6.2 Pythagoras’ Theorem, PT3 Focus Practice


6.2 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,
BD = BC + CD
  = 3 + 6
  = 9
AD2= 42 + 92
   = 16 + 81
   = 97
AD = √97
  = 9.849
  = 9.85 cm


Question 2:
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 = √169
  = 13 cm

Perimeter of the whole diagram
= AB + BC + CE+ DE + AD
= 6 + 8 + 13 + 12 + 5
= 44 cm


Question 3:
Diagram below shows a triangle ACD and ABC is a straight line.
Calculate the length, in cm, of AD.

Solution:
In ∆ DBC,
BC2= 252 – 242
   = 625 – 576
   = 49
BC = √49
  = 7 cm
AB= 17 – 7 = 10 cm

In ∆ DAB,
AD2= 102 + 242
   = 100 + 576
   = 676
AD = √676
  = 26 cm


Question 4:
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


Question 5:
In the diagram, ABCD is a trapezium and AED is a right-angled triangle.
 
Calculate the area, in cm2, of the shaded region.

Solution:
AD2= 52 + 122
   = 25 + 144
   = 169
AD = √169
  = 13 cm
Area of trapezium ABDC
= ½ (13 + 15) × 9
= 126 cm2

Area of triangle AED
= ½ × 5 × 12
= 30 cm2

Area of the shaded region
= 126 – 30
= 96 cm2

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