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Solving Problems in Geometry Facts & Worksheets

In this lesson, we will solve real-life and mathematical problems involving angle measure, area, surface area, and volume. We will also cover the formulas for the area and circumference of a circle and use them to solve problems.

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Worksheets /Math /Grade 7 (ages 12-13) /Solving Problems in Geometry Facts & Worksheets
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In this lesson, we will solve real-life and mathematical problems involving angle measure, area, surface area, and volume. We will also cover the formulas for the area and circumference of a circle and use them to solve problems. Moreover, we will use facts about angle relationships to solve multi-step problems.

See the fact file below for more information on the solving problems in geometry or alternatively, you can download our 42-page Solving Problems in Geometry worksheet pack to utilise within the classroom or home environment.

Key Facts & Information

INTRODUCTION TO CIRCLES

  • A circle is a geometric figure that needs only two parts to identify and classify it: its center and its radius, which is the distance from the center to any point on the circle.
  • A circle is the set of all points equidistant, or equal distances, from the center point P. Twice the radius r is called the diameter.

CIRCUMFERENCE OF A CIRCLE

  • As with triangles and rectangles, we can attempt to derive formulas for the area and “perimeter” of a circle. Unlike triangles, rectangles, and other shapes, the distance around the outside of the circle is called the circumference rather than the perimeter – the concept, however, is almost the same.
  • Solving for the circumference of a circle is not as simple as solving for the perimeter of a rectangle or triangle, however. Given a circular object, one approach might be to wrap a string exactly once around the object and then straighten the string and measure its length.
  • As we increase the diameter or radius of a circle, its circumference also gets bigger.
  • If we measure the circumference and diameter of a circle, the latter is always slightly more than thrice the diameter. Below is a representation of this claim, where D is the diameter and C is the circumference of each circle.
  • If we divide the circumference C of any circle by its diameter D, we get a constant number. This constant, known as ℼ (pi), is an irrational non-repeating decimal, which is approximately 3.14. This can be expressed as: C/D = ℼ.
  • We can derive an expression for the circumference in terms of the diameter by multiplying both sides of the expression (C/D = ℼ) by D, thereby isolating C.
  • Because the diameter is two times the radius (in other words, D = 2r), we can substitute 2r for D in the previous expression.
  • Therefore, we can solve for the circumference of the circle given the radius or the diameter. For most calculations that need a decimal answer, estimating ℼ14 is often used.
  • For example, if a circle has a radius of 6 meters, then its circumference C is 12ℼ
  • The answer above is exact. If an approximate numerical answer is required, we can estimate ℼ as 3.14.

AREA OF A CIRCLE

  • Let’s try to get an estimate of the area of a circle by drawing a circle inside a square as shown below. The area of the circle is shaded.
  • Draw a vertical and horizontal diameter in the circle and label them D. Let’s utilize the square by having its sides of length D, as well.
  • We know that a square with sides of length D has the following area Asquare: A = DxD.
  • Because the circle of diameter D apparently has a smaller area than the square with sides of length D, we can conclude that the circle’s area must be less than D². By inspection, we can guess that the area Acircle of the circle is approximately three-fourths that of the square.
  • Through some more complicated mathematics that is beyond the scope tutorial, it can be shown that the area of a circle is exactly the following:
  • We will rearrange the expression, keeping in mind that the radius (r) is equal to half the diameter (D). Thus, in other words, D = 2r.
  • Let’s substitute this value for r into the expression for the area of the circle. We must make the substitution twice.
  • For example, a circle has a diameter of 6 cm. Find its area.

ANGLE RELATIONSHIPS

  • Adjacent angles are two angles that share a common side and common vertex, and they do not overlap.
  • ∠1 and ∠2 are adjacent angles.
  • ∠ABC and ∠2 are NOT adjacent angles.
  • Two adjacent angles whose noncommon sides form opposite rays make up a linear pair.
  • ∠1 and ∠2 are form linear pairs.
  • The line through points W, X, and Y is a straight line.
  • ∠1 and ∠2 are supplementary angles.
  • Supplementary angles are two angles that form a linear pair.
  • A linear pair forms a straight angle that measures 180°. Thus, there are two angles whose measures add up to 180°, which suggests they are supplementary angles.
  • Right angles are two congruent angles that form a linear pair.
  • When two congruent angles whose sum adds up to 180°, each measuring 90°, forms a right triangle.
  • Vertical angles are two angles whose sides form two pairs of opposite rays. We can think of these as opposite angles formed by intersecting lines.
  • Angle pairs ∠1 and ∠2, and ∠3 and ∠4 are vertical angles.
  • Vertical angles are NOT adjacent. Therefore, ∠1 and ∠3 are not vertical angles. However, they are linear pairs.
  • Vertical angles are always equal in measure.
  • Vertical angles, such as ∠1 and ∠2, make linear pairs with the same angle, ∠4, resulting in m∠1 + m∠4 = 180° and m∠2 + m∠4 = 180°. Therefore, we can conclude that m∠1 = m∠2, so they are congruent.
  • Complementary angles are two angles with a sum of 90°. They can be placed so that they create perpendicular lines, or they may be two separate angles.
  • ∠1 and ∠2 are complementary angles.
  • ∠X and ∠Y are complementary angles.
  • Line segment AB is perpendicular to line segment BC.
  • Complements of the same angle are congruent.
  • If m∠x is complementary to m∠y, and m∠z is complementary to m∠y, then we can conclude that m∠x = m∠ Take note of the following: m∠x = 60°, m∠y = 30°, and m∠z = 60°.
  • Two acute angles in a right triangle are complementary.
  • Angles in a triangle add up to 180°. After subtracting 90° for the right angle, there are 90° left for the remaining two acute angles, thus making them complementary angles.
  • Supplementary angles are two angles with a sum of 180°. They can be placed so that they create a linear pair, or they may be two separate angles.
  • ∠1 and ∠2 are supplementary angles.
  • ∠X and ∠Y are supplementary angles.
  • Points A, B, and C make a straight line.
  • Supplements of the same angle are congruent.
  • If m∠x is supplementary to m∠y, and m∠z is supplementary to m∠y, then we can conclude that m∠x = m∠ Take note of the following: m∠x = 60°, m∠y = 120°, and m∠z = 60°.

SOLVING FOR THE AREA OF TWO-DIMENSIONAL FIGURES

  • Triangles could be of various types, but the formula for the area of all kinds of triangles is the same.
  • To find the the area of a parallelogram, we use the formula b x h, where b represents the base and h represents the height (vertical distance between the base and the top).
  • We can obtain the area of a rhombus, given the length of its diagonals.
  • The area of a kite uses the same formula as the area of a rhombus. The area of a kite is equal to half the product of the diagonals.
  • To get the area of a trapezoid, we add the length of the parallel sides and multiply that with half of the height. Take note that the height needs to be perpendicular to the parallel sides.
  • Just to review, below are the formulas for the area of a rectangle and a square.
  • A = s x s = s²; where s is the length of one side

SOLVING FOR THE AREA OF THREE-DIMENSIONAL FIGURES

  • A cube is a three-dimensional figure with six matching square sides.
  • V = s x s x s = s³; where s is the length of one of its sides
  • A rectangular solid is also known as a rectangular prism or a cuboid. In a rectangular solid, all of its angles are right angles and opposite faces are equal.
  • The length, width, and height of rectangular solids may be of different lengths. A cube is a special case of a cuboid in which all six faces are squares.
  • V = lwh; where l is the length, w is the width, and h is the height
  • A prism is a solid that has two parallel faces that are congruent polygons at both ends. These faces form the bases of the prism. A prism is named after the shape of its base.
  • The other faces are in the shape of parallelograms. These are called lateral faces. The diagrams below show a triangular prism and a rectangular
  • V = Ah; where A is the area of the base and h is the height or length of the prism
  • A pyramid is a solid with a polygon base that is connected by triangular faces to its vertex. The lateral faces meet at a common vertex. The height of the pyramid is the perpendicular distance from the base to the vertex.
  • A pyramid is named after the shape of its base. A rectangular pyramid has a rectangle base, while a triangular pyramid has a triangle base.
  • V = 1/3Ah; where A is the area of the base and h is the height of the pyramid

SOLVING FOR THE SURFACE AREA OF SOLIDS

  • The surface area of a cube is the sum of the area of the six squares that cover it.
  • SA = 6s²; where s is the length of one of its sides
  • To calculate the surface area of a cuboid, we need to calculate first the area of each face and add up all the areas to get the surface area.
  • The surface area of a prism is the total area of all its external faces by figuring out the shape of its base, solving for the area of each face, and adding up all the areas to get the total surface area.
  • SA = 2A+ph; where A is the area of the base, p is the perimeter of the base, and h is the height
  • If the pyramid is a square pyramid, we can use the formula for the surface area of a square pyramid.
  • SA = b² + 2bs; where b is the length of the base and s is the slant height

Solving Problems in Geometry Worksheets

This is a fantastic bundle which includes everything you need to know about the solving problems in geometry across 42 in-depth pages. These are ready-to-use Solving Problems in Geometry worksheets that are perfect for teaching students how to solve real-life and mathematical problems involving angle measure, area, surface area, and volume. We will also cover the formulas for the area and circumference of a circle and use them to solve problems. Moreover, we will use facts about angle relationships to solve multi-step problems.

Complete List Of Included Worksheets

  • Lesson Plan
  • Solving Problems in Geometry
  • Parts of a Circle
  • Circumference of a Circle
  • Circle Word Problems
  • Multiple Rays
  • Angle Relationships
  • Complementary Angles
  • Supplementary Angles
  • Area Word Problems
  • Surface Area Word Problems
  • Volume Word Problems

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Use With Any Curriculum

These worksheets have been specifically designed for use with any international curriculum. You can use these worksheets as-is, or edit them using Google Slides to make them more specific to your own student ability levels and curriculum standards.

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