In plain English: HSG.GPE.B.7 is the Common Core geometry standard that asks students to use coordinates to compute perimeters of polygons and areas of triangles and rectangles. Students find each side length with the distance formula, add the sides for perimeter, and find areas from a base and height or by enclosing a triangle in a rectangle. It is usually taught in high school Geometry.
Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula.
Common Core State Standards for Mathematics · Domain: Expressing Geometric Properties with Equations (GPE) · Cluster: Use coordinates to prove simple geometric theorems algebraically Also written as HSG-GPE.B.7 or G-GPE.7 · Official standard
Students use the coordinates of vertices to compute the perimeter of polygons and the area of triangles and rectangles. The distance formula, which is the Pythagorean Theorem written in coordinates, turns every side into a length, and the lengths add up to the perimeter.
For area, students use a horizontal or vertical side as a base when there is one, enclose other triangles in a rectangle and subtract the corner triangles, and use slopes to confirm a slanted rectangle before multiplying its side lengths. Map scales connect the work to real distances and areas.
Learning Objectives
By the end of this lesson, students will be able to:
Use the distance formula to find the length of any segment between two points on the coordinate plane
Compute the perimeter of a polygon from the coordinates of its vertices, giving exact and rounded answers
Compute the area of a triangle from coordinates, using a horizontal or vertical base or the box method
Verify with slopes that a quadrilateral is a rectangle and compute its area from its side lengths
Apply a map scale correctly to lengths and to areas
Prior Knowledge Required
Students should already be comfortable with:
Plotting polygons in the coordinate plane and finding horizontal and vertical lengths 6.G.A.3
The Pythagorean Theorem for distance between two points 8.G.B.8
Area of right triangles, other triangles and rectangles 6.G.A.1
Slopes of parallel and perpendicular lines HSG.GPE.B.5
Put a coordinate grid on the board and plot three pairs of points. Ask students to find each distance without a formula sheet:
Warm-Up Prompt
"How far apart are (1, 2) and (1, 9)? How far apart are (-3, 4) and (5, 4)? Now find the distance from (1, 2) to (5, 5). What is different about the last pair, and how did you handle it?"
The first two pairs share a coordinate, so students count or subtract: 7 and 8. For the third pair, draw the right triangle with legs 4 and 3 and let students use the Pythagorean Theorem to get 5. Name the move out loud: every slanted segment on a grid is the hypotenuse of a right triangle whose legs are the changes in x and in y. That is the distance formula, and it is the tool for the rest of the lesson.
Direct Instruction20 minutes
Part 1: Perimeter of any polygon. Write the distance formula next to the right triangle from the warm-up: the distance between (x₁, y₁) and (x₂, y₂) is √((x₂ - x₁)² + (y₂ - y₁)²). Then give the procedure for perimeter:
List the vertices in order around the polygon, so each side joins two consecutive vertices and the last vertex joins back to the first.
Find each side length. Subtract when the side is horizontal or vertical; use the distance formula when it is slanted.
Keep exact values such as √29 until the end, and simplify radicals where possible (√20 = 2√5).
Add all the sides and round only the final answer, with units if the grid has a scale.
Part 2: Area of triangles and rectangles. For a triangle with a horizontal or vertical side, use that side as the base and count the perpendicular distance to the opposite vertex as the height. For any other triangle, enclose it in a rectangle with horizontal and vertical sides and subtract the right triangles in the corners (Diagram 1). For a rectangle whose sides are slanted, first check with slopes that consecutive sides are perpendicular, then multiply the two side lengths found with the distance formula (Diagram 2).
Perimeter of a triangle
Find the perimeter of the triangle with vertices A(1, 1), B(7, 1) and C(4, 5).
Equation: AB = 6, BC = √(3² + 4²) = 5, CA = √(3² + 4²) = 5, so P = 16 units
Area of a triangle, box method
Find the area of the triangle with vertices P(-2, 1), Q(4, 3) and R(1, 7). No side is horizontal or vertical.
Show that A(0, 3), B(4, 1), C(7, 7), D(3, 9) is a rectangle, then find its area and perimeter.
Equation: Slopes -½ and 2 multiply to -1; AB = 2√5, BC = 3√5; area = 30, P = 10√5 ≈ 22.36
Perimeter in context
A park map uses 1 grid unit = 50 m. A walking loop has corners at (0, 0), (8, 0), (11, 4) and (5, 12). How long is one lap?
Equation: 8 + 5 + 10 + 13 = 36 units; 36 × 50 m = 1,800 m
Triangle with a vertical base
Find the area and perimeter of the triangle with vertices A(2, -3), B(2, 5) and C(-4, 1).
Equation: Base AB = 8, height = 6, area = 24; P = 8 + 2√52 = 8 + 4√13 ≈ 22.42
After the last example, point out that the height of a triangle is always measured perpendicular to the base. When the base is vertical, the height is a horizontal distance: here from x = -4 to x = 2. Stress the scale step in the context example: a length on the map is multiplied by 50, but an area on the same map would be multiplied by 50² = 2,500.
Guided Practice15 minutes
Pairs work three problems on graph paper. They plot the points first, then decide which method fits before computing:
The triangle with vertices (-2, -3), (10, -3) and (10, 2): perimeter 12 + 5 + 13 = 30 units and area ½(12)(5) = 30 square units.
The triangle with vertices (0, 0), (5, 2) and (2, 6), by the box method: 30 - (5 + 6 + 6) = 13 square units.
The quadrilateral (-1, 1), (2, -2), (7, 3), (4, 6): show it is a rectangle with slopes -1 and 1, then find its sides 3√2 and 5√2, its area 30 and its perimeter 16√2 ≈ 22.63.
Circulate and listen for these errors: mixing coordinates inside one difference (such as x₂ - y₁), adding the legs instead of using the Pythagorean Theorem, forgetting the ½ for a triangle, and using a slanted side as the height.
Independent Practice10-15 minutes
Students work alone on four problems and round decimals to the nearest hundredth:
The perimeter of the quadrilateral (-3, -1), (1, 2), (6, 2), (3, -2): 5 + 5 + 5 + √37 ≈ 21.08 units.
The area of the triangle (1, -2), (6, -2), (4, 5): ½(5)(7) = 17.5 square units.
The perimeter of the pentagon (0, 0), (4, 0), (6, 3), (3, 6), (-1, 3): 4 + √13 + 3√2 + 5 + √10 ≈ 20.01 units.
The area and perimeter of the rectangle (1, 0), (4, 1), (2, 7), (-1, 6): sides √10 and 2√10, area 20, perimeter 6√10 ≈ 18.97.
Closure5 minutes
Exit ticket: for the triangle with vertices (0, 0), (6, 0) and (2, 5), find (1) the area and (2) the perimeter to the nearest hundredth. (Answers: area ½(6)(5) = 15; perimeter 6 + √41 + √29 ≈ 17.79.) (3) In one sentence, explain why the perimeter needs the distance formula but the area here does not.
Differentiation Strategies
For Struggling Students
Have students draw the right triangle under every slanted side and label the legs Δx and Δy before using the formula
Give a vertex table with columns for the point, the next point, Δx, Δy and the side length, so no side is skipped
Start with triangles that have a horizontal or vertical side before moving to the box method
For Advanced Students
Show that the box method always gives the same result as the formula ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| (an extension beyond this standard)
Find the height of a triangle with no horizontal side by finding the foot of the perpendicular from a vertex, and compare with the box method
Find all points C on the line y = 4 so that the triangle with A(0, 0) and B(6, 0) has perimeter 16
Assessment Guidance
What to Look For
Check that students list vertices in order and include the closing side from the last vertex back to the first. Look for exact radicals carried to the end, with rounding only in the final answer. For area, students should name their base and explain why their height is perpendicular to it; a student who multiplies two slanted sides of a triangle, or two sides of a quadrilateral without checking the right angles, has not yet separated triangles and rectangles from general shapes. In scale problems, check that areas use the square of the scale factor.
02
Classroom Activities
3 Activities
1
Fence the Community Garden
20 minGroups of 3-4
Groups plan a garden on a grid where 1 unit = 1 meter. They find how much fence to buy (perimeter) and how much ground to prepare (area), then see that moving one corner can change the perimeter without changing the area.
Procedure
Plot the garden with corners (0, 0), (12, 0), (12, 5), (6, 13) and (0, 5)
Find the five side lengths: 12, 5, 10, 10 and 5 meters, so the fence is 42 m long
At $18 per meter of fence, find the cost of the fence ($756)
Split the garden into the 12 m by 5 m rectangle and the triangle on top with base 12 and height 8, and find the area: 60 + 48 = 108 m²
Discussion Questions
Which side lengths could you find without the distance formula? Which needed it?
Why do you split the shape into a rectangle and a triangle instead of multiplying two sides?
A bag of compost covers 4 m². How many bags does the garden need? (27)
Challenge Variation
Move the top corner from (6, 13) to (8, 13). The area stays 108 m² because the triangle keeps base 12 and height 8, but the fence becomes 12 + 5 + √80 + 8√2 + 5 ≈ 42.26 m. Ask groups to explain which top corner on the line y = 13 gives the shortest fence.
2
Three Ways to One Area
20 minPairs
Pairs find the area of the same triangle, with vertices A(1, 1), B(9, 5) and C(3, 7), in three different ways and confirm that all three give 20 square units. Comparing the methods builds judgment about which one fits a given figure.
Procedure
Method 1, box: the rectangle from (1, 1) to (9, 7) has area 48. Subtract the corner triangles 16, 6 and 6 to get 20
Method 2, split: the vertical line x = 3 through C meets AB at (3, 2). This cuts the triangle into two triangles with the vertical base 5: ½(5)(2) + ½(5)(6) = 5 + 15 = 20
Method 3, base and height: AB = √80 = 4√5. The perpendicular from C meets AB at F(5, 3), and CF = √20 = 2√5. Area = ½(4√5)(2√5) = 20
Check that CF is perpendicular to AB: the slopes are -2 and ½
Discussion Questions
Which method needed the fewest steps? Which one would you use on a test?
In Method 3, why can you not use CA as the height?
Why does Method 2 work only after you find where the vertical line meets AB?
Modification for Distance Learning
Use a free online graphing tool. Each partner builds one method on a shared graph with the polygon and segment tools, and the pair compares results in a short video call or a shared document.
3
Build to Order
20 minPairs
Pairs receive 6 challenge cards. Each card asks for a polygon with a given perimeter or area, and pairs must place the vertices on grid points and prove the measure with coordinates. Working backward makes students think about how coordinates control length and area.
The 6 Cards, with One Sample Answer Each
Card 1: a triangle with area 12 and no horizontal or vertical side. Sample: (0, 0), (6, 2), (3, 5)
Card 2: a rectangle with no horizontal side and area 10. Sample: (0, 0), (2, 1), (0, 5), (-2, 4), with sides √5 and 2√5
Card 3: a rhombus that is not a square, with perimeter 20 and no side on a grid line. Sample: (0, 0), (3, 4), (7, 7), (4, 3)
Card 4: two triangles with the same area and different perimeters. Sample: (0, 0), (8, 0), (4, 3) with perimeter 18, and (0, 0), (8, 0), (0, 3) with perimeter 11 + √73 ≈ 19.54, both with area 12
Card 5: a rectangle with perimeter 12√2. Sample: (0, 0), (2, 2), (-2, 6), (-4, 4), with sides 2√2 and 4√2 and area 16
Card 6: a right triangle with area 10 and no leg on a grid line. Sample: (0, 0), (4, 2), (2, 6), with the right angle at (4, 2)
Procedure
Partners take turns: one proposes vertices, the other checks the measure with the distance formula, slopes or the box method
Each pair writes its answer and the proof on the back of the card
Pairs trade cards with another pair and check each other's proofs
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Box Method for the Area of a Triangle
Triangle PQR with P(-2, 1), Q(4, 3) and R(1, 7), drawn to scale. The dashed 6-by-6 rectangle has area 36. The three shaded corner triangles have areas 6, 6 and 9, so the triangle has area 36 - 21 = 15 square units.
Diagram 2: A Rectangle with Slanted Sides
Rectangle ABCD with A(0, 3), B(4, 1), C(7, 7) and D(3, 9), drawn to scale. The slopes of AB and BC multiply to -1, so the sides meet at a right angle. The distance formula gives the side lengths 2√5 and 3√5, so the area is 30 square units and the perimeter is 10√5 ≈ 22.36 units.
04
Homework Assignment
~30 min
HSG.GPE.B.7 Homework: Perimeter and Area with Coordinates
Directions: Plot every figure on graph paper. Show each side length or base and height with the coordinates you used. Give exact answers first, then round decimals to the nearest hundredth, and include units when a scale is given.
Part 1: Perimeter (Problems 1-2)
Find the perimeter of the triangle with vertices J(-4, -1), K(4, -1) and L(0, 2). What kind of triangle is it?
A surveyor's map uses 1 grid unit = 10 meters. A lot has corners at (0, 0), (10, 0), (13, 4), (7, 12) and (0, 6). Find the length of fence needed to enclose the lot, to the nearest meter.
Part 2: Area (Problems 3-4)
Use the box method to find the area of the triangle with vertices A(-3, 2), B(5, -2) and C(1, 6). Then use the distance formula to decide whether the triangle is isosceles.
Show that E(-2, -1), F(4, -4), G(8, 4) and H(2, 7) are the vertices of a rectangle. Then find its area and its perimeter.
Part 3: Modeling and Reasoning (Problems 5-6)
A plan for a triangular garden bed uses 1 unit = 1 foot, with corners at (2, 1), (14, 4) and (6, 10). (a) Find the area of the bed. (b) Plastic edging is sold in 8-foot pieces. How many pieces are needed to go around the bed? (c) One bag of mulch covers 12 square feet. How many bags are needed?
Every triangle with vertices (0, 0), (10, 0) and (x, 4) has the same area. Explain why, and give the area. Then find the perimeter when x = 0, x = 5 and x = 10. Which of these three triangles has the smallest perimeter?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Side Lengths
Every side found correctly with the distance formula or subtraction
One side missing or one arithmetic error
Most sides incorrect
Area Method
Base and perpendicular height, box method or slope check used correctly
Correct method with an error in one step
Slanted sides multiplied without justification
Exact and Rounded Answers
Exact values kept, rounded only at the end, units and scale correct
Correct values with early rounding or missing units
Answers missing or incorrect
Reasoning in Context
Pieces, bags and comparisons justified from the computed values
Correct values, conclusion not explained
No conclusion
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
What is the distance between (-1, 4) and (5, -4)?
Answer: B
Δx = 5 - (-1) = 6 and Δy = -4 - 4 = -8, so d = √(6² + (-8)²) = √100 = 10. Choice A adds the legs instead of using the Pythagorean Theorem. Choice C forgets to square the differences, and choice D forgets the square root.
Question 2 of 20 · Multiple Choice
What is the perimeter of the triangle with vertices (0, 0), (0, 9) and (12, 0)?
Answer: D
The legs are 9 and 12, and the hypotenuse is √(12² + 9²) = √225 = 15, so the perimeter is 9 + 12 + 15 = 36. Choice A leaves out the slanted side. Choice B is the area, ½(12)(9). Choice C multiplies the legs.
Question 3 of 20 · Multiple Choice
What is the area of the triangle with vertices (-2, -3), (6, -3) and (1, 4)?
Answer: A
The base from (-2, -3) to (6, -3) is horizontal with length 8. The height is the vertical distance from y = -3 to y = 4, which is 7. Area = ½(8)(7) = 28. Choice B forgets the ½. Choice C uses 4 as the height, the y-coordinate of the top vertex, instead of the distance from the base. Choice D uses only part of the base, from x = 1 to x = 6.
Question 4 of 20 · Multiple Choice
What is the perimeter of the rectangle with vertices (-4, 1), (3, 1), (3, -5) and (-4, -5)?
Answer: C
The width is 3 - (-4) = 7 and the height is 1 - (-5) = 6, so P = 2(7 + 6) = 26. Choice A is the area. Choice B adds only one width and one height. Choice D uses 5 - 1 = 4 for the height, dropping the sign of -5.
Question 5 of 20 · Multiple Choice
The points (1, 1), (3, -1), (6, 2) and (4, 4) are the vertices of a rectangle. What is its area?
Answer: C
The sides are √(2² + 2²) = 2√2 and √(3² + 3²) = 3√2, and the slopes -1 and 1 show the right angles. Area = 2√2 · 3√2 = 12. Choice B is the perimeter, 2(2√2 + 3√2). Choice D is the area of the 5-by-5 box around the rectangle, and choice A halves the correct area as if it were a triangle.
Question 6 of 20 · Multiple Choice
Use the box method to find the area of the triangle with vertices (0, 1), (4, 0) and (3, 5).
Answer: A
The box from x = 0 to 4 and y = 0 to 5 has area 20. The corner triangles have areas ½(4)(1) = 2, ½(1)(5) = 2.5 and ½(3)(4) = 6, a total of 10.5. Area = 20 - 10.5 = 9.5. Choice B is the box alone, choice C is the corner triangles alone, and choice D doubles the correct area (a missing ½).
Question 7 of 20 · Multiple Choice
A student finds the distance from (2, 7) to (5, 3) by writing √(3 + 4) = √7. What is the correct distance?
Answer: D
Δx = 3 and Δy = -4, so d = √(3² + (-4)²) = √(9 + 16) = 5. The student did not square the differences (choice A). Choice B adds the legs. Choice C subtracts the squares: the squares are always added, and (-4)² = 16 is positive.
Question 8 of 20 · Multiple Choice
What is the perimeter of the square with vertices (0, 0), (3, 1), (2, 4) and (-1, 3)?
Answer: B
Each side is √(3² + 1²) = √10, so P = 4√10 ≈ 12.65. Choice A is the area, (√10)² = 10. Choice C adds 3 + 1 for each side instead of using the distance formula. Choice D multiplies 4 by 10 and forgets the square root.
Question 9 of 20 · Multiple Choice
Triangle ABC has A(0, 0), B(6, 0) and C(2, k) with k > 0. Its area is 21 square units. What is k?
Answer: A
AB lies on the x-axis with length 6, so the height is k. Then ½(6)k = 21, so 3k = 21 and k = 7. Choice B solves 6k = 21 and forgets the ½. Choice C uses 3, half of the base, as the base and then the ½ again. Choice D solves 2k = 21, using the x-coordinate 2 in place of ½(6) = 3.
Question 10 of 20 · Multiple Choice
A map uses 1 grid unit = 20 m. A triangular field has corners at (0, 0), (15, 0) and (15, 8). How much fence is needed to go around the field?
Answer: B
The sides are 15, 8 and √(15² + 8²) = 17 units, a total of 40 units. Each unit is 20 m, so the fence is 40 × 20 = 800 m. Choice A forgets the scale. Choice C leaves out the slanted side: (15 + 8) × 20. Choice D multiplies the area in grid units, 60, by 20.
Question 11 of 20 · Multiple Choice
For the same field (1 unit = 20 m, corners (0, 0), (15, 0) and (15, 8)), what is the area in square meters?
Answer: D
The area on the grid is ½(15)(8) = 60 square units. One square unit is 20 m by 20 m = 400 m², so the area is 60 × 400 = 24,000 m². Choice A multiplies by 20 instead of 20², a common error with scales. Choice B forgets the ½, and choice C forgets the scale.
Question 12 of 20 · Multiple Choice
The points A(-1, -1), B(3, 0), C(1, 8) and D(-3, 7) are the vertices of a rectangle. What is its area?
Answer: C
AB = √(4² + 1²) = √17 and BC = √((-2)² + 8²) = √68 = 2√17. The slopes ¼ and -4 multiply to -1. Area = √17 · 2√17 = 34. Choice A squares only one side. Choice B is the perimeter, and choice D is the area of the 6-by-9 box around the rectangle.
Question 13 of 20 · Multiple Choice
What is the perimeter of the quadrilateral with vertices (0, 0), (6, 0), (9, 4) and (3, 4)?
Answer: B
The horizontal sides are 6 each. The slanted sides are √(3² + 4²) = 5 each. P = 6 + 5 + 6 + 5 = 22. Choice A uses the height 4 as the slanted side. Choice C treats all four sides as 6. Choice D uses the horizontal shift 3 as the slanted side.
Question 14 of 20 · Multiple Choice
The diagonal from (0, 0) to (9, 4) splits the quadrilateral (0, 0), (6, 0), (9, 4), (3, 4) into two triangles. What is the total area?
Answer: D
The triangle (0, 0), (6, 0), (9, 4) has base 6 and height 4, area 12. The triangle (0, 0), (9, 4), (3, 4) has horizontal base 6 from (3, 4) to (9, 4) and height 4, area 12. Total 24. Choice A multiplies the base 6 by the slanted side 5. Choice B is only one triangle, and choice C forgets the ½ in both triangles.
Question 15 of 20 · Short Answer
Find the perimeter and the area of the triangle with vertices (-1, -2), (5, 6) and (-1, 6).
The side from (-1, 6) to (5, 6) is 6, the side from (-1, -2) to (-1, 6) is 8, and the slanted side is √(6² + 8²) = 10. Perimeter = 24 units. The two legs meet at a right angle at (-1, 6), so area = ½(6)(8) = 24 square units.
Question 16 of 20 · Short Answer
Find the area of the triangle with vertices (-4, 1), (2, -3) and (3, 4).
Box: x from -4 to 3 (width 7) and y from -3 to 4 (height 7), area 49. Corner triangles: ½(6)(4) = 12, ½(1)(7) = 3.5 and ½(7)(3) = 10.5, a total of 26. Area = 49 - 26 = 23 square units.
Question 17 of 20 · Short Answer
Find the exact perimeter of the pentagon with vertices (-2, 0), (2, -3), (6, 0), (4, 4) and (0, 4), then round it to the nearest hundredth.
A triangular lot is drawn on a grid where 1 unit = 5 m. Its corners are (0, 0), (12, 0) and (4, 9). Find the area of the lot in square meters.
The base on the x-axis is 12 units and the height is 9 units, so the area is ½(12)(9) = 54 square units. Each square unit is 5 m × 5 m = 25 m², so the lot is 54 × 25 = 1,350 m². Multiplying by 5 instead of 25 would give 270, which is wrong.
Question 19 of 20 · Short Answer
A student says the square with vertices (0, 0), (3, 4), (-1, 7) and (-4, 3) has area 49, because it fits in a box that is 7 units wide and 7 units tall. Explain the error and find the correct area.
The box around the square also contains four right triangles outside the square, so 49 is too large. Each side of the square is √(3² + 4²) = 5, and the slopes 4/3 and -3/4 show the right angles, so the area is 5² = 25 square units. Check with the box: 49 - 4 · ½(3)(4) = 49 - 24 = 25.
Question 20 of 20 · Short Answer
Show that W(-2, 3), X(1, -3), Y(3, -2) and Z(0, 4) are the vertices of a rectangle. Then find its perimeter and area.
Slopes: WX = -6/3 = -2, XY = 1/2, YZ = -2, ZW = 1/2. Opposite sides are parallel and -2 · ½ = -1, so every angle is a right angle. WX = √(3² + 6²) = 3√5 and XY = √(2² + 1²) = √5. Perimeter = 2(3√5 + √5) = 8√5 ≈ 17.89 units; area = 3√5 · √5 = 15 square units.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GPE.B.7 mean?
HSG.GPE.B.7 means students use the coordinates of the vertices of a figure to compute its perimeter, or its area when the figure is a triangle or a rectangle. The main tool is the distance formula, which gives the length of any side, including slanted ones.
Is HSG.GPE.B.7 taught in Geometry or Algebra?
HSG.GPE.B.7 is usually taught in high school Geometry, in the unit on coordinate geometry. It uses algebra skills (square roots and slopes) but the content is geometric: lengths, perimeters and areas.
What is the distance formula, and where does it come from?
The distance between (x₁, y₁) and (x₂, y₂) is √((x₂ - x₁)² + (y₂ - y₁)²). It is the Pythagorean Theorem: the horizontal change and the vertical change are the legs of a right triangle, and the segment is its hypotenuse. Students who remember the right triangle can rebuild the formula at any time.
How do you find the area of a triangle when no side is horizontal or vertical?
Enclose the triangle in the smallest rectangle with horizontal and vertical sides, then subtract the right triangles in the corners. Each corner triangle has horizontal and vertical legs, so its area is easy. Another option is to find a base with the distance formula and the perpendicular height to it, which takes more work.
Does the standard include the area of other polygons, like trapezoids?
No. The standard names perimeters of polygons, but areas only of triangles and rectangles. Other shapes can still appear when they split into triangles and rectangles, as in the garden activity, and the shoelace formula is an extension beyond the standard.
Why is my area wrong when I multiply two sides of a triangle?
The height of a triangle must be perpendicular to the base. Two slanted sides of a triangle usually do not meet at a right angle, so their product, halved, is not the area. Check the angle with slopes, or use the box method instead.
How do I know a quadrilateral is a rectangle before multiplying its sides?
Find the slopes of all four sides. Opposite sides must have equal slopes, and consecutive sides must have slopes whose product is -1 (or one side horizontal and the other vertical). Only then is area = length × width. Proving shapes with slopes is the related standard HSG.GPE.B.4.
Should answers be exact or rounded?
Keep exact values such as 2√5 while you work, and round only the final answer. Rounding each side first can change the last digit of a perimeter. Many teachers ask for both the exact value and a decimal to the nearest hundredth.
How does a map scale change perimeter and area?
If 1 grid unit stands for k meters, every length is multiplied by k and every area by k². A perimeter of 30 units at 10 m per unit is 300 m, but an area of 50 square units is 50 × 100 = 5,000 m².
Is coordinate geometry on the SAT?
Yes. Distances, areas and figures in the coordinate plane appear in the Geometry and Trigonometry domain of the digital SAT Math section. The skills in this standard, especially the distance formula and the area of triangles, apply directly.
07
Related Standards
5 standards
These standards connect to HSG.GPE.B.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
6.G.A.3Prerequisite
Draw polygons in the coordinate plane and find horizontal and vertical side lengths
