Prealgebra · Lesson 11.4

Measuring Segments

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You have spent three lessons on angles, which measure the turn at a corner. Sides are the other half of a shape. A segment is the straight path between two points, and its length is the distance from one end to the other. Over the next few lessons you will split a segment into pieces, add those pieces back together, and add up all the sides of a shape to find the distance around it.

Problem
AMB7AB = ?
In the diagram above, M is the midpoint of AB and the tick marks show the two halves are equal. Given AM=7, find AB. Give the number of units.
Show a hint
  • A midpoint splits a segment into two equal pieces, so MB has the same length as AM.
  • The whole AB is those two equal halves added together. Double the 7.
Show the full solution
The midpoint M cuts AB into two equal halves, so MB=AM=7. The whole segment is both halves together, which means AB=2×7. So AB=14.
Problem
AMNBAB = 24
In the diagram above, M is the midpoint of AB with AB=24, and N is the midpoint of MB. Find AN. Give the number of units.
Show a hint
  • Start with M. A midpoint splits the whole into two equal halves, so what does that make AM and MB?
  • Now N sits in the middle of MB, which gives you MN. Then AN is AM and MN added together.
Show the full solution
Since M is the midpoint of AB, AM=MB=12. Then N is the midpoint of MB, so MN=6. Going from A to N covers AM then MN, so AN=AM+MN=12+6=18. Each midpoint only halves the segment it belongs to, so N cuts MB in half, not AB.
Problem
ABC8?20
In the diagram above, B lies on AC with AB=8, and the bracket shows the whole segment AC=20. Find BC. Give the number of units.
Show a hint
  • The two pieces AB and BC together make up the whole segment AC, so AB+BC=AC.
  • Put in the lengths you know and subtract to solve for BC.
Show the full solution
The two pieces AB and BC add up to the whole, so AB+BC=AC. That reads 8+BC=20, which gives BC=208=12. Whenever a point sits between the two ends, a missing piece is the whole minus the piece you know.
Problem
ABCD?27
In the diagram above, AB, BC, and CD each carry one tick mark, so the three pieces are equal, and AD=27. Find the length of AB. Give the number of units.
Show a hint
  • The tick marks tell you AB, BC, and CD are all the same length, and together they make up AD.
  • Three equal pieces fill AD=27, so split that whole into three.
Show the full solution
The tick marks mean AB, BC, and CD are equal, and end to end they make up AD=27. So each piece is 27÷3=9.
Problem
85
The diagram above shows a rectangle that is 8 units wide and 5 units tall. Find its perimeter. Give the number of units.
Show a hint
  • The perimeter is the distance all the way around, so add up the lengths of all four sides.
  • A rectangle has two widths and two heights. You can add 8+5+8+5, or take 2×(8+5).
Show the full solution
A rectangle has two sides of length 8 and two of length 5, so the perimeter is 8+5+8+5=26. A faster route is to add one width and one height, then double, 2×(8+5)=26.
Problem
perimeter = 66
The diagram above shows a regular hexagon. All six sides are equal, marked by the ticks, and its perimeter is 66. Find the length of one side. Give the number of units.
Show a hint
  • The perimeter is the sum of all the side lengths. Since the ticks tell you every side is equal, the six equal sides add up to 66.
  • Six equal sides summing to 66 means one side is 66 split into 6 equal parts. Divide to find it.
Show the full solution
A regular hexagon has 6 equal sides, so the perimeter is one side length added six times. To get a single side, split the perimeter into 6 equal parts. 66÷6=11 So one side is 11.
Problem
624325
The diagram above shows an L-shaped figure with its six side lengths marked 6, 2, 4, 3, 2, and 5. Find its perimeter. Give the number of units.
Show a hint
  • The perimeter is the distance all the way around, so it is the sum of every side. An L-shape has six sides, and each one is marked in the diagram.
  • Add all six lengths together in one running total, making sure you use each marked number exactly once.
Show the full solution
The perimeter is the sum of all the sides, so add the six marked lengths. 6+2+4+3+2+5=22 The perimeter is 22.
Problem
2xx2x
The diagram above shows an isosceles triangle whose two equal legs, marked with ticks and labelled 2x, are each twice its base, labelled x. The perimeter is 40. Find the base. Give the number of units.
Show a hint
  • The perimeter is the sum of all three sides, so add the two legs and the base and set that equal to 40.
  • The two legs are each 2x and the base is x, so combine the like terms into one x term before solving.
Show the full solution
The perimeter is the sum of the three sides, so 2x+2x+x=40. Combining the like terms gives 5x=40, so x=8. The base is x, which is 8.
Problem
7?9
In the diagram above, two sides of a triangle are 7 and 9, and the third side (marked with a question mark) is a whole number. Find the largest the third side could be. Give the number of units.
Show a hint
  • The third side of a triangle is always shorter than the other two sides added together. Add the two given sides.
  • Once you have that sum, the third side has to be less than it. Since the side is a whole number, take the biggest whole number below that sum.
Show the full solution
The two given sides add to 7+9=16, and the third side has to come in under that, so the biggest whole number it can be is 15. Any side of a triangle is less than the sum of the other two. At exactly 16 the triangle flattens into a straight line.
Problem
7?9
In the diagram above, two sides of a triangle are 7 and 9, and the third side (marked with a question mark) is a whole number. What is the smallest the third side could be? Give the number of units.
Show a hint
  • The triangle inequality says any two sides must sum to more than the third. Turn that around. The third side must be more than the difference of the other two.
  • Find 97, then pick the smallest whole number that is strictly larger than it.
Show the full solution
By the triangle inequality, the third side has to be more than the difference of the other two, or the triangle collapses flat. That difference is 97=2, so the third side must be more than 2. The smallest whole number bigger than 2 is 3.
Problem
4?9
In the diagram above, two sides of a triangle are 4 and 9, and the third side (the question mark) is a whole number. How many different whole-number lengths are possible for it? Give the number of whole-number lengths.
Show a hint
  • The triangle inequality says the third side has to be less than the sum of the other two and more than their difference. Work out both of those bounds from 4 and 9.
  • The third side must be more than 94 and less than 9+4. Count the whole numbers strictly between those two bounds.
Show the full solution
The triangle inequality traps the third side between the sum and the difference of the other two sides. It has to be less than 9+4=13 and more than 94=5, so 5<third side<13. The whole numbers strictly between 5 and 13 are 6,7,8,9,10,11,12. That is 7 possible lengths.
Problem
131514
In the diagram above is a triangular garden with sides 13, 14, and 15 feet. Fencing costs $4 per foot. Find the total cost to fence it. Give the number of dollars.
Show a hint
  • The fence runs all the way around, so start with the perimeter, the sum of the three sides.
  • Once you have the perimeter in feet, multiply by the cost of a single foot.
Show the full solution
The fence runs the whole border, so start with the perimeter, 13+14+15=42 feet. Each foot costs $4, so the total is 42×4=168 dollars.
Problem
ABCD6ACCD
In the diagram above, points B and C lie on segment AD with AC and CD in the ratio 3:1, and B is the midpoint of AC. The bracket shows BC=6. Find AD. Give the number of units.
Show a hint
  • Since B is the midpoint of AC, the bracket BC=6 is exactly half of AC. Use that to find the whole of AC first.
  • Once you know AC, the ratio AC:CD=3:1 tells you CD. Then AD is just AC+CD.
Show the full solution
The midpoint B splits AC into two equal halves, so BC is half of AC. That gives AC=2×6=12. The ratio AC:CD=3:1 means CD is a third of AC, so CD=12÷3=4. Then AD=AC+CD=12+4=16.

Practice these ideas

Practice
PMQ11PQ = ?
In the diagram above, M is the midpoint of PQ and PM=11. Find PQ. Give the number of units.
Show the solution
The midpoint splits PQ into two equal halves, so PQ is twice PM. PQ=2×11=22
Practice
ABC13?30
In the diagram above, B lies on AC with AB=13, and the bracket marks the whole length AC=30. Find BC. Give the number of units.
Show the solution
The point B splits AC into two pieces, AB and BC, that add up to the whole. So BC=ACAB. Reading the diagram, AC=30 and AB=13, which gives BC=3013=17.
Practice
XYZWV?32
In the diagram above, the tick marks show that XY, YZ, ZW, and WV are all equal, and the whole length XV=32. Find XY. Give the number of units.
Show the solution
The four tick marks mean XY, YZ, ZW, and WV are equal pieces that add up to XV. So one piece is a quarter of the whole. XY=324=8
Practice
ANMBAB = 20
In the diagram above, M is the midpoint of AB with AB=20, and N is the midpoint of AM. Find NB. Give the number of units.
Show the solution
Each midpoint cuts a segment in half. Since M is the midpoint of AB, we get AM=202=10. Since N is the midpoint of AM, we get AN=102=5. Then N sits 5 units from A, so NB=ABAN=205=15.
Practice
96
The diagram above shows a rectangle that is 9 wide and 6 tall. Find its perimeter. Give the number of units.
Show the solution
The perimeter is the sum of all four sides. A rectangle has two sides of length 9 and two of length 6, so add one of each and double it. 2×(9+6)=2×15=30 The perimeter is 30.
Practice
724335
The diagram above shows an L-shaped figure with its six sides labelled. Find its perimeter. Give the number of units.
Show the solution
The perimeter is the sum of every side, so add the six labelled lengths. 7+2+4+3+3+5=24
Practice
2xx2x
In the diagram above, the two equal legs of the isosceles triangle each measure 2x, the base measures x, and the perimeter is 35. Find the base. Give the number of units.
Show the solution
The perimeter is the sum of the three sides, so add the two legs and the base. Each leg is 2x and the base is x, which gives 2x+2x+x=35. Combine the like terms to get 5x=35, so x=7. The base is x, so its length is 7.
Practice
perimeter = 72
The diagram above shows a regular octagon with perimeter 72. Since all eight sides are equal, find the length of one side. Give the number of units.
Show the solution
All eight sides of a regular octagon are equal, so one side is the perimeter split into 8 equal parts. 72÷8=9 One side has length 9.
Practice
8?11
In the diagram above, two sides of the triangle measure 8 and 11, and the third side is a whole number. What is the largest length the third side can be? Give the number of units.
Show the solution
By the triangle inequality, the two known sides must sum to more than the third side, so the third side is less than 8+11=19. The largest whole number below 19 is 18.
Practice
8?11
In the diagram above, a triangle has two sides of length 8 and 11, and its third side is a whole number of units. What is the smallest length that third side can be? Give the number of units.
Show the solution
For a triangle to close up, the third side must be longer than the difference of the other two, so 118=3. The side has to be more than 3, and the smallest whole number that beats 3 is 4.
Practice
6?10
In the diagram above, two sides of a triangle are 6 and 10, and the third side is a whole number. How many different whole-number lengths could the third side be? Give the number of whole-number lengths.
Show the solution
By the triangle inequality, the third side must be less than the sum of the other two and more than their difference. So it is more than 106=4 and less than 10+6=16, giving 4<third<16. The whole numbers strictly between 4 and 16 run from 5 to 15, which is 155+1=11 values. The answer is 11.
Practice
1414
The diagram above shows a square garden that measures 14 feet on each side. Fencing costs $5 per foot, and you want to fence the whole garden. How much will the fencing cost? Give the number of dollars.
Show the solution
Fencing follows the border, so start with the perimeter. A square has four equal sides, so the perimeter is 4×14=56 feet. Each foot costs $5, so the fencing costs 56×5=280 dollars.
Practice
ABCD?AD = 36
In the diagram above, B and C lie on segment AD with AB:BC:CD=3:4:5, and AD=36. Find BC. Give the number of units.
Show the solution
The ratio splits AD into 3+4+5=12 equal pieces. So one piece is 36÷12=3. Now BC is worth 4 of those pieces, so BC=4×3=12.