T Madas Squaring a Sum T Madas Squaring

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© T Madas

© T Madas

Squaring a Sum © T Madas

Squaring a Sum © T Madas

Squaring a Sum © T Madas

Squaring a Sum © T Madas

Squaring a Difference © T Madas

Squaring a Difference © T Madas

Squaring a Difference © T Madas

Squaring a Difference © T Madas

Difference of Squares © T Madas

Difference of Squares © T Madas

Difference of Squares © T Madas

Difference of Squares © T Madas

Exam Question No Calculator © T Madas

Exam Question No Calculator © T Madas

Calculate 3. 222 + 2 x 3. 22 x 1. 78 + 1. 782

Calculate 3. 222 + 2 x 3. 22 x 1. 78 + 1. 782 © T Madas

© T Madas

© T Madas

Calculate 1012 – 992 © T Madas

Calculate 1012 – 992 © T Madas

© T Madas

© T Madas

© T Madas

© T Madas

13 + 5 The right angled triangle shown below has the lengths of two

13 + 5 The right angled triangle shown below has the lengths of two of its sides given in terms of surds. 1. Show that the area of this triangle is 4 square units 2. Show that the hypotenuse of the triangle is 6 units long. 13 – 5 © T Madas

The right angled triangle shown below has the lengths of two of its sides

The right angled triangle shown below has the lengths of two of its sides given in terms of surds. 1. Show that the area of this triangle is 4 square units 2. Show that the hypotenuse of the triangle is 6 units long. We could use an identity on the numerator (a + b) (a – b) 13 + 5 A=b x 2 h = 13 + 5 x a 2 – b 2 13 – 5 = 2 = 4 square units 13 – 5 © T Madas

The right angled triangle shown below has the lengths of two of its sides

The right angled triangle shown below has the lengths of two of its sides given in terms of surds. 1. Show that the area of this triangle is 4 square units 2. Show that the hypotenuse of the triangle is 6 units long. 2 2 d = 13 + 5 + 13 – 5 2 13 + 5 d 2= 13 + 5 + 2 x 13 x 5 + 13 + 5 – 2 x 13 x 5 d d 2= 36 d = 6 units Using the identity: 13 – 5 (a ± b )2 a 2 ± 2 a b + b 2 © T Madas

© T Madas

© T Madas

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