Pythagorean Theorem worksheet

1. Leg b: 4

1. Start from the rule a² + b² = c² (legs a and b, hypotenuse c) and move a² to the other side, so the missing leg is on its own

   $b^2$$=c^2-a^2$

2. Put in the hypotenuse and the leg you know, then square them

   $b^2$$=5^2-3^2$$=25-9$$=16$

3. Take the square root to undo the squaring and get leg b

   $b$$=\sqrt{16}$$=4$

4. Find the area: the two legs act as the base and the height, so multiply them and halve it

   $\frac{3\times 4}{2}$$=6$

5. Find the perimeter, the distance all the way around, by adding the three sides

   $3+4+5$$=12$

6. Find the angle across from leg a. Divide a by b, then use the tan⁻¹ (inverse tangent) button to turn that into an angle

   $A$$=\text{tan⁻¹}\left(\frac{3}{4}\right)$$=36.8699\text{°}$

7. The two smaller angles always add up to 90°, so subtract to get the angle across from leg b

   $90-36.8699$$=53.1301\text{°}$

2. Right triangle? No, it is an obtuse triangle

1. The longest side, 5, would have to be the hypotenuse

   $3\text{,}3\text{,}5$

2. Square the two shorter sides (multiply each by itself) and add them

   $3^2+3^2$$=9+9$$=18$

3. Square the longest side, so you can compare it with that sum

   $5^2$$=25$

4. The sum of the squares is smaller than the longest side squared, so the largest angle is more than 90° (obtuse)

   $18<25$

5. So the answer is

   $\text{No, it is an obtuse triangle}$

3. Hypotenuse c: 5

1. Start from the Pythagorean rule: the two legs (a and b) squared add up to the hypotenuse (c) squared

   $c^2$$=a^2+b^2$

2. Put in the two legs and square each one (multiply it by itself)

   $c^2$$=3^2+4^2$$=9+16$$=25$

3. Take the square root to undo the squaring and get the hypotenuse

   $c$$=\sqrt{25}$$=5$

4. Find the area: the two legs act as the base and the height, so multiply them and halve it

   $\frac{3\times 4}{2}$$=6$

5. Find the perimeter, the distance all the way around, by adding the three sides

   $3+4+5$$=12$

6. Find the angle across from leg a. Divide a by b, then use the tan⁻¹ (inverse tangent) button to turn that into an angle

   $A$$=\text{tan⁻¹}\left(\frac{3}{4}\right)$$=36.8699\text{°}$

7. The two smaller angles always add up to 90°, so subtract to get the angle across from leg b

   $90-36.8699$$=53.1301\text{°}$

4. Hypotenuse c: 5.66

1. Start from the Pythagorean rule: the two legs (a and b) squared add up to the hypotenuse (c) squared

   $c^2$$=a^2+b^2$

2. Put in the two legs and square each one (multiply it by itself)

   $c^2$$=4^2+4^2$$=16+16$$=32$

3. Take the square root to undo the squaring and get the hypotenuse

   $c$$=\sqrt{32}$$=5.656854$

4. For the exact answer, pull the biggest perfect square (like 4, 9 or 25) out of 32

   $\sqrt{32}$$=\text{4√2}$

5. Find the area: the two legs act as the base and the height, so multiply them and halve it

   $\frac{4\times 4}{2}$$=8$

6. Find the perimeter, the distance all the way around, by adding the three sides

   $4+4+5.656854$$=13.656854$

7. Find the angle across from leg a. Divide a by b, then use the tan⁻¹ (inverse tangent) button to turn that into an angle

   $A$$=\text{tan⁻¹}\left(\frac{4}{4}\right)$$=45\text{°}$

8. The two smaller angles always add up to 90°, so subtract to get the angle across from leg b

   $90-45$$=45\text{°}$

5. Right triangle? No, it is an obtuse triangle

1. The longest side, 7, would have to be the hypotenuse

   $4\text{,}5\text{,}7$

2. Square the two shorter sides (multiply each by itself) and add them

   $4^2+5^2$$=16+25$$=41$

3. Square the longest side, so you can compare it with that sum

   $7^2$$=49$

4. The sum of the squares is smaller than the longest side squared, so the largest angle is more than 90° (obtuse)

   $41<49$

5. So the answer is

   $\text{No, it is an obtuse triangle}$

6. Hypotenuse c: 5.83

1. Start from the Pythagorean rule: the two legs (a and b) squared add up to the hypotenuse (c) squared

   $c^2$$=a^2+b^2$

2. Put in the two legs and square each one (multiply it by itself)

   $c^2$$=3^2+5^2$$=9+25$$=34$

3. Take the square root to undo the squaring and get the hypotenuse

   $c$$=\sqrt{34}$$=5.830952$

4. For the exact answer, pull the biggest perfect square (like 4, 9 or 25) out of 34

   $\sqrt{34}$$=\text{√34}$

5. Find the area: the two legs act as the base and the height, so multiply them and halve it

   $\frac{3\times 5}{2}$$=7.5$

6. Find the perimeter, the distance all the way around, by adding the three sides

   $3+5+5.830952$$=13.830952$

7. Find the angle across from leg a. Divide a by b, then use the tan⁻¹ (inverse tangent) button to turn that into an angle

   $A$$=\text{tan⁻¹}\left(\frac{3}{5}\right)$$=30.9638\text{°}$

8. The two smaller angles always add up to 90°, so subtract to get the angle across from leg b

   $90-30.9638$$=59.0362\text{°}$

7. Right triangle? Yes, it is a right triangle

1. Sort the sides so the longest comes last. The longest side, 5, would have to be the hypotenuse

   $3\text{,}4\text{,}5$

2. Square the two shorter sides (multiply each by itself) and add them

   $3^2+4^2$$=9+16$$=25$

3. Square the longest side, so you can compare it with that sum

   $5^2$$=25$

4. The sum of the squares equals the square of the longest side, so the Pythagorean rule works backwards here: the angle across from the longest side is exactly 90°

   $25$$=25$

5. So the answer is

   $\text{Yes, it is a right triangle}$

8. Right triangle? No, it is an acute triangle

1. The longest side, 4, would have to be the hypotenuse

   $3\text{,}4\text{,}4$

2. Square the two shorter sides (multiply each by itself) and add them

   $3^2+4^2$$=9+16$$=25$

3. Square the longest side, so you can compare it with that sum

   $4^2$$=16$

4. The sum of the squares is bigger than the longest side squared, so the largest angle is less than 90° (acute)

   $25>16$

5. So the answer is

   $\text{No, it is an acute triangle}$

9. Hypotenuse c: 6.4

1. Start from the Pythagorean rule: the two legs (a and b) squared add up to the hypotenuse (c) squared

   $c^2$$=a^2+b^2$

2. Put in the two legs and square each one (multiply it by itself)

   $c^2$$=5^2+4^2$$=25+16$$=41$

3. Take the square root to undo the squaring and get the hypotenuse

   $c$$=\sqrt{41}$$=6.403124$

4. For the exact answer, pull the biggest perfect square (like 4, 9 or 25) out of 41

   $\sqrt{41}$$=\text{√41}$

5. Find the area: the two legs act as the base and the height, so multiply them and halve it

   $\frac{5\times 4}{2}$$=10$

6. Find the perimeter, the distance all the way around, by adding the three sides

   $5+4+6.403124$$=15.403124$

7. Find the angle across from leg a. Divide a by b, then use the tan⁻¹ (inverse tangent) button to turn that into an angle

   $A$$=\text{tan⁻¹}\left(\frac{5}{4}\right)$$=51.3402\text{°}$

8. The two smaller angles always add up to 90°, so subtract to get the angle across from leg b

   $90-51.3402$$=38.6598\text{°}$

10. Right triangle? No, it is an acute triangle

1. The longest side, 5, would have to be the hypotenuse

   $4\text{,}4\text{,}5$

2. Square the two shorter sides (multiply each by itself) and add them

   $4^2+4^2$$=16+16$$=32$

3. Square the longest side, so you can compare it with that sum

   $5^2$$=25$

4. The sum of the squares is bigger than the longest side squared, so the largest angle is less than 90° (acute)

   $32>25$

5. So the answer is

   $\text{No, it is an acute triangle}$