QuanticaPrealgebraOpen in the courseRight triangles come with a rule that ties their three sides together, and it holds exactly, not just close enough. This lesson does more than hand you the formula. You will see why it has to be true using nothing but the areas of squares, and then use it to find lengths you cannot measure directly.
Problem
In the diagram above is a right triangle with legs \(3\) and \(4\). Find the hypotenuse. Give the number of units.
Show a hint
- The hypotenuse is the side opposite the right angle. The Pythagorean Theorem says the squares of the two legs add up to the square of the hypotenuse.
- Square each leg and add, then take the square root of that total to get the hypotenuse.
Show the full solution
Square the legs and add. \(c^2 = 3^2 + 4^2 = 9 + 16 = 25\), so \(c = \sqrt{25} = \boxed{5}\). That is the 3-4-5 right triangle, the first one worth memorizing. Note the last step is a square root, not the sum itself.
Problem
In the diagram above is a right triangle with legs \(6\) and \(8\). Find the hypotenuse. Give the number of units.
Show a hint
- The hypotenuse is the side across from the right angle. The Pythagorean Theorem says its square equals the sum of the squares of the two legs.
- Square each leg and add them to get \(c^2\). Then take the square root to find \(c\).
Show the full solution
Add the squares of the legs. $$c^2 = 6^2 + 8^2 = 36 + 64 = 100$$ Take the square root, \(c = \sqrt{100} = \boxed{10}\). This is the 3-4-5 triangle with every side doubled, so you could have read the answer off without squaring anything.
Problem
In the diagram above is a right triangle with legs \(5\) and \(12\). Find the hypotenuse. Give the number of units.
Show a hint
- The hypotenuse \(c\) is the side opposite the right angle. By the Pythagorean Theorem, \(a^2 + b^2 = c^2\), so square each leg and add.
- You have \(c^2 = 5^2 + 12^2\). Work out the two squares, add them, then take the square root to get \(c\).
Show the full solution
Add the squares of the legs. $$c^2 = 5^2 + 12^2 = 25 + 144 = 169$$ Since \(13^2 = 169\), the hypotenuse is \(\boxed{13}\). The 5-12-13 triangle is worth memorizing right alongside 3-4-5.
Problem
In the diagram above is a right triangle with hypotenuse \(13\) and one leg \(5\). The other leg is the question mark. Find the missing leg. Give the number of units.
Show a hint
- The Pythagorean Theorem says \(a^2 + b^2 = c^2\). Here the hypotenuse is known and one leg is missing, so rearrange to \(b^2 = c^2 - a^2\) and subtract instead of add.
- Compute \(13^2\) and \(5^2\), subtract, then take the square root of what is left.
Show the full solution
The hypotenuse is known here, so subtract. $$b^2 = 13^2 - 5^2 = 169 - 25 = 144$$ Take the square root, \(b = \sqrt{144} = \boxed{12}\). Looking for a leg always means subtracting the squares. These sides are the 5-12-13 triple again.
Problem
In the diagram above is a right triangle with legs \(12\) and \(16\). Find the hypotenuse. Give the number of units.
Show a hint
- The Pythagorean Theorem says \(a^2 + b^2 = c^2\). Square both legs and add to get \(c^2\), then take the square root.
- Look closer at \(12\) and \(16\). They are \(4\) times \(3\) and \(4\), so this is a 3-4-5 triangle scaled up by \(4\).
Show the full solution
Add the squares of the legs. $$12^2 + 16^2 = 144 + 256 = 400$$ Take the square root, \(c = \sqrt{400} = \boxed{20}\). Faster route, \(12\) and \(16\) are \(4\) times \(3\) and \(4\), so this is a 3-4-5 triangle scaled by \(4\) and the hypotenuse is \(4 \times 5\).
Problem
In the diagram above is a right triangle with legs \(8\) and \(15\). Find the hypotenuse. Give the number of units.
Show a hint
- The hypotenuse squared equals the sum of the squares of the legs, so \(c^2 = 8^2 + 15^2\).
- Add \(64 + 225\), then take the square root of that total.
Show the full solution
Add the squares of the legs. $$c^2 = 8^2 + 15^2 = 64 + 225 = 289$$ Take the square root, \(c = \sqrt{289} = \boxed{17}\). Add 8-15-17 to your list of whole-number triples, next to 3-4-5 and 5-12-13.
Problem
In the diagram above is a right triangle with hypotenuse \(17\) and one leg \(8\). Find the missing leg. Give the number of units.
Show a hint
- This time the hypotenuse is known and a leg is missing, so rearrange the Pythagorean Theorem to \(b^2 = c^2 - a^2\).
- Square the hypotenuse and the known leg, subtract, then take the square root of what is left.
Show the full solution
The hypotenuse is given, so subtract. $$b^2 = 17^2 - 8^2 = 289 - 64 = 225$$ Take the square root, \(b = \sqrt{225} = \boxed{15}\). That fills in the 8-15-17 triple from the last problem, read backwards.
Problem
In the diagram above, a ladder leans against a wall. Its foot sits \(7\) feet from the wall, and the top reaches \(24\) feet up the wall. How long is the ladder? Give the number of units.
Show a hint
- The wall and the ground meet at a right angle, so the ladder, the ground, and the wall form a right triangle. The ladder leans across from the right angle, so it is the hypotenuse. The two legs are the \(7\) feet along the ground and the \(24\) feet up the wall.
- By the Pythagorean Theorem, square each leg and add them to get the square of the ladder's length, \(c^2 = 7^2 + 24^2\). Work that sum out, then take the square root to get \(c\).
Show the full solution
The ladder is the hypotenuse, and the legs are the \(7\) feet along the ground and the \(24\) feet up the wall. $$c^2 = 7^2 + 24^2 = 49 + 576 = 625$$ Taking the square root, the ladder is \(\boxed{25}\) feet long. A wall meets the ground at a right angle, so a leaning ladder is always the hypotenuse. 7-24-25 is another triple worth knowing.
Problem
In the diagram above, the tops of two poles are joined by a straight wire. The tops differ in height by \(20\) feet, and the poles stand \(21\) feet apart. Find the length of the wire. Give the number of units.
Show a hint
- The wire, the horizontal gap, and the height difference form a right triangle, with the wire as the hypotenuse. Which two lengths are the legs?
- By the Pythagorean Theorem, square the two legs and add them, then take the square root to get the wire.
Show the full solution
The wire is the hypotenuse, with legs \(20\) for the height difference and \(21\) for the gap between the poles. $$c^2 = 20^2 + 21^2 = 400 + 441 = 841$$ Taking the square root, the wire is \(\boxed{29}\) feet long. The poles do not need equal heights for this to work, since only the difference between the tops matters.
Problem
In the diagram above, two points sit \(9\) units apart horizontally and \(40\) units apart vertically. The straight line between them is the hypotenuse of a right triangle whose legs are those two gaps. Find the straight distance between the points. Give the number of units.
Show a hint
- The horizontal and vertical gaps are the two legs, and the straight distance is the hypotenuse \(c\). Use \(a^2 + b^2 = c^2\) with the legs \(9\) and \(40\).
- Square each leg and add. \(9^2 = 81\) and \(40^2 = 1600\), so \(c^2 = 1681\). Now take the square root.
Show the full solution
The straight distance is the hypotenuse, with the horizontal and vertical gaps as the legs. By the Pythagorean Theorem, $$c^2 = 9^2 + 40^2 = 81 + 1600 = 1681.$$ Taking the square root, \(c = \sqrt{1681} = 41\). So the distance is \(\boxed{41}\).
Problem
In the diagram above, a rectangle has a diagonal of length \(26\) and one side of length \(10\). Find the length of the other side. Give the number of units.
Show a hint
- The diagonal cuts the rectangle into two right triangles. In each one the two sides of the rectangle are the legs and the diagonal is the hypotenuse.
- You know the hypotenuse and one leg, so use \(b^2 = c^2 - a^2\) to find the missing leg.
Show the full solution
The diagonal is the hypotenuse of a right triangle whose legs are the two sides of the rectangle. To find a missing leg you subtract the squares and take the square root, $$\sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24.$$ So the other side is \(\boxed{24}\).
Problem
In the diagram above is a right triangle with legs \(35\) and \(12\). Find the hypotenuse. Give the number of units.
Show a hint
- The Pythagorean Theorem says the squares of the two legs add up to the square of the hypotenuse, \(a^2 + b^2 = c^2\). Square each leg and add.
- You should have \(35^2 + 12^2 = 1369\). The hypotenuse is the square root of that. Try to spot which whole number squares to \(1369\).
Show the full solution
Add the squares of the legs. $$c^2 = 35^2 + 12^2 = 1225 + 144 = 1369$$ Take the square root, \(c = \sqrt{1369} = \boxed{37}\). 12-35-37 is one more whole-number triple. The two legs can be wildly different in size and the hypotenuse still comes out whole.
Problem
In the diagram above is a right triangle with hypotenuse \(25\) and one leg \(20\). The other leg is unknown. Find the area of the triangle. Give the number of square units.
Show a hint
- To find the area you need a base and a height, and the two legs work perfectly. But you only know one leg, so use the Pythagorean Theorem first to get the missing one.
- Once you have both legs, the area of a right triangle is half the product of the legs, since the legs are the base and height.
Show the full solution
Find the missing leg first. $$b = \sqrt{25^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15$$ In a right triangle the legs are the base and height, so the area is $$\frac{1}{2}\times 20 \times 15 = \boxed{150}.$$ The hypotenuse is never the base or the height, which is why the missing leg has to come first.
Practice these ideas
Practice
In the diagram above is a right triangle with legs \(16\) and \(12\). Find the hypotenuse. Give the number of units.
Show the solution
The two legs are \(16\) and \(12\), so the hypotenuse is the square root of the sum of their squares.$$c = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$$So the hypotenuse is \(\boxed{20}\).
Practice
In the diagram above is a right triangle with legs \(24\) and \(7\). Find the hypotenuse. Give the number of units.
Show the solution
Add the squares of the legs. $$c^2 = 24^2 + 7^2 = 576 + 49 = 625$$ Taking the square root, \(c = \sqrt{625} = \boxed{25}\). This is the 7-24-25 triangle, another whole-number triple worth recognizing on sight.
Practice
In the diagram above is a right triangle with legs \(21\) and \(20\). Find the hypotenuse. Give the number of units.
Show the solution
By the Pythagorean Theorem, the hypotenuse is the square root of the sum of the squares of the legs. $$c = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841}$$ Since \(29^2 = 841\), the hypotenuse is \(\boxed{29}\).
Practice
In the diagram above is a right triangle with hypotenuse \(25\) and one leg \(24\). Find the other leg. Give the number of units.
Show the solution
Subtract to isolate the missing leg. $$b^2 = 25^2 - 24^2 = 625 - 576 = 49$$ Take the square root, \(b = \sqrt{49} = \boxed{7}\). Same 7-24-25 triangle as before, only now you are handed the hypotenuse and asked for a leg.
Practice
In the diagram above is a right triangle with hypotenuse \(15\) and one leg \(9\). Find the missing leg. Give the number of units.
Show the solution
Since we know the hypotenuse and want a leg, we subtract. By the Pythagorean Theorem, \(b^2 = c^2 - a^2\), so $$b^2 = 15^2 - 9^2 = 225 - 81 = 144.$$ Taking the square root, \(b = \sqrt{144} = 12\). The missing leg is \(\boxed{12}\).
Practice
In the diagram above is a right triangle with hypotenuse \(41\) and one leg \(40\). Find the length of the other leg. Give the number of units.
Show the solution
Subtract the squares. $$b^2 = 41^2 - 40^2 = 1681 - 1600 = 81$$ Take the square root, so the other leg is \(\boxed{9}\) units. The sides \(9\), \(40\), \(41\) form a Pythagorean triple, and it turned up earlier in this lesson too.
Practice
In the diagram above is a right triangle with legs \(24\) and \(10\). Find the hypotenuse. Give the number of units.
Show the solution
The hypotenuse is the longest side, opposite the right angle. By the Pythagorean Theorem, \(c^2 = 24^2 + 10^2\), so $$c = \sqrt{576 + 100} = \sqrt{676} = 26.$$ The hypotenuse is \(\boxed{26}\).
Practice
In the diagram above, a ladder leans against a wall. Its foot sits \(5\) feet from the wall and its top reaches \(12\) feet up. The ladder itself is the hypotenuse of the right triangle. How long is the ladder? Give the number of units.
Show the solution
The ladder is the hypotenuse, with legs \(5\) along the ground and \(12\) up the wall. $$c^2 = 5^2 + 12^2 = 25 + 144 = 169$$ Taking the square root, the ladder is \(\boxed{13}\) feet long. The wall meets the ground at a right angle, so a leaning ladder is always the hypotenuse. These sides are the 5-12-13 triple.
Practice
In the diagram above the two points are \(8\) apart horizontally and \(15\) apart vertically. The straight line joining them is the hypotenuse of a right triangle whose legs are those two gaps. Find the distance between the points. Give the number of units.
Show the solution
The two gaps are the legs and the straight distance is the hypotenuse. $$\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$$ So the distance is \(\boxed{17}\) units. This is the 8-15-17 triple again, showing up as a distance instead of a triangle side.
Practice
In the diagram above, a wire runs from the top of one pole to the top of another. The tops differ in height by \(24\) feet, and the poles stand \(18\) feet apart. Find the length of the wire. Give the number of units.
Show the solution
The wire is the hypotenuse of a right triangle with legs \(18\) and \(24\), so by the Pythagorean Theorem $$18^2 + 24^2 = 324 + 576 = 900.$$ Take the square root, \(\sqrt{900} = 30\). The wire is \(\boxed{30}\) feet long.
Practice
The diagram above shows a rectangle with a diagonal of \(37\) and one side of \(12\). The diagonal cuts the rectangle into a right triangle, with the diagonal as the hypotenuse. Find the length of the other side. Give the number of units.
Show the solution
The diagonal is the hypotenuse, so the missing side is a leg. To find a leg, subtract the squares. $$b^2 = 37^2 - 12^2 = 1369 - 144 = 1225.$$ Take the square root. \(b = \sqrt{1225} = 35\). The other side is \(\boxed{35}\) units.
Practice
In the diagram above is a right triangle with legs \(40\) and \(9\). Find the hypotenuse. Give the number of units.
Show the solution
The hypotenuse squared equals the sum of the squares of the legs, so $$c^2 = 40^2 + 9^2 = 1600 + 81 = 1681.$$ Take the square root, \(c = \sqrt{1681} = 41\). The hypotenuse is \(\boxed{41}\).
Practice
In the diagram above is a right triangle with hypotenuse \(17\) and one leg \(15\). Find the area of the triangle. Give the number of square units.
Show the solution
Find the missing leg first. $$b^2 = 17^2 - 15^2 = 289 - 225 = 64,\quad b = \sqrt{64} = 8$$ The legs \(15\) and \(8\) are the base and height, so the area is $$\tfrac{1}{2}\times 15 \times 8 = \boxed{60}$$ square units. Only the legs serve as base and height in a right triangle, so the hypotenuse never enters the area.