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This right triangle is drawn with its right angle at the top. Click the hypotenuse.
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Find the small square that marks the right angle first. Which side does not touch that corner?
Step-by-step solution
The small square at the top vertex marks the right angle.
The two slanted sides both touch that top corner, so they are the legs.
The hypotenuse is the side opposite the right angle — the side that does not touch it. Here that is the bottom side.
It is also the longest side, which you can see: the bottom side is longer than either slanted side.
The hypotenuse is the bottom side.
Check: the hypotenuse must be the longest side. In the figure, the bottom side is clearly longer than either slanted side ✓.
Tip: "Hypotenuse" does not mean "slanted side". It means the side across from the right angle. Turn the page in your mind: the labels never depend on how the triangle is drawn.
Question 2 of 10 · Triangles and rectangles: area and perimeter
A triangular piece of sheet metal is shown below. The dashed line shows the height.
What is the area of the triangle?
Hint
The height does not have to be inside the triangle. Look for the segment that meets the (extended) base at a right angle.
Step-by-step solution
Use A = \tfrac{1}{2}bh (on the formula sheet).
The base is the bottom side, b = 8 m.
The height is the dashed segment that meets the extended base at a right angle: h = 6 m. In an obtuse triangle the height can fall outside the triangle.
The slanted sides (12.5 m and 6.7 m) are not used for area.
Compute: A = \tfrac{1}{2}(8)(6) = \tfrac{1}{2}(48) = 24.
The area is 24 m².
Check: a parallelogram with base 8 and height 6 has area 48; two copies of this triangle make that parallelogram, so one triangle is 48 \div 2 = 24 ✓.
Tip: The height is always measured straight up from the line of the base (look for the right-angle mark). It may sit outside the triangle — that is still the height to use.
Question 3 of 10 · Circles: area and circumference
A large pizza is 16 inches in diameter. To the nearest square inch, what is the area of the top of the pizza? Use \pi \approx 3.14.
Hint
Pizza sizes give the distance straight across. What does the area formula need?
Step-by-step solution
"16 inches in diameter" → d = 16, so the radius is r = 16 \div 2 = 8 in.
Area: A = \pi r^2 = 3.14 \times 8^2.
8^2 = 64, and 3.14 \times 64 = 200.96.
Round to the nearest square inch: 200.96 \approx 201.
The area is about 201 in².
Check: estimate 3 \times 64 = 192; the answer is a bit more ✓. (With the calculator's \pi key: \pi \times 64 \approx 201.06, which also rounds to 201 ✓.)
Tip: Always halve a diameter before using A = \pi r^2. Because the radius is squared, using the diameter makes the area 4 times too large (804 instead of 201).
Question 4 of 10 · Circles: area and circumference
Fill in
A round backyard trampoline has a diameter of 14 feet. What is the area of the jumping surface? Use \pi \approx 3.14.
ft²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The area formula uses the radius. What do you need to do with the diameter first?
Step-by-step solution
Area of a circle: A = \pi r^2 (on the formula sheet).
The radius is half the diameter: r = 14 \div 2 = 7 ft.
Square the radius: 7^2 = 49.
Multiply by \pi: A = 3.14 \times 49 = 153.86.
The area is 153.86 ft².
Check:3 \times 49 = 147, and the answer should be a little more than that because \pi is a little more than 3 ✓.
Tip: Square the radius, not the diameter. Using 14 instead of 7 gives 3.14 \times 196 = 615.44 — four times too big.
Question 5 of 10 · Pythagorean theorem
Drop-down
A 15-foot ladder leans against a building and touches the wall 12 feet above the ground. Let x be the distance from the bottom of the ladder to the building.
The equation can be used to find x. The bottom of the ladder is feet from the building.
Choose an answer from each menu.
Hint
Which length is across from the right angle where the wall meets the ground? That one goes alone on one side of the equation.
Step-by-step solution
The wall and ground form the right angle. The legs are x (ground) and 12 (wall). The ladder, 15 ft, is the hypotenuse.
Equation: x^2 + 12^2 = 15^2.
Solve: x^2 + 144 = 225, so x^2 = 225 - 144 = 81.
x = \sqrt{81} = 9.
Equation x^2 + 12^2 = 15^2; the ladder's base is 9 ft from the building.
Check:9^2 + 12^2 = 81 + 144 = 225 = 15^2 ✓. 9-12-15 is the 3-4-5 triple times 3 ✓.
Tip: Write the equation with the hypotenuse alone: (\text{leg})^2 + (\text{leg})^2 = (\text{hypotenuse})^2. Then solve for the unknown.
Question 6 of 10 · Polygons: area and perimeter
The floor of a gazebo is a regular hexagon with sides of 6 feet. The apothem (the distance from the center to the middle of a side) is about 5.2 feet.
What is the approximate area of the floor?
Hint
Split the hexagon into triangles from its center. How many triangles, and what are each one's base and height?
Step-by-step solution
Lines from the center to the six corners split the hexagon into 6 identical triangles.
Each triangle: base 6 ft (one side), height 5.2 ft (the apothem).
One triangle: \tfrac{1}{2}(6)(5.2) = 15.6\text{ ft}^2.
Check: a circle of radius 6 (through the corners) would have area 3.14 \times 36 \approx 113; the hexagon fits inside that circle, so it must be smaller — 93.6 < 113 ✓.
Tip: Regular polygon area = n triangles × \tfrac{1}{2} × side × apothem. This is not on the formula sheet, but the triangle method always works.
Question 7 of 10 · Composite 2-D figures
Fill in
A concrete walkway 3 ft wide is built around all four sides of a rectangular pool that measures 20 ft by 10 ft, as shown.
What is the area of the walkway?
ft²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Find the outside length and width of the pool plus the walkway. The walkway adds 3 ft on each side.
Step-by-step solution
The walkway adds 3 ft on the left and 3 ft on the right: outer length = 20 + 3 + 3 = 26 ft.
It adds 3 ft on the top and bottom: outer width = 10 + 3 + 3 = 16 ft.
Outer rectangle: 26 \times 16 = 416 ft².
Pool: 20 \times 10 = 200 ft².
Walkway = outer − pool: 416 - 200 = 216.
The walkway's area is 216 ft².
Check: split the walkway into strips: two long strips 26 \times 3 = 78 each, two short strips 10 \times 3 = 30 each: 78 + 78 + 30 + 30 = 216 ✓.
Tip: For a border of width x, add 2x to each dimension. Adding only x (getting 23 \times 13) is the most common error in frame and walkway problems.
Question 8 of 10 · Composite 2-D figures
Fill in
A machine shop starts with a rectangular metal sheet. It cuts a right triangle from one corner and a square hole from the middle, as shown. The shaded part is the metal that is left.
What is the area of the metal that is left?
cm²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Start with the full sheet. Two pieces are removed, and one of them is a triangle.
Step-by-step solution
Full sheet:20 \times 15 = 300\text{ cm}^2.
Triangle cut from the corner: legs 8 cm and 6 cm. The legs are the base and height, so A = \tfrac{1}{2} \times 8 \times 6 = 24\text{ cm}^2.
Square hole:4 \times 4 = 16\text{ cm}^2.
Metal left:300 - 24 - 16 = 260\text{ cm}^2.
260 cm² of metal is left.
Check:260 + 24 + 16 = 300 ✓. Common wrong results: 300 - 48 - 16 = 236 (forgot the \tfrac{1}{2} for the triangle) and 300 - 24 = 276 (forgot the square hole).
Tip: A right triangle's two legs are its base and height — no extra height is needed. Subtract each removed piece from the original shape.
Question 9 of 10 · Triangles and rectangles: area and perimeter
A rectangular room measures 15 ft 6 in by 11 ft 9 in. Baseboard trim will go around the whole room (ignore the doorway). Trim is sold in 8-foot pieces.
What is the smallest number of pieces that must be bought?
Hint
Write each measurement in feet only (6 in is what part of a foot?). Then find the distance around.
Step-by-step solution
Convert to feet: 6\text{ in} = \tfrac{6}{12} = 0.5 ft, so 15\text{ ft }6\text{ in} = 15.5 ft. 9\text{ in} = \tfrac{9}{12} = 0.75 ft, so 11\text{ ft }9\text{ in} = 11.75 ft.
Perimeter: P = 2(15.5) + 2(11.75) = 31 + 23.5 = 54.5 ft.
Pieces: 54.5 \div 8 = 6.8125, so round up to 7 pieces.
7 pieces must be bought.
Check: 6 pieces give 48 ft (too short), 7 pieces give 56 ft \ge 54.5 ft ✓.
Tip: Inches are twelfths of a foot, not tenths: 9 in is 0.75 ft, not0.9 ft. Convert mixed units before you add or multiply.
Question 10 of 10 · Polygons: area and perimeter
No calculator
The base of a parallelogram is twice its height. The area of the parallelogram is 98 square inches.
What is the length of the base?
Hint
Let the height be h and write the base using h. Put both into A = bh.
Step-by-step solution
Let the height be h. Then the base is 2h.
Area: A = bh = (2h)(h) = 2h^2.
Set equal to 98: 2h^2 = 98, so h^2 = 49.
h = \sqrt{49} = 7 in.
Base: 2h = 14 in.
The base is 14 inches.
Check:14 \times 7 = 98 ✓, and 14 is twice 7 ✓. Estimate: a square with area 98 would have sides near 10; a base longer than 10 and a height shorter than 10 fits a figure twice as long as it is tall ✓.
Tip: Do not stop at h^2 = 49 or h = 7 — re-read what the question asks. Here it wants the base.