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# 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.

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

c2  = 122 + 192

c  =

=

=  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

y2  =  62  +  92

y  =

=

=  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 =

=

=  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)

26  =  2 (8 + b)

8  +  b  =  13

b  =  5

So the width of the rectangle is 5 cm.

c2  =  a2  +  b2

c  =

c  =

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

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

192  =  BD2  +  142

BD  =

=

=

=  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.

x  =  short side

=

=

=

=  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

x2  =  81 + 12.25

x =

=  9.7 m (to one decimal point).

The length of the ladder is 9.7 m.