## 13.2.4 The Pythagoras’ Theorem, PT3 Focus Practice

Question 8: In diagram below ABDE is a square and EDC is a straight line. The area of the square ABDE is 144 cm2. Calculate the length, in cm, of BC. Solution: BD = √144   = 12 cm BC2 = 122 + 92 = 144 + 81 = 225 BC = √225   = 15 cm … Read more

## 13.2.3 The Pythagoras’ Theorem, PT3 Focus Practice

Question 6: In the diagram shows two right-angled triangles. Calculate the perimeter of the whole diagram. Solution: In ∆ ABC, AC2 = 62 + 82 = 36 + 64 = 100 AC = √100   = 10 cm AD = 5 cm In ∆ EDC, EC2 = 122 + 52 = 144 + 25 = 169 EC = … Read more

## 13.2.2 The Pythagoras’ Theorem, PT3 Focus Practice

Question 3: In diagram below, ABCH is a square and DEFG is a rectangle. HCDE is a straight line and HC = CD. Find the perimeter, in cm, of the whole diagram. Solution: G H 2 =D H 2 +D G 2    = 24 2 + 7 2    =576+49    =625 GH=25 cm Perimeter of the whole diagram =12+12+12+26+7+14+25 =108 cm Question … Read more

## 13.2.1 Pythagoras’ Theorem, PT3 Focus Practice

13.2.1 Pythagoras’ Theorem, PT3 Focus Practice   Question 1: In the diagram, ABC is a right-angled triangle and BCD is a straight line. Calculate the length of AD, correct to two decimal places. Solution: In ∆ ABC, BC2 = 52 – 42 = 25 – 16 = 9 BC = √9   = 3 cm In ∆ ABD, … Read more

## 13.1 The Pythagoras’ Theorem

13.1 The Pythagoras’ Theorem 13.1.1 The Pythagoras’ Theorem 1. In a right-angled triangle, the hypotenuse is the longest side of the triangle. 2. Pythagoras’ Theorem:   In a right-angled triangle, the square   of the hypotenuse is equal to the sum   of the squares of the other two sides. Example 1:   Solution: x 2 = 5 2 + … Read more