Week 9 HW – Right Triangles, Surface Area & Volume
Name: ______January 18, 2017
1. John leaves school to go home. He walks 6 blocks North and then 8 blocks west. How far is John from the school?
2. A 26 foot long ladder is leaning up against a house with its base 10 feet away from the house on the ground. If the side of the house is perfectly vertical, how far up the house would you be if you climbed to the top of the ladder?
3. If you walk 50 yards south, then 40 yards east, and finally 20 yards north, how far are you from your starting point? Express your answer in yards.
4. Find the area of this right triangle.
5. A rectangle 4 cm by 12 cm is divided into four triangles, as shown. The areas of three of the four triangles are stated in the diagram. Find the number of sq. cm in the area of the shaded triangle.
6. Find the area of rectangle ABCD.
7. A twenty-foot high spotlight is being shined towards the top of your head, creating a shadow. If you are five feet tall, and you are standing twelve feet away from the spotlight, how long is your shadow?
BONUS PROBLEMS
8. The following geometric design is constructed by adding new squares to each rectangle. To the nearest whole number, what would be the length of the longest diagonal, if the two smallest squares have side-lengths of three?
9. What is the length of the hypotenuse for the longest triangle in the figure? (All triangles are right triangles)
10. A "Pythagorean Triple" is a set of positiveintegers,a,bandcthat fits the rule: a² + b² = c². The smallest Pythagorean Triple is 3, 4 and 5 because 3² + 4² = 5². Name at least 2 other Pythagorean Triples.
SURFACE AREA & VOLUME
1. A rectangular box is 2 cm high, 4 cm wide and 6 cm deep. Michelle packs the box with cubes, each 2 cm by 2 cm by 2 cm, with no space left over. How many cubes does she fit into the box?
2. A group of toy makers want to wrap the world's biggest toy block with wrapping paper. The block has dimensions of 8 feet by 6 feet by 12 feet. How much wrapping paper, in square feet, will they need in order to wrap the entire toy block? Assume there is no overlap in wrapping paper.
3. A rectangular prism with a square base has a volume of 144 cubic centimeters. If the height is four centimeters, what is the length of the base in centimeters?
4. A supermarket clerk makes a solid pyramid out of identical cereal boxes. The top five layers are shown. What is the total number of cereal boxes in these top five layers?
5. Thirty cubes are placed in a line such that they are joined face to face. The edges of each cube are 1 cm long. Find the surface area of the resulting solid, in sq. cm.
6. A large cube, 5 cm by 5 cm by 5 cm, is painted orange on all six faces. Then it is cut into 125 small cubes, each 1 cm by 1 cm by 1 cm. How many of the small cubes are not painted orange on any face?
BONUS PROBLEMS
7. Anna wants to cover the outside of a rectangular box with colored paper. The box has a square base with area of 16 square inches. The volume of the box is 80 cubic inches. How many square inches of paper will Anna need to completely cover the box, including the top and bottom, with no paper left over?
8. A cubical box without a top is 4 cm on each edge. It contains 64 identical 1 cm cubes that exactly fill the box. How many of these small cubes actually touch the box?
9. Twenty unit cubes are glued together to form this figure, with “holes” which you can see through. The total figure measures 3 x 3 x 3. If the figure is fully dipped in a bucket of paint, how many square units of surface area would be painted?
RIGHT TRIANGLES - SOLUTIONS
1. 10 blocks
The distance from school to home is the length of the hypotenuse.
Let c be the missing distance from school to home and a = 6 and b = 8
c2= a2+ b2
c2= 62+ 82
c2= 36 + 64
c2= 100
c = 100
c = 10
The distance from school to home is 10 blocks.
2. 24 feet
3. 50 yards
4. 24 cm2
First find the length of the missing side: 100 – 64 = 36 = a². The missing side is 6cm long, so the area is ½ * 8 * 6 or 24 cm²
5. 6 cm²
The area of a rectangle is obtained by multiplying the length by the width. If the rectangle measures 4cm by 12 cm, the total area is 48 cm². The sum of the areas of the other 3 triangles is 16 + 18 + 8 or 42 cm². The area of the shaded triangle is 48 – 42 or 6 cm².
6. 12
Using the Pythagorean Theorem, we can find that side (DC)² = 3² + 4² = 25, so DC =
5. We also know that the area of the triangle is 1/2 base x height. In this case,
the base is 5, and the height would be the same as AD or BC. The area of the rectangle would be simply base x height, or 5 x either AD or BC. However, since we don’t know the measure of AD or BC, we need to find the area of the triangle a different way, knowing that it will be equal to ½ the area of the rectangle ABCD. Imagine rotating the rectangle so that the measurement of 3 becomes the base, and 4 becomes the height of the triangle. Now we can calculate the area of the triangle: ½ · 3 · 4 = 6. Therefore, the area of the rectangle is 12.
7. 14 ft
8. 55
Following the geometric pattern, we can see that the smallest squares have side lengths of 3, the next has a length of 6, the next is 9, then 15, then 24, then 39. So the length of the largest square’s diagonal would be 392+392 = 3042 = 55.
9. 3
Drawing a chart will help reveal the pattern.
Leg 1 / Leg 2 / Hypotenuse1 / 1 / 2
1 / 2 / 3
1 / 3 / 4 = 2
1 / 4 / 5
1 / 5 / 6
1 / 6 / 7
1 / 7 / 8
1 / 8 / 9 = 3
10. Possible answers include (5,12,13), (9,12,15), (6,8,10), (10,24,26) or
Cascade Ridge PTSA Math Club 5 Show your work on a separate sheet of paper.
Week 9 HW – Right Triangles, Surface Area & Volume
Name: ______January 18, 2017
SURFACE AREA & VOLUME - SOLUTIONS:
1. 6 cubes
Compare the volumes of a cube and the box. The volume of one cube = 2 x 2 x 2 = 8 cm³. The volume of the box = 2 x 4 x 6 = 48 cm³. Therefore, Michelle can fit 48 ÷ 8 = 6 cubes into the box.
2. 432 ft2
Two of the faces of the block are 8 × 6 ft, so each of these faces has an area of 48 ft2.
Two of the faces of the block are 8 × 12 ft, so each of these faces has an area of 96 ft2.
Two of the faces of the block are 6 × 12 ft, so each of these faces has an area of 72 ft2.
Adding them all up, 48 + 48 + 96 + 96 + 72 + 72 = 432 ft2.
3. 6 cm
If the volume is 144, and the height is 4, then the area of the base is 144 ÷ 4 = 36 cm2. Since the base is square, each side must be 6 cm.
4. 100 boxes
Break up the problem into layers. The top layer has 4 boxes. The next layer has 2 x 5 = 10 boxes. The third has 3 x 6 = 18 boxes. The fourth, 4 x 7 = 28 boxes, and the fifth, 5 x 8 = 40 boxes. Adding them all up: 4 + 10 + 18 + 28 + 40 = 100.
5. 122 cm²
Find the area of each face of the entire solid. The front face of the entire solid is a rectangle with length 30 cm and width 1 cm. Its area is 30 x 1 or 30 cm². The top, back and bottom faces are congruent to the front face. Each has an area of 30 cm². The left and right ends of the solid are squares 1 cm by 1 cm. Each has an area of 1 cm². The surface area of the solid is 4 x 30 + 2 x 1 = 122 cm².
6. 27
Separate the problem into parts:
Painted on 3 faces: 8 (all the corner cubes)
Painted on 2 faces: 36 (3 cubes on each of the 12 edges)
Painted on 1 face: 54 (9 cubes on each of the 6 faces)
8 + 36 + 54 = 98 = total # of cubes painted on one or more faces. Then 125 – 98 = 27 small cubes which are not painted orange on any face.
7. 112 in2
Find the dimensions of the box. Then add the areas of the faces. The base is a square of area 16 in² so each of the base’s sides is 4 inches long. The given volume of 80 in³ = base x height, so the height is 5 inches. The top and bottom each have an area of 16 in². Each side has an area of 4 x 5 = 20 in². The amount of paper needed is 16 + 16 + 20 + 20 + 20 + 20 = 112 in².
8. 52 cubes
Count the cubes that touch the box in each horizontal layer. In the bottom layer, all 16 cubes touch the box. In each of the three other layers, all the outer cubes touch either one or two sides of the box. That is 12 outer cubes per layer. Thus there are a total of 16 + 3 x 12 = 52 small cubes that touch the bottom or a side of the box.
9. 72
Count in an organized way: exterior vs. interior squares. Each face of the figure has 8 painted squares for a total of 6 x 8 = 48 painted faces. Inside each “hole” 4 squares are painted for a total of 6 x 4 = 24 painted faces. The total number of square units of painted surface is 24 + 48 = 72.
Cascade Ridge PTSA Math Club 5 Show your work on a separate sheet of paper.