HOW TO FIND MISSING PYTHAGOREAN TRIPLE

Find k if the following are Pythagorean triples:

Example 1 :

{8, 15, k}

Solution :

From the given numbers, the largest number is 15.

k²  =  8² + 15²

k²  =  64 + 225

k²  =  289

k  =  17

So, {8, 15, 17} is a Pythagorean triple.

Example 2 :

{k, 24, 26}

Solution :

Let k² + 24² =  26²

k² + 576  =  676

k²  =  676-576

 k  =  √100

 k  =  10

Example 3 :

{14, k, 50}

Solution :

Let 14² + k² =  50²

196 + k²  =  2500

k²  =  2500-196

 k  =  √2304

 k  =  48

Example 4 :

For what values of n does {n, n+1, n+2} form a Pythagorean triple?

Solution :

Let n² + (n+1)² = (n+2)²

n²+n²+2n+1  =  n²+4n+4

2n²-n²+2n-4n+1-4  =  0

n²-2n-3  =  0

(n-3)(n+1)  =  0

n  =  3 and n  =  -1

So, for n  =  3 the given set will create Pythagorean triple.

Example 5 :

Show that {n, n+1, n+3} cannot form a Pythagorean triple.

Solution :

Let n² + (n+1)² = (n+3)²

n²+n²+2n+1  =  n²+6n+9

2n²-n²+2n-5n+1-9  =  0

n²-3n-8  =  0

It is not factorable. So, by solving this quadratic equation, we will not get integer value for n. 

Therefore the given set of numbers will not create Pythagorean triple.

Example 6 :

A rectangle with diagonals of length 20 cm has sides in the ratio 2 : 1.

Find the :

a) perimeter

b) area of the rectangle

Solution :

Let 2x and x be the length and width of the rectangle.

In a rectangle, the diagonal will bisect the rectangle into two right triangles.

Here 2x, x and 20 are Pythagorean triples. 

So, (2x)² + x²  =  20²

4x²+x²  =  400

5x²  =  400

x²  =  400/5

x²  =  80

x  =  √80

Perimeter of the rectangle  =  2(2x+x)

=  2(3x)

=  6x

Perimeter of the rectangle  =  6√80

=  24√5

Area of the rectangle  =  2x(x)

=  2x²

=  2(80)

=  160 cm²

Example 7 :

A rhombus has sides of length 6 cm. One of its diagonals is 10 cm long. Find the length of the other diagonal.

Solution :

In a rhombus, the diagonals will bisect each other at right angle and all the sides will have equal length.

Length of diagonal  =  10 cm

Half of the diagonal  =  5 cm

Let x be the half length of another diagonal.

So, {6, x and 5} is Pythagorean triple.

6²  =  x² + 5²

x²  =  36-25

x²  =  11

x  =  √11 ==>  3.31

2x  =  3.31(2)

2x  =  6.62

Length of another diagonal is 6.62 cm.

Example 8 :

A square has diagonals of length 10 cm. Find the length of its sides.

Solution :

Length of each side be x.

x² + x²  =  10²

2x²  =  100

x²  =  50

x  =  √50

x  =  5√2

So, side length of square is 5√2 cm.

Example 9 :

A 9 m ladder is placed against a wall. The foot of the ladder is 1.5m from the foot of the wall. How far up the wall does the ladder rea

Solution :

From the picture given above, using Pythagorean theorem

9² = x² + 1.5²

81 = x² + 2.25

81 - 2.25 = x²

x² = 78.75

x = √78.75

x = 8.87 m

So, the required height of the wall is 8.87 m.

Example 10 :

Shown is a square with side length 5 cm. Find the length of the diagonal

Solution :

Since it is square, all sides are equal.

5² + 5² = x²

x² = 25 + 25

= 50

x = √50

x = √(2 x 5 x 5)

x = 5√2

So, the missing side is 5√2 cm.

Example 11 :

Shown is a right angle triangle. Calculate:

(a) the perimeter of the triangle.

(b) the area of the triangle

Solution :

Let x be the missing side,

25² = 7² + x²

625 = 49 + x² 

x² = 625 - 49

x² = 576

x = √576

= √(24 x 24)

= 24 cm

So, the missing side is 24 cm.

a) Perimeter of the triangle = 25 + 7 + 24

= 56 cm

b) Area of triangle = (1/2) x base x height

base = 24 cm and height = 7 cm

= (1/2) x 24 x 7

= 12 x 7

= 84 cm²

Example 12 :

A rectangle is 20 cm long and 8 cm wide. Find the length of the diagonal of the rectangle.

Solution :

Let the diagonal of the rectangle be x.

x² = 20² + 8²

= 400 + 64

x² = 464

x = √464

= 21.54 cm

So, the length of the rectangle is 21.54 cm.

Example 13 :

An airplane is lying from Redville to Leek. The airplane lies 50 miles East and then 180 miles South. How far is Leek from Redville directly?

Solution :

Let x be the unknown side.

x² = 180² + 50²

= 32400 + 2500

= 34900

x = √34900

= 186.81 miles.

Example 14 :

A frame is made from wire. The frame is a trapezium Calculate the total amount of wire needed to make the frame. Give your answer to 1 decimal place.

Solution :

AB = 14 cm

EB = CD = 8 cm

AE = AB - BE

= 14 - 8

AE = 6 cm

ED = BC = 12 cm

In triangle AED,

AD² = AE² + ED²

= 6² + 12²

= 36 + 144

= 180

AD = √180

= 13.41 cm

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