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Heron's Formula: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Geometry. You can skip it.

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Problem 1 of 10

Problem 1 The longest side and the semiperimeter

For each proposed set of side lengths, find the semiperimeter minus the longest length. Use that difference to explain whether the lengths give a nondegenerate triangle, a flat arrangement, or a gap. All lengths are in centimeters.

Case A. The lengths are 14, 23, and 31.

Case B. The lengths are 15, 25, and 40.

Case C. The lengths are 17, 24, and 44.
1. Hint 1
  The longest length gives the smallest of the three differences from the semiperimeter.
2. Hint 2
  For positive lengths, twice the difference for the longest length is the sum of the two shorter lengths minus the longest.
Answer

$\begin{gathered}\text{Case A. Triangle, }3\text{ cm}\\ \text{Case B. Flat, }0\text{ cm}\\ \text{Case C. Gap, }-1.5\text{ cm}\end{gathered}$

Full solution

Case A.
The semiperimeter is 34 cm.

$34-31=3\text{ cm}.$

The difference is positive, so the two shorter lengths total more than the longest.
The triangle is nondegenerate.

Case B.
The semiperimeter is 40 cm.

$40-40=0\text{ cm}.$

The two shorter lengths exactly reach across the longest.
The arrangement is flat.

Case C.
The semiperimeter is 42.5 cm.

$42.5-44=-1.5\text{ cm}.$

The two shorter lengths together fall short of the longest, so a gap remains.

Answer

$\begin{gathered}\text{Case A. Triangle, }3\text{ cm}\\ \text{Case B. Flat, }0\text{ cm}\\ \text{Case C. Gap, }-1.5\text{ cm}\end{gathered}$

Key idea

With positive proposed lengths, the sign of the smallest difference tells you whether they can enclose a region.
Problem 2 A radical from decimal lengths

A triangular template has sides 6.5 cm, 8.5 cm, and 10 cm. Use Heron's formula to find its exact area in simplified radical form.
1. Hint 1
  Check the shorter two lengths against the longest. Then find half the perimeter.
2. Hint 2
  Subtract each side from the semiperimeter. After multiplying, look for a perfect square in the product.
Answer

$5\sqrt{30}\text{ cm}^2$ .

Full solution

The positive shorter lengths add to 15 cm, which is greater than 10 cm.
The semiperimeter is 12.5 cm.

Subtracting each side in order gives 6 cm, 4 cm, and 2.5 cm.

$\begin{aligned} K^2&=12.5\cdot6\cdot4\cdot2.5,\\ K^2&=750. \end{aligned}$

Write 750 as 25 times 30 and take the positive root.

$K=5\sqrt{30}\text{ cm}^2.$

Answer

$5\sqrt{30}\text{ cm}^2$ .

Key idea

Exact decimal side lengths can give an exact radical area.
Problem 3 Round the area of a cut tile

A triangular tile has sides 18 cm, 23 cm, and 33 cm. Find its area to the nearest hundredth of a square centimeter.
1. Hint 1
  Add the two shorter lengths to check the data, then halve the perimeter.
2. Hint 2
  Keep the product under the square root exact before evaluating and rounding.
Answer

Approximately $198.41\text{ cm}^2$ .

Full solution

All three lengths are positive, and 41 cm exceeds 33 cm.
The semiperimeter is 37 cm.

The three differences from 37 cm are 19 cm, 14 cm, and 4 cm.

$\begin{aligned} K^2&=37\cdot19\cdot14\cdot4,\\ K^2&=39368. \end{aligned}$

The square root of 39368 is about 198.4137.
The third digit after the decimal point is 3.

$K\approx198.41\text{ cm}^2.$

Answer

Approximately $198.41\text{ cm}^2$ .

Key idea

Round the evaluated square root to the requested place.
Problem 4 Two frames with the same length of trim

Each proposed triangular frame would use all 96 cm of straight trim. Find the third length for each frame. Check whether each can form a nondegenerate triangle, and find the exact area of each that can.

Frame A. Two pieces measure 26 cm and 34 cm.

Frame B. Two pieces measure 19 cm and 22 cm.
1. Hint 1
  Subtract the two given lengths from the full perimeter for each frame.
2. Hint 2
  Test the recovered triple before using the common semiperimeter. A positive recovered third length alone is insufficient.
Answer

$\begin{gathered}\text{Frame A. Third side }36\text{ cm}\\ \text{Area }48\sqrt{77}\text{ cm}^2\\ \text{Frame B. Third piece }55\text{ cm}\\ \text{The frame cannot close.}\end{gathered}$

Full solution

The semiperimeter for either proposed frame is 48 cm.

Frame A.
Subtract the two known lengths to find the third.

$96-26-34=36\text{ cm}.$

All sides are positive and the shorter pair totals 60 cm, which exceeds 36 cm.
The differences from 48 cm are 22 cm, 14 cm, and 12 cm.

$\begin{aligned} K^2&=48\cdot22\cdot14\cdot12,\\ K^2&=177408. \end{aligned}$

The product is $48^2$ times 77.
Take the positive square root.

$K=48\sqrt{77}\text{ cm}^2.$

Frame B.
Recover its third length the same way.

$96-19-22=55\text{ cm}.$

The shorter pieces total 41 cm, which falls short of 55 cm.
This frame cannot close.

Answer

$\begin{gathered}\text{Frame A. Third side }36\text{ cm}\\ \text{Area }48\sqrt{77}\text{ cm}^2\\ \text{Frame B. Third piece }55\text{ cm}\\ \text{The frame cannot close.}\end{gathered}$

Key idea

A known perimeter gives a missing length, but the complete triple still has to pass the triangle inequality.
Problem 5 Height across a triangular fabric panel

A triangular fabric panel has sides 13 m, 20 m, and 25 m. Find the exact perpendicular height from the opposite vertex to the 25 m side.

Use the 25 m side as the base for h.

Text description of this figure

The triangle has sides of 13, 20, and 25 meters. The 25-meter side is horizontal. From the vertex opposite that side, a dashed perpendicular segment labeled h meets the base at a right angle.
1. Hint 1
  Heron's formula gives the area from the three sides without a measured height.
2. Hint 2
  Once you have the area, double it and divide by the chosen base.
Answer

$\frac{48\sqrt{29}}{25}\text{ m}$ .

Full solution

All three lengths are positive.
The two shorter sides total 33 m, more than the 25 m base.
The semiperimeter is 29 m.

The differences from 29 m are 16 m, 9 m, and 4 m.

$\begin{aligned} K^2&=29\cdot16\cdot9\cdot4,\\ K^2&=29\cdot24^2. \end{aligned}$

$K=24\sqrt{29}\text{ m}^2.$

Half of 25 times the height equals the area in square meters.
Solve for the height in meters.

$\begin{aligned} \tfrac12\cdot25h&=24\sqrt{29},\\ h&=\frac{48\sqrt{29}}{25}. \end{aligned}$

Answer

$\frac{48\sqrt{29}}{25}\text{ m}$ .

Key idea

Use the base named in the question when recovering a height from area.
Problem 6 Do the two panels need equal paint?

A maker will paint one face of each triangular panel with a uniform coat. Use an exact area comparison to decide whether the two faces need the same amount of paint.

Panel A. The sides measure 19 cm, 22 cm, and 25 cm.

Panel B. The sides measure 11 cm, 37 cm, and 40 cm.

Both panels are drawn at the same scale. All labels are in centimeters.

Text description of this figure

At a common scale, panel A has a solid outline, 25-centimeter horizontal side, 19-centimeter left side and 22-centimeter right side. Panel B has a dashed outline, 40-centimeter horizontal side, 11-centimeter left side and 37-centimeter right side.
1. Hint 1
  Compare the two products under the square roots to compare the exact areas.
2. Hint 2
  Compute a separate semiperimeter and three differences for each panel.
Answer

Yes. Both faces have the same area and need the same amount of paint for the same coverage.

Full solution

Panel A.
All lengths are positive, and 41 cm exceeds 25 cm.
The semiperimeter is 33 cm.
Its differences from the sides are 14 cm, 11 cm, and 8 cm.

$\begin{aligned} K_A^2&=33\cdot14\cdot11\cdot8,\\ K_A^2&=40656. \end{aligned}$

Panel B.
All lengths are positive, and 48 cm exceeds 40 cm.
The semiperimeter is 44 cm.
Its differences from the sides are 33 cm, 7 cm, and 4 cm.

$\begin{aligned} K_B^2&=44\cdot33\cdot7\cdot4,\\ K_B^2&=40656. \end{aligned}$

The squared areas are equal.
The areas are positive, so they are equal as well.

Answer

Yes. Both faces have the same area and need the same amount of paint for the same coverage.

Key idea

Triangles can have different perimeters and the same exact area.
Problem 7 Match the factors in two calculations

A triangle has side lengths 13, 16, and 21 units, assigned in order to $a$ , $b$ , and $c$ . The sine area formula and cosine law give the following equation.
$16K^2=(2ab-D)(2ab+D).$
Here $D$ stands for $a^2+b^2-c^2$ . Find the value of each factor on the right. Show how each factor becomes a product of two doubled terms from Heron's formula, and give the exact area.
1. Hint 1
  Compute $D$ and $2ab$ separately. Keep the sign of $D$ .
2. Hint 2
  Rewrite the first factor as a difference involving $c^2$ and the square of the difference of $a$ and $b$ . For the second, use the square of their sum and $c^2$ .
3. Hint 3
  Match the four small factors to $2s$ and the three doubled side differences.
Answer

$\begin{gathered}2ab-D=24\cdot18\\ 2ab+D=50\cdot8\end{gathered}$
The factors are 432 and 400, respectively.
$K=60\sqrt{3}\text{ square units}.$

Full solution

All sides are positive and 29 exceeds 21, so the triangle is nondegenerate.
The semiperimeter is 25 units.
Its differences are 12, 9, and 4 units.

$\begin{aligned} D&=13^2+16^2-21^2,\\ D&=-16. \end{aligned}$

$2ab=416.$

Subtracting $D$ gives 432.
Factor the equivalent difference of squares.

$\begin{aligned} 2ab-D&=21^2-(16-13)^2,\\ 2ab-D&=(21+3)(21-3),\\ 2ab-D&=24\cdot18. \end{aligned}$

The numbers 24 and 18 are twice 12 and twice 9, respectively.
Adding $D$ gives 400.
Factor the other difference of squares.

$\begin{aligned} 2ab+D&=(13+16)^2-21^2,\\ 2ab+D&=(29+21)(29-21),\\ 2ab+D&=50\cdot8. \end{aligned}$

The numbers 50 and 8 are twice 25 and twice 4, respectively.
The four factors of 2 cancel the 16 in front of the squared area.

$\begin{aligned} K^2&=25\cdot12\cdot9\cdot4,\\ K^2&=10800. \end{aligned}$

$K=60\sqrt{3}\text{ square units}.$

Answer

$\begin{gathered}2ab-D=24\cdot18\\ 2ab+D=50\cdot8\end{gathered}$
The factors are 432 and 400, respectively.
$K=60\sqrt{3}\text{ square units}.$

Key idea

Each difference of squares reveals two doubled semiperimeter terms.
Problem 8 A field record rounds the sides first

A triangular section of fabric has measured lengths 10.2 cm, 18.2 cm, and 28.2 cm. Treat these as exact for this problem. A second record lists each side rounded to the nearest centimeter.

Case A. Check the original lengths. If they can enclose a triangle, find the area to the nearest tenth of a square centimeter.

Case B. Check the lengths in the second record. Can you use them as the sides of a nondegenerate triangle?
1. Hint 1
  Compare the original shorter-side sum with the original longest side before evaluating an area.
2. Hint 2
  Round the three lengths separately and repeat the strict inequality check.
Answer

$\begin{gathered}\text{Case A. Area }\approx22.7\text{ cm}^2\\ \text{Case B. }10,\ 18,\ 28\text{ cm}\\ \text{Flat arrangement.}\end{gathered}$

Full solution

Case A.
The positive shorter lengths add to 28.4 cm, more than 28.2 cm.
The semiperimeter is 28.3 cm.

The three differences are 18.1 cm, 10.1 cm, and 0.1 cm.

$K^2=28.3\cdot18.1\cdot10.1\cdot0.1.$

The exact product is 517.3523.
Take its square root, then round.

$K\approx22.7\text{ cm}^2.$

Case B.
The second record lists 10 cm, 18 cm, and 28 cm.

$10+18=28.$

Those lengths can only make a flat arrangement.

Answer

$\begin{gathered}\text{Case A. Area }\approx22.7\text{ cm}^2\\ \text{Case B. }10,\ 18,\ 28\text{ cm}\\ \text{Flat arrangement.}\end{gathered}$

Key idea

Rounding the sides first can change whether the triangle inequality holds.
Problem 9 Rebuild the sides from the three differences

For a nondegenerate triangle, the three differences $s-a$ , $s-b$ , and $s-c$ are 5 cm, 8 cm, and 16 cm, respectively. Find the sides and the exact area.
1. Hint 1
  Add the three differences in symbols and use the fact that twice the semiperimeter equals the perimeter.
2. Hint 2
  Subtract each known difference from the semiperimeter to recover its side. Then check the recovered lengths before applying Heron's formula.
Answer

The sides are 24 cm, 21 cm, and 13 cm, respectively. The area is $8\sqrt{290}\text{ cm}^2$ .

Full solution

Group the three copies of $s$ and the three side lengths.

$\begin{aligned} &(s-a)+(s-b)+(s-c)\\ &\qquad=3s-(a+b+c). \end{aligned}$

The parenthesized sum is the perimeter, or $2s$ .
Subtracting it from $3s$ leaves $s$ .

Their numerical sum is 29 cm.
Recover the sides by subtracting.

$\begin{aligned} a&=29-5,\\ a&=24. \end{aligned}$

$\begin{aligned} b&=29-8,\\ b&=21. \end{aligned}$

$\begin{aligned} c&=29-16,\\ c&=13. \end{aligned}$

All three are positive, and the shorter sides total 34 cm, more than 24 cm.
The squared area uses the three supplied differences.

$\begin{aligned} K^2&=29\cdot5\cdot8\cdot16,\\ K^2&=64\cdot290. \end{aligned}$

$K=8\sqrt{290}\text{ cm}^2.$

Answer

The sides are 24 cm, 21 cm, and 13 cm, respectively. The area is $8\sqrt{290}\text{ cm}^2$ .

Key idea

The three side differences add to the semiperimeter.
Problem 10 Why the sign of the product detects closure

Let $a$ , $b$ , and $c$ be positive proposed side lengths, with $c$ a largest length. Let $s$ be half their sum, and call the product under the square root in Heron's formula $P$ .
$P=s(s-a)(s-b)(s-c).$
Case A. Prove that $P$ is positive exactly when the lengths form a nondegenerate triangle.

Case B. Prove that $P$ is zero exactly when the lengths give a flat arrangement.

Case C. Prove that $P$ is negative exactly when the shorter lengths fall short of the longest.
1. Hint 1
  First decide which factors are positive regardless of whether a triangle exists.
2. Hint 2
  In twice the factor $s-a$ , group $c$ with the subtracted $a$ . Use the fact that $c$ is at least $a$ . Do the corresponding check for $s-b$ .
3. Hint 3
  The sign of $a+b-c$ decides whether the three lengths close with room to bend, meet flat, or fall short.
Answer

$\begin{gathered}\text{Case A. }P>0\text{ exactly if }a+b>c.\\ \text{Case B. }P=0\text{ exactly if }a+b=c.\\ \text{Case C. }P<0\text{ exactly if }a+b<c.\end{gathered}$

Full solution

The semiperimeter is positive because all three proposed side lengths are positive.

$2(s-a)=b+(c-a).$

Since $b$ is positive and $c$ is at least $a$ , the right side is positive.
Thus $s-a$ is positive.

$2(s-b)=a+(c-b).$

Since $a$ is positive and $c$ is at least $b$ , this right side is positive as well.
Thus $s-b$ is positive.

Multiplying $s-c$ by the other three factors preserves its sign.

$2(s-c)=a+b-c.$

Case A.
The product is positive exactly when $a+b>c$ .
With three positive lengths and $c$ a largest one, this is the strict triangle condition.

Case B.
The product is zero exactly when $a+b=c$ .
The two shorter lengths must stretch straight to meet.

Case C.
The product is negative exactly when $a+b<c$ .
The two shorter lengths cannot reach across the longest, even in a straight line.

Answer

$\begin{gathered}\text{Case A. }P>0\text{ exactly if }a+b>c.\\ \text{Case B. }P=0\text{ exactly if }a+b=c.\\ \text{Case C. }P<0\text{ exactly if }a+b<c.\end{gathered}$

Key idea

For positive proposed lengths, only the difference for the longest side can be zero or negative.

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