LEAVE A COMMENT FOR US

Home > Measurement > Pythagoras Theorem > Advanced problems using Pythagoras Theorem

Advanced problems using Pythagoras Theorem

Here are some advanced problems using Pythagoras theorem and other concepts of measurement of shapes.

Example 1 : What is the length of the diagonal in this rectangle (correct to one decimal place)?

The rectangle can be split into two right-angled triangles. We can then apply Pythagoras theorem to obtain the required dimensions in the shape. Advanced Problems_Pythagoras theorem_1

Here the lengths of two short sides are given, we need to find the hypotenuse. So,

c2  = 122 + 192

c  =  sqrt{144~+~361}

=  sqrt{505}

=  22.5 mm

 

Example 2 : Find the values of x and y in the triangle here.

This triangle is comprised of two right-angled triangles (1 and 2 respectively). So if we find the hypotenuse of the two triangles, we would have obtained the values of x and y.

For triangle 2  Advanced Problems_Pythagoras theorem_2

y2  =  62  +  92

y  =  sqrt{36~+~81}

=  sqrt{117}

=  10.8 cm

For triangle 1, the length of the short side is the difference between the entire length of the triangle and the short side of the triangle 2.

=  18  –  6  =  12 cm

x2  =  92  +  122

x =  sqrt{81~+~144}

=  sqrt{225}

=  15 cm

So the values of x and y are 15 cm and 10.8 cm respectively.

 

Example 3 : The perimeter of the rectangle is 26 cm. Find its width and the length of the diagonal.

Perimeter of the rectangle  =  2 (l + b)  Advanced Problems_Pythagoras theorem_3

26  =  2 (8 + b)

8  +  b  =  13

b  =  5

So the width of the rectangle is 5 cm.

c2  =  a2  +  b2

c  =  sqrt{8^2~+~5^2}

c  =  sqrt{89}

c  =  9.4 cm (correct to one decimal place).

 

Example 4 : The isosceles triangle ABC has a perimeter of 66 cm. Find the lengths of AB and BD.

Perimeter of triangle ABC  =  AB  +  BC  +  AC  Advanced Problems_Pythagoras theorem_4

66  =  2AB  +  28 (since AB = AC)

2AB  =  66  –  28

2AB  =  38

AB  =  19 cm

Since triangle ABC is an isosceles triangle, BD divides it into two equal right angled triangles.

So AD  =  DC  =  14 cm. Now applying Pythagoras theorem

AB2  =  BD2  +  AD2

192  =  BD2  +  142

BD  =  sqrt{19^2~-~14^2}

=  sqrt{361~-~196}

=  sqrt{165}

=  12.8 cm

 

Example 5 : Find the value of k in this shape.

This shape is comprised of a rectangle (in the bottom) and a right-angled triangle at the top. Looking at the right-angled triangle and using Pythagoras theorem, we can find the length of its short side.  Advanced Problems_Pythagoras theorem_5

x  =  short side

=  sqrt{13^2~-~11^2}

=  sqrt{169~-~121}

=  sqrt{48}

=  6.9 cm

So the length of k is the sum of the short side of the right-angled triangle and the side of the rectangle (7 cm).

So k  =  6.9 + 7  =  13.9 cm

 

Example 6 : The foot of a ladder is 3.5 m away from the base of a wall. If the ladder reaches 9 m up the wall, find the length of the ladder.

x2  =  92  +  3.52 Advanced Problems_Pythagoras theorem_6

x2  =  81 + 12.25

x =  sqrt{93.25}

=  9.7 m (to one decimal point).

The length of the ladder is 9.7 m.