQ and RS. The length of AC is therefore 40 cm.
2. To calculate the base of the hexagon, calculate the distance between the points N and Q. Since MN is the median of triangle PQM, NQ is equal to half the side length of PQ. Therefore, NQ = 40 cm/2 = 20 cm.
3. Now that we have the base (NQ) and height (AC) of the hexagon, we can calculate its area using the formula for the area of a hexagon:
Area = (3 * √3 * s^2) / 2,
where s is the length of one side of the hexagon.
4. In our case, the length of one side of the hexagon is equal to NQ, so s = 20 cm.
5. Plugging in the values into the formula, we get:
Area = (3 * √3 * (20 cm)^2) / 2
= (3 * √3 * 400 cm^2) / 2
= (3 * √3 * 400 cm^2) / 2
Therefore, the area of hexagon NMQTUS is (3 * √3 * 400 cm^2) / 2.
Moving on to part b:
To determine the length of AC in the rectangular piece of paper PQRS:
1. Draw a rectangle with sides PQ = 30 cm and PS = 40 cm.
2. Fold the rectangle so that the diagonally opposite corners P and R coincide.
3. This creates a crease along the line segment AC, where A is on side PS and C is on side QR.
4. Note that since P and R are diagonally opposite corners of a rectangle, PR is equal to the diagonal of the rectangle.
5. Use the Pythagorean theorem to calculate the length of PR:
PR^2 = PQ^2 + QR^2
PR^2 = (30 cm)^2 + (40 cm)^2
PR^2 = 900 cm^2 + 1600 cm^2
PR^2 = 2500 cm^2
PR = √2500 cm
PR = 50 cm
6. Since AC is the folded crease and coincides with PR, the length of AC is also 50 cm.
Therefore, the length of AC in the rectangular piece of paper PQRS is 50 cm.