Lesson Library> Pythagoras' Theorem
Pythagoras' Theorem
Introduction
In the middle of the campus, there is a large, rectangular lawn $150$ x $80$ meters with a sign saying “Do not walk on the grass”. Sarah is at one of the corners of the lawn and is heading to the study hall located at the opposite corner.
How much shorter is the distance if she walks straight across the lawn compared to if she goes around it?
Many problems in geometry involve diagonal distances. But how do we actually determine such diagonal distances? In this lesson, you will learn a formula called Pythagoras' theorem that will help you to do just that!
Some Definitions
Before we move on, you should know that the two shorter sides in a right-angled triangle are called the legs. The longest side in the triangle opposite the right-angle is called the hypotenuse.
Keep these definitions in mind. You will see them used many times again throughout this lesson.
Discovering Pythagoras' Theorem
Both squares below have a side length of $7$. Inside the squares, there are four right-angled triangles with legs of length $3$ and $4$ and hypotenuse of length $x$.
The triangles are placed so that two smaller squares are formed in the left figure and one larger square is formed in the right figure.
Exercise 1
What is the area of the grey and the white region in the left square above?
The area of the white region is the sum of the two smaller squares, which is $3^2\ +$$4^2 = \pmb{25}$. The area of the grey region is the sum of the four triangles, which is $4\cdot \frac{3 \ \cdot \ 4}{2}=\pmb{24}$. We could also say that the area of the grey region is everything that is not white, which is $7^2-25=24$.
Exercise 2
What is the area of the grey and the white region in the right square above? What then is $x$?
Both the left and the right square have the same total area, $7^2=49$, and since the grey triangles cover same area in both figures, $4\cdot \frac{3 \ \cdot \ 4}{2}=24$, this means that the remaining white area must be the same in both squares. The white area is $3^2+$$4^2 = 25$ in the left square, so it must be $\pmb{25}$ in the right square too. We now have:
$$x^2=25$$
$$\sqrt{x^2} = \pm \sqrt{25}$$
$$x= \pm 5$$
Since we are dealing with distances, only the positive solution applies. This means that $x=\pmb{5}$.
Exercise 3
Find a general formula for determining the length of a right-angled triangle’s hypotenuse if the length of the triangle’s legs are $a$ and $b$.
Let us draw the same two squares as in the above exercise, but this time let the squares have side length $(a+b)$ and the legs of the grey triangles have lengths $a$ and $b$. Let the length of the hypotenuse be $c$.
The grey triangles cover the same area, so the remaining white area must be the same in both figures. This means that:
$$\mathbf{a^2 + b^2 = c^2}$$
This relationship between a right-angled triangle's legs and its hypotenuse is what's called Pythagoras' theorem. Pythagoras' theorem basically states that the sum of the legs squared is equal to the hypotenuse squared. The length of the hypotenuse $c$ is thus:
$$\mathbf{\sqrt{a^2 + b^2} = c}$$
A great way to visualize Pythagoras' theorem is with the following figure. The area of the two smaller squares (the square of the legs) is then equal to the area of the larger square (the square of the hypotenuse).
Use the formula you found in Exercise 3 to solve the question we posed at the beginning of this lesson.
Exercise 4
In the middle of the campus, there is a large, rectangular lawn $150$ x $80$ meters with a sign saying “Do not walk on the grass”. Sarah is at one of the corners of the lawn and is heading to the study hall located at the opposite corner. How much shorter is the distance if she walks straight across the lawn compared to if she goes around it?
Let the distance across the lawn be $x$ meters. According to Pythagoras' theorem, it follows that:
$$80^2 + 150^2 = x^2$$
$$\sqrt{80^2 + 150^2} = x$$
$$\sqrt{28900} = x$$
$$170 = x$$
The distance around the lawn is $80+150=230$ meters. So the distance straight across the lawn is $230-170=\mathbf{50}$ meters shorter compared to going around it.
A Common Mistake
A common mistake many students make when working with Pythagoras' theorem is to confuse $(a + b)^2$ with $a^2 + b^2$. These expressions are not equal to each other! The first expression is equal to $a^2 + 2ab + b^2$, which is $2ab$ greater than $a^2 + b^2$. We can illustrate this with the help of the following figure, where $(a + b)^2$ is the area of the whole square while $a^2 + b^2$ is only the area of the white region.
We can take the square root out of $(a + b)^2$ and get $a + b$. However, we can not do that with $a^2 + b^2$.
$$ \sqrt{(a + b)^2} \ = \ a + b $$
$$ \mathbf{\sqrt{a^2 + b^2} \ \neq \ a + b} $$
Make sure you really understand this so that you won't fall into this trap yourself.
The Inverse Theorem
The inversion of Pythagoras' theorem also holds true. If we have three positive integers $a$, $b$ and $c$ for which $a^2 + b^2 = c^2$, they form the sides of a right-angled triangle.
We can prove this by drawing a triangle with the sides $a$, $b$ and $c$ for which $a^2 + b^2 = c^2$. We will also draw a right-angle triangle with the sides $a$, $b$ and $d$.
Since the second triangle is right-angled, we know that $a^2 + b^2 = d^2$ according to Pythagoras' theorem. It then follows that:
$a^2 + b^2 = c^2 = d^2$
$c = d$
The two triangles must therefore have the exact same side lengths and shape, which means that $a$, $b$ and $c$ is also right-angled.
Exercise 1
Determine if the following triangles are right-angled.
a)
$100=100$ so the triangle is right-angled.
b)
$7^2+10^2=149$
$12^2=144$
$149 \neq 144$ so the triangle is not right-angled.
Pythagoras' Theorem in Problem Solving
Pythagoras' theorem is one of the most useful theorems in geometry, because we only need to know the length of two of the sides of a right-angled triangle to determine the length of the third side.
Exercise 1
Determine $x$ in the figure below.
According to Pythagoras' theorem, we have:
$$ 5^2 + x^2 = 13^2 $$
$$ 25 + x^2 = 169 $$
$$ x^2 = 144 $$
The side lengths of a triangle must be positive, so it follows that $\mathbf{x = 12}$.
Exercise 2
A man walks $5$ meters north, $8$ meters east and then $10$ meters north again. How far has he moved from his starting point?
We can illustrate the man's walk with the following figure:
The man has walked a total of $15$ meters north and $8$ meters east. The distance from start to finish is then according to Pythagoras' theorem $\sqrt{8^2 + 15^2} = \mathbf{17}$ meters.
Exercise 3
In the figure below, the side $AB$ is twice as long as $BC$ and $ACDE$ is a square with the area $64$. What is the area of the triangle $ABC$?
Let $x = BC$ so that $2x = AC$. Since $AC = \sqrt{64} = 8$ it follows according to Pythagoras' theorem that:
$$AB^2+BC^2=AC^2$$
$$ (2x)^2 + x^2 = 8^2 $$
$$ 5x^2 = 64 $$
$$ x^2 = \frac{64}{5} $$
$$ x = \frac{8}{\sqrt{5}} $$
The area of $ABC$ is thus:
$\dfrac{AB \cdot BC}{2} \ = \ \dfrac{2 \bigg(\dfrac{8}{\sqrt{5}} \bigg) \cdot \dfrac{8}{\sqrt{5}}}{2} \ = \ $$ \mathbf{\dfrac{64}{5}} $
Pythagorean Triples
When three integers $a$, $b$ and $c$ satisfy Pythagoras' theorem $a^2 + b^2 = c^2$, they are called a Pythagorean triple. There are infinitely many such triples, but the four most common ones are:
$(3,4,5) \ \ \ \ \ \ (5,12,13)$
$(8,15,17) \ \ \ \ \ \ (7,24,25)$
These Pythagorean triples appear in problems frequently, so if you know these triples by heart you can perform your calculations faster and identify right-angled triangles more easily. For example, if two sides in a figure are of length $7$ and $25$, the chance is high that there is a right-angled triangle with sides $7$, $24$ and $25$ hiding somewhere.
We can also create new triples from old ones by multiplying all three integers by the same number. For example, $(9, 12, 15)$ is an extension of $(3,4,5)$, where we have multiplied by $3$, and $(10, 24, 26)$ is an extension of $(5, 12, 13)$, where we have multiplied by $2$. This can be compared to drawing a triangle with the same dimensions as the original triple and then doubling or tripling the length of its sides. Since the triangles are similar, the larger one must also be right-angled. The four triples we introduced at the beginning are what’s called primitive triples, because they can’t be simplified any further.
Exercise 1
Determine if the following triangles are right-angled or not by only performing the calculations in your head.
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We can solve all of these three exercises by scaling down the sides of the triangles to their smallest possible length where all sides are still integers. We will then check if the numbers match any of the Pythagorean triples. To scale down the sides, we divide the three number by their greatest common factor (GCF).
a) $\text{GCF}(15,20,25)=5$ $\dfrac{(15,20,25)}{5} \ =(3,4,5)$
$(3,4,5)$ is a Pythagorean triple so the triangle is right-angled.
b) $\text{GCF}(72,108,132)=12$ $\dfrac{(72,108,132)}{12} \ =(6,9,11)$
$(6,9,11)$ is not a Pythagorean triple so the triangle is not right-angled.
c) $\text{GCF}(1400,4800,5000)=200$ $\dfrac{(1400,4800,5000)}{200} \ =(7,24,25)$
$(7,24,25)$ is a Pythagorean triple so the triangle is right-angled.
Finding All Pythagorean Triples
As mentioned, there are infinitely many Pythagorean triples and they can generally be written as:
$$ \mathbf{a = k (m^2-n^2)} \\ \mathbf{b = 2kmn} \\ \mathbf{c = k (m^2 + n^2)}$$
where $k$, $m$ and $n$ are natural numbers and $m>n$.
The proof of this consists of two parts: sufficiency (that these triples meet the definition) and necessity (that all triples that meet the definition are written in the above form). For now, we will only cover the first part, to prove that such $a, b, c$ meet the definition, and leave the second part for a future lesson:
$$ a^2 + b^2 = (k (m^2-n^2))^2+ (2kmn)^2 $$
$$ a^2 + b^2 = k^2 ((m^2-n^2)^2 + 4 (mn)^2 ) $$
$$ a^2 + b^2 = k^2 (m^4-2 (mn)^2 + n^4 + 4 (mn)^2) $$
$$ a^2 + b^2 = k^2 (m^4 + 2 (mn)^2 + n^4) $$
$$ a^2 + b^2 = (k (m^2 + n^2))^2 $$
$$ a^2 + b^2 = c^2 $$
So the statement holds.
Unfolding Figures to Right-angled Triangles
In some problems the right-angled triangle won't be obviously visible at first sight. To solve these questions we must unfold, unwrap or shift a few sides in the figure to create a right-angled triangle which we can apply Pythagoras' theorem on.
Example 1
Just as the spider was about to devour the fly, it flew up to the opposite corner of the room. The dimensions of the room are $a = 5$ m, $b = 4$ m and $c = 3$ m. How long is the shortest path that the spider can crawl to get to the fly?
Solution
In the figure below, we have unfolded the room into a 2D-plane. The shortest way the spider can crawl to the opposite corner is going to be along one of the dotted lines.
The distances along these dotted lines are:
$\sqrt{5^2+7}^2=\sqrt{74}$
$\sqrt{9^2+3}^2=\sqrt{90}$
So the shortest distance the spider can crawl is $\mathbf{\sqrt{74}}$ meters.
Exercise 1
Christoffer is kayaking along a river. However, instead of paddling in a straight line he goes along the river in a zig-zag pattern, always maintaining the same angle towards the bank of the river. Christoffer paddles a total distance of $1700$ meters and crosses the river four times during his trip. If the distance between his start and his finish point was $1500$ meters, how wide is the river?
Let us "unfold" Christoffer's trip along the river. Instead of crossing the river four times, we can imagine him paddling straight across a rectangular lake that is four rivers wide and $1500$ meters long.
Let the river be $r$ meters wide. Applying Pythagoras' theorem gives us:
$$(4r)^2+1500 \ ^2=1700 \ ^2$$
$$16r^2+2 \ 250 \ 000=2 \ 890 \ 000$$
$$16r^2=640 \ 000$$
$$r^2=40 \ 000$$
$$r=200$$
So the river is $\mathbf{200}$ meters wide.
Exercise 2
A rope is attached to one end of a $12$ meter long cylindrical tube. The rope is wound four times around the tube before being fastened at the tube's opposite end. If the circumference of the tube is $4$ meters, how long is the shortest rope that can be used?
We start by cutting up the cylinder along its height and unfolding it. We then get a rectangle with a width equal to the circumference of the tube ($4$ meters) and a length of $12$ meters. Each diagonal stretch in the picture corresponds to the rope we have wrapped around the tube.
Since the rope is wrapped four times around the tube, we will have wound the rope a total of $4 \cdot 4 =16$ meters sideways and $12$ meters up along the tube.
Let the length of the rope be $x$. Applying Pythagoras' theorem gives us:
$$16^2+12^2=x^2$$
$$400=x^2$$
$$20=x$$
So the shortest rope that can be used would be $\mathbf{20}$ meters long.
Exercise 3
What is the length of the diagonal $x$ in the figure below?
We can solve this problem easily by shifting the length of the sides to create one larger right-angled triangle. The sum of all the horizontal lengths become the new triangle's base and the sum of all the vertical lengths become the new triangle's height.
The new, larger triangle's base is $4+6+3=13$ and its height is $2+4+3=9$. Using Pythagoras' theorem, we determine that triangle's hypotenuse $x=\sqrt{13^2+9^2}=$$\sqrt{250}=$$\mathbf{5\sqrt{10}}$.
Space Diagonals
Pythagoras’ theorem can also be applied to find distances in 3D. Such distances are often called space diagonals.
Exercise 1
The cuboid below has length $a$, width $b$ and height $c$. What is its space diagonal?
Let’s draw the diagonal of the rectangular base of the cuboid. We can now construct a right-angled triangle that has the base diagonal and the height $c$ as its legs. The hypotenuse of this right-angled triangle is the space diagonal we are looking for.
The length of the base diagonal is $\sqrt{a^2+b^2}$. Let the length of the space diagonal be $x$. Applying Pythagoras' theorem gives us:
$$(\sqrt{a^2+b^2} \ )^2 + c^2 \ = \ x^2$$
$$a^2+b^2 + c^2 \ = \ x^2$$
$$\mathbf{\sqrt{a^2+b^2 + c^2} \ = \ x}$$
This formula holds true generally. To find the space diagonal of a cuboid we just square the cuboid’s length, width and height, add them together and finally take the square root of the result.
Use the general formula you found in Exercise 1 to solve the problems below.
Exercise 2
In the prism below, $AC=BC=10$, $CD=15$ and $M$ is the midpoint on $AB$. Determine $DM$.
$DM$ is the hypotenuse of the right-angled triangle $\triangle DCM$. We already know the length of one of the legs $(CD=15)$, so now we just need the length of $CM$.
Since $AC=BC$, $\triangle ABC$ is an isosceles triangle with $CM$ as its height. This means that $\triangle BMC$ is right-angled. To determine $|CM|$, we first need to find the length of $MB$.
$MB$ is half of $AB$, and since $AB=$$\sqrt{10^2+10^2}=$$10\sqrt{2}$, it follows that $MB=5\sqrt{2}$. This means that $CM=\sqrt{10^2+5\sqrt{2}^2}=$$5\sqrt{6}$ which finally gives us:
$$DM \ = \sqrt{15^2+(5\sqrt{6} \ )^2}$$
$$DM \ = \sqrt{325}$$
$$\mathbf{DM \ = 5\sqrt{15}}$$
Exercise 3
The figure below has the side lengths $4$, $7$ and $5$ and a space diagonal of length $9$. Could the figure be a cuboid where all angles are $90^{\circ}$?
If the figure is a cuboid, it must fulfill the general space diagonal formula $d=\sqrt{a^2+b^2 + c^2}$. However, if we input the lengths of the figure we find that:
$$\sqrt{7^2+4^2 + 5^2} \ = \ \sqrt{90} \ \neq \ 9$$
So the figure can not be a cuboid.
Exercise 4
A cube sits tightly inside a sphere such that all eight corners touch the sphere. If the area of the cross sectional plane of the sphere is $4\pi$, what is the side of the cube?
Let the side of the cube be $x$ and the radius of the sphere be $r$. The diameter of the sphere is also the cube's space diagonal, which has the length $\sqrt{x^2+x^2+x^2} \ =$$\sqrt{3x^2}=$$x\sqrt{3}$. So if we can just find the radius of the sphere we are done. Since the area of the sphere's cross sectional plane is $4\pi$, we have:
$$4\pi \ = \ r^2 \pi$$
$$4 \ = \ r^2 $$
$$ \pm 2 \ = \ r $$
Since only positive solutions apply, the radius of the sphere must be $2$. The diameter is thus $4$, which gives us:
$$\text{Diameter} \ = \ \text{Space diagonal}$$
$$4 \ = \ x\sqrt{3}$$
$$\mathbf{\dfrac{4}{\sqrt{3}} \ = \ x}$$
Problems to Solve
We have finally come to the end of the lesson. The problems below will test you on everything that has been covered so far. If you can solve all of them, you will have fully mastered Pythagoras' theorem!
Exercise 1
Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown below. What is $DE$?
The area of the rectangle is $5 \cdot 6=30$. If we let $CE=x$, then the area of the triangle is $\dfrac{5x}{2}=30$ which means that $x=12.$ According to the Pythagoras' theorem, it follows that $DE=\sqrt{5^2+12^2}=$$\mathbf{13}$.
Exercise 2
In the figure below, $AC = 24$ and $CE = 7$. Furthermore, the triangle $ABC$ is similar to $CDE$. How long is the distance $AE$?
Do you remember the Pythagorean triple $(7, 24, 25)$? Since this problem contains two of these integers, it’s a clue that we should use Pythagoras' theorem to solve the question. Let's draw the distance $AE$ in the figure.
If the triangle $ACE$ were right-angled, we could use Pythagoras' theorem to calculate the length of $AE$. But since $\angle{CAB} + \angle{ACB} = 90^{\circ}$ and $\angle{CAB} = \angle{DCE}$ it follows that:
$$ \angle{DCE } + \angle{ACB} = \angle{C} = 90^{\circ} $$
So the triangle $ACE$ is right-angled! It then follows that:
$$ 24^2 + 7^2 = AE^2 $$
$$ \sqrt{625} = AE $$
$$ \mathbf{25 = AE} $$
Exercise 3
Rectangle $ABCD$ is inscribed in a semicircle with diameter $FE$, as shown in the figure. Let $DA=16$ and $FD=AE=9$. What is the area of $ABCD$?
Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is the midpoint of $DA$, so $OD=OA=\dfrac{16}{2}= 8$. Applying Pythagoras' theorem in right-angled triangle $ODC$, we have that $CD$ is $\sqrt{17^2-8^2}=15$. The area of $ABCD$ is therefore $16\cdot 15=\mathbf{240}$.
Exercise 4
Determine the height $h$ and the distance $x$ in the triangle below.
Observe that the lengths of the sides of the triangle satisfy Pythagoras' theorem: $(6^2 + 8^2 = 10^2)$. This means that the large triangle is right-angled.
All three triangles $ABC$, $ABD$ and $BCD$ are similar, since their angles have the same measure. This means that:
$$ \frac{AC}{BC} = \frac{BC}{CD} $$
$$ \frac{10}{6} = \frac{6}{x} $$
$$ x = 3.6 $$
and that:
$$ \frac{AC}{AB} = \frac{BC}{BD} $$
$$ \frac{10}{8} = \frac{8}{h} $$
$$ h = 4.8 $$
The answer is thus $\mathbf{x = 3.6}$ and $\mathbf{h = 4.8}$.
Exercise 5
In parallelogram $ABCD$, $DE$ is the altitude to the base $AB$ and $DF$ is the altitude to the base $BC$. If $DC=12$, $EB=4$, and $DE=6$, then what is $DF$?
Note that $(DE) \ \cdot \ (AB)=$$(DF) \ \cdot \ (BC)=$$Area(ABCD)$. If we can find $BC$, $AB$ and $DE$, we will be able to determine $DF$.
First, note that $DE=6$ and $AB=CD=12$, by properties of a parallelogram. Also, $AD=BC$.
Since $\angle DEA$ is a right angle, we can use Pythagoras' theorem:
$$(AE)^2+(ED)^2=(AD)^2$$
$$\sqrt{(AB-4)^2+6^2} \ \ = \ AD$$
$$\sqrt{8^2+36} \ \ = \ AD$$
$$\sqrt{100} \ = \ AD$$
$$10=AD=BC$$
Now we can finally substitute:
$$(6)(12)=(DF)(10)$$
$$DF=\frac{72}{10}$$
$$\mathbf{DF=7.2}$$
Exercise 6
What is $x$, the length of the square's side?
Let us draw the diagonal of the square.
If we now "extract" the three perpendicular lines and the diagonal, we can create a new right-angled triangle with legs $3$ and $12+9=21$. The hypotenuse of this triangle is the diagonal of the square.
Applying Pythagoras' theorem, we find the length of the square's diagonal to be $\sqrt{21^2+3^3}=$$\sqrt{450}=$$15\sqrt{2}$. We can also express the square's diagonal as $\sqrt{x^2+x^2}=$$x\sqrt{2}$. This means that:
$$15\sqrt{2} \ = \ x\sqrt{2}$$
$$\mathbf{15\ \ = \ x}$$
Exercise 7
What is the area of the innermost circle divided by the area of the outermost circle?
The side of the first (innermost) square is equal to the first circle's diameter. Furthermore, the diagonal of the first square is equal to the second circle's diameter. According to Pythagora's theorem, the ratio between a square and its diagonal is $1 : \sqrt{1^2+1^2}=\sqrt{2}$. Let the first circle's radius be $r$. Then the second circle's radius will be $r\sqrt{2}$. The ratio between the area of the first and the second circle is then:
$r^2 \pi \ : \ (r\sqrt{2})^2 \pi$
$r^2 \pi \ : \ 2r^2 \pi$
$1 \ : 2$
So the second circle's area is twice as large as the first. By the same reasoning, the third circle is twice as large as the second, the fourth twice as large as the third, etc. The ratio between the area of the circles is thereby:
$$1 \ : \ 2 \ : \ 4 \ : \ 8$$
So the area of the fourth (outermost) circle divided by the area of the first (innermost) circle is $\mathbf{\dfrac{1}{8}}$.
Exercise 8
In the cube $ABCDEFGH$, $J$ and $I$ are the midpoints of segments $FB$ and $HD$, respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2$?
Note that $EJCI$ is a rhombus by symmetry. Let the side length of the cube be $s$. Applying Pythagoras' theorem, we have $EC= \sqrt{s^2+s^2+s^2} \ $$ = s\sqrt 3$ and $JI= \sqrt{s^2+s^2} =$$ s\sqrt 2$. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is $EC \ \cdot \ JI = \dfrac{s^2\sqrt 6}{2}$. This gives $R = \dfrac{\sqrt 6}2$. Thus $R^2 = \mathbf{\dfrac{3}{2}}$
Exercise 9
Triangle $ABC$ is a right-angled triangle. $O$ is the centre of the circle with diameter $AB$. $M1$ and $M2$ are the mid points of $BC$ and $AC$ respectively. $M1$ and $M2$ are also the centres of the circles of diameter $BC$ and $AC$ respectively. If the area of triangle $ABC$ is $5$, what is the value of the shaded area?
If we add the areas of the two smaller semicircles and $\triangle ABC$, and then subtract the area of the largest semicircle, we are left with the area of the shaded region. Our task is therefore to find the value of:
$\text{area(Shaded region)} \ =$
$\text{area(Small semicircles)} \ + \ $$\text{area(} \triangle ABC \text{)} \ - \ $$ \text{area(Large semicircle)}$
We have seen that Pythagoras' theorem can be visualized by turning the sides of a right-angled triangle into three squares. The sum of the two smaller squares' area is then equal to the larger square's area: $a^2+b^2=c^2$.
However, Pythagoras' theorem doesn't only apply to squares. We can actually draw any type of figure from the sides of the right-angled triangle. As long as these figures are similar, the theorem still holds. Imagine that we would turn the sides of the right-angled triangle into semicircles instead of squares. There must exist some number $x$, so that if we multiply a square's area $a^2$ by $x$, the new value $xa^2$ is the area of a semicircle with the same diameter as the square's side.
If we extend this reasoning to all three squares we can draw the following figure:
$$xa^2+xb^2=xc^2$$
We can extend this reasoning to any type of figure. As long as the three figures are similar, the sum of the areas of two smaller ones will always equal the area of the largest figure.
This means that:
$\text{area(Small semicircles)} \ = $$ \text{area(Large semicircle)}$
Substituting this into our original equation gives us:
$\text{area(Shaded region)} \ =$
$\text{area(Small semicircles)} \ + \ $$\text{area(} \triangle ABC \text{)} \ - \ $$ \text{area(Small semicircles)} \ =$
$\text{area(} \triangle ABC \text{)} \ = \ 5$
So the area of the shaded region is $\pmb{5}$.