How to Work with Segment Lengths

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

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 $AB = 2 \times 7$. So $AB = \boxed{14}$.

Problem

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 $$AN = AM + MN = 12 + 6 = \boxed{18}.$$ Each midpoint only halves the segment it belongs to, so $N$ cuts $MB$ in half, not $AB$.

Problem

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 $$BC = 20 - 8 = \boxed{12}.$$ Whenever a point sits between the two ends, a missing piece is the whole minus the piece you know.

Problem

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 $$27 \div 3 = \boxed{9}.$$

Problem

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 $2\times(8+5)$.

Show the full solution

A rectangle has two sides of length $8$ and two of length $5$, so the perimeter is $8 + 5 + 8 + 5 = \boxed{26}$. A faster route is to add one width and one height, then double, $2\times(8+5) = 26$.

Problem

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. $$66 \div 6 = 11$$ So one side is $\boxed{11}$.

Problem

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 $\boxed{22}$.

Problem

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 $\boxed{8}$.

Problem

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 $\boxed{15}$. 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

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 $9 - 7$, 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 $9 - 7 = 2$, so the third side must be more than $2$. The smallest whole number bigger than $2$ is $\boxed{3}$.

Problem

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 $9 - 4$ 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 $9 - 4 = 5$, so $$5 < \text{third side} < 13.$$ The whole numbers strictly between $5$ and $13$ are $6, 7, 8, 9, 10, 11, 12$. That is $\boxed{7}$ possible lengths.

Problem

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 $42 \times 4 = \boxed{168}$ dollars.

Problem

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 $AC = 2 \times 6 = 12$. The ratio $AC : CD = 3 : 1$ means $CD$ is a third of $AC$, so $CD = 12 \div 3 = 4$. Then $AD = AC + CD = 12 + 4 = \boxed{16}$.

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Practice

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 $PM$. $$PQ = 2 \times 11 = \boxed{22}$$

Practice

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 $BC = AC - AB$. Reading the diagram, $AC = 30$ and $AB = 13$, which gives $$BC = 30 - 13 = \boxed{17}.$$

Practice

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. $$XY = \frac{32}{4} = \boxed{8}$$

Practice

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 $AM = \frac{20}{2} = 10$. Since $N$ is the midpoint of $AM$, we get $AN = \frac{10}{2} = 5$. Then $N$ sits $5$ units from $A$, so $$NB = AB - AN = 20 - 5 = \boxed{15}.$$

Practice

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. $$2\times(9+6) = 2\times 15 = 30$$ The perimeter is $\boxed{30}$.

Practice

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. $$7 + 2 + 4 + 3 + 3 + 5 = \boxed{24}$$

Practice

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 $\boxed{7}$.

Practice

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. $$72 \div 8 = 9$$ One side has length $\boxed{9}$.

Practice

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 $\boxed{18}$.

Practice

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 $11 - 8 = 3$. The side has to be more than $3$, and the smallest whole number that beats $3$ is $\boxed{4}$.

Practice

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 $10-6=4$ and less than $10+6=16$, giving $4 < \text{third} < 16$. The whole numbers strictly between $4$ and $16$ run from $5$ to $15$, which is $15-5+1=11$ values. The answer is $\boxed{11}$.

Practice

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 $4 \times 14 = 56$ feet. Each foot costs $\$5$, so the fencing costs $$56 \times 5 = \boxed{280}$$ dollars.

Practice

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 $$36 \div 12 = 3.$$ Now $BC$ is worth $4$ of those pieces, so $BC = 4 \times 3 = \boxed{12}$.

Measuring Segments

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