# High School Geometry Word Problems with Solutions

TR Smith is a product designer and former teacher who uses math in her work every day.

Geometry and trigonometry problems are usually presented with diagrams to make it easier to visualize spatial relations. More challenging geometry problems require you to figure out the picture for yourself. Here are six geometry word problems suitable for high school or advanced middle school students.

## (1) Finding a Shape

A regular polygon has N sides, where N is an even number. If you cut the shape in half along a line that connects opposite vertices, you get two identical shapes, each of whose angles sum to 900 degrees. What is the value of N?

Solution: This is a two-part problem. First, the sum of the interior angles of any M-sided polygon is 180(M - 2) degrees. For example, if M = 4, the sum of the angles of any quadrilateral is 180(4 - 2) - 360. If a shape's angle sum is 900, then we must solve the equation 180(M - 2) = 900 to find M. With a little algebra we get M = 7. Therefore, the half-shape looks like this:

If you put two of these shapes together along the longest side, you get a 12-sided polygon called a dodecagon.

## (2) Inverted Cone Cup

A cup in the shape of an inverted cone is filled with water such that the height of the liquid is 2/3 of the height of the cup. What fraction of the cone's capacity is filled with water?

Solution: In this problem it doesn't matter what the dimensions of the cone are, so for simplicity let's say the cone has a radius of 3 units and a height of 3 units. If it is filled to 2/3 of its capacity, the liquid forms a smaller cone with a radius of 2 units and a height of 2 units.

The capacity of the cone with a height of H and a radius of R is (π/3)(H)(R^2). Thus, the cone cup as a volume of 9π and the water has a volume of 8π/3.

Dividing 8π/3 by 9π gives us 8/27. Therefore, the water takes up 8/27 of the cone's capacity.

A square has a side length of 4. Can you fit a quadrilateral with a perimeter of 18 inside the square? The quadrilateral cannot be self-intersecting but the corners and edges of the quadrilateral may lie on the perimeter of the square. If you can, demonstrate it. If you can't, explain why it's impossible.

Solution: First off, a square with a side length of 4 has a perimeter of 16. Here is a 4-by-4 square.

A 4-by-4 square not only has a perimeter of 16 but an area of 16 as well. It is impossible to fit a quadrilateral with a larger area inside the square, but totally possible to fit a quadrilateral with a longer perimeter inside. For example, this tweezers shape:

The side lengths are 5, 5, 4, and 4, for a total of 18. There are many other possible solutions, all of them non-convex. If you make the tweezers shape even narrower while still keeping three of its points on the perimeter of the square, you can make its perimeter even larger. The perimeter of a quadrilateral that fits inside a 4-by-4 square cannot exceed 16*sqrt(2), which is about 22.627.

## (4) Unit Conversion

If Karen measures the edge length of a cube in a unit of length called zaps, the surface area is twice the volume. If she measures it in a different unit of length called zops, the surface area is half the volume. Knowing this, how do you convert between zaps and zops?

Solution: In zaps, let's call the side length of the cube x. The surface area is 6x^2 and the volume is x^3. If the surface area is twice the volume then 6x^2 = 2x^3. Solving this for x gives you x = 3 zaps.

In zops, let's say the side length of the cube is y. Again, the surface area is 6y^2 and the volume is y^3. If the surface area is half the volume, then 6y^2 = (1/2)y^3. Solving this for y gives you y = 12 zops.

If 3 zaps = 12 zops, then 1 zap = 4 zops. Equivalently, a zop is 1/4 of a zap.

## (5) Open Ended Problem

Find 5 shapes whose perimeter and area are equal.

Solution: There are infinitely many shapes whose perimeter and area are equal. Here is a sample of common shapes that fit the bill.

• Rectangles: A 4-by-4 square and 6-by-3 rectangle have areas equal to their perimeters
• Right triangles: The 6-8-10 right triangle and the 5-12-13 right triangle have areas equal to their perimeters.
• Circle: A circle with a diameter of 4 has perimeter and radius equal to 4π.
• Regular hexagon: A regular hexagon with a side length of 4/sqrt(3) has perimeter and area equal to 8*sqrt(3).
• Equilateral triangle: An equilateral triangle with a side length of 4*sqrt(3) has an area and perimeter of 12*sqrt(3).
• L-shapes: There are infinitely many ways to construct L-shapes with equal area and perimeter. Here are some examples where the side lengths are all integers.

## (6) Concrete Volume

Jenny and John are paving a concrete patio and they need to know the total volume of wet concrete to be poured to buy enough bags of concrete. The shape of the patio is a semi-circle with a diameter of 16 feet 6 inches, and concrete is to be poured 4 inches deep. How many cubic feet of wet concrete do they need? If one bag makes 4 cubic feet, how many bags do they need?

Solution: The radius of the patio in feet is 33/4, and the depth in feet is 1/3. The shape is half of a very flat cylinder. The volume of a half cylinder is (π/6)hr^2. Plugging in h = 1/3 and r = 33/4 gives us

volume = (π/6)(1/3)(33/4)^2
= 121π/32
≈ 11.879 cubic feet

Since 11.879 is a little under 12 cubic feet, Jenny and John need 3 bags.

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