FIND THE MISSING SIDE OF A RIGHT TRIANGLE

Find the values of the unknowns in the following triangles :

Problem 1 :

Solution :

Using Pythagorean theorem.

x² = 6² + 8²

x² = 36 + 64

x² = 100

x = 10

So, the unknown value is 10.

Problem 2 :

Solution :

(25)² = (15)² + x²

625 = 225 + x²

625 – 225 = x²

400 = x²

20 = x

So, the unknown value is 20.

Problem 3 :

Solution :

(30)² = (18)² + x²

900 = 324 + x²

900 – 324 = x²

576 = x²

24 = x

So, the unknown value is 24.

Problem 4 :

Solution :

x² = 2² + (1.5)²

x² = 4 + 2.25

x² = 6.25

x = 2.5

So, the unknown value is 2.5.

Problem 5 :

Solution :

x² = 2² + 2²

x² = 4 + 4

x² = 8

x = 2.828

So, the unknown value is 2.828.

Problem 6 :

Solution :

x² = 3² + 3²

x² = 9 + 9

x² = 81

x = 9

So, the unknown value is 9`.

Problem 7 :

Solution :

By observing the figure.

Using Pythagorean theorem.

x² = (10)² + (10)²

x² = 100 + 100

x² = 200

x = 14.14

So, the unknown value is 14.14.

Problem 8 :

Solution :

x² = (√2)² + (√2)²

x² = 2 + 2

x² = 4

x = 2

So, the unknown value is 2.

Problem 9 :

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Solution :

Using Pythagorean theorem, 

14² = 7² + x²

196 = 49 + x²

x² = 196 - 49

x² = 147

x = √147

x = √(3 x 7 x 7)

x = 7 √3

Problem 10 :

The skyscrapers shown are connected by a skywalk with support beams. Use the Pythagorean Theorem to approximate the length of each support beam

Solution :

x² = 23.26² + 47.57²

x² = 541.02 + 2262.90

= 2803.92

x = √2803.92

x = 52.95

x = 53 m approximately

Problem 11 :

An anemometer is a device used to measure wind speed. The anemometer shown is attached to the top of a pole. Support wires are attached to the pole 5 feet above the ground. Each support wire is 6 feet long. How far from the base of the pole is each wire attached to the ground?

Solution :

6² = d² + 5²

36 = d² + 25

d² = 36 - 25

d² = 11

d = √11

= 3.31 ft

So, the required distance is 3.31 ft

Problem 12 :

Determine the area of the indicated square in each of the following diagrams.

Solution :

Area of square at the left = 25 m²

Side length of square at left = √25

= 5 m

Area of square at the bottom = 144 m²

Side length of square at the bottom = √144

= 12 m

Let x be the side length of square at the top.

Inside the triangle, using Pythagorean theorem

x² = 5² + 12²

x² = 25 + 144

x² = 169

x = √169

x = 13 m

Area of square at the top is 169 m²

Problem 13 :

For each of the following, use the given areas to determine side lengths a, b and c.

Solution :

Area of square at the left = 144 cm²

Side length of square at left = √144

b = 12 cm

Area of square at the bottom = 81 cm²

Side length of square at the bottom = √81

a = 9 cm

Area of square at the top = 225 cm²

Side length of square at the top = √225

c = 15 cm

Problem 14 :

You are making a kite and need to figure out how much binding to buy. You need the binding for the perimeter of the kite. The binding comes in packages of two yards. How many packages should you buy

Solution :

Let x be the the length of hypotenuse of green triangle

x² = 12² + 15²

x² = 144 + 225

x² = 369

x = √369

x = 19.2 inches

Let x be the the length of hypotenuse of yellow triangle

x² = 12² + 20²

x² = 144 + 400

x² = 544

x = √544

x = 23.3 inches

Perimeter of kite = 2(19.2) + 2(23.3)

= 2(19.2 + 23.3)

= 2(42.5)

= 85 inches

1 yard = 36 inches

1 inch = 1/36 yard

85 inches = 85/36

= 2.4 yards

More than 1 package is needed.

Problem 15 :

Describe and correct the error in using the Pythagorean Theorem

Solution :

Using Pythagorean theorem, finding the missing side.

x² = 7² + 24²

x² = 49 + 576

x² = 625

x = √625

x = 25

This is the error.