6.2.2 Pythagoras’ Theorem, PT3 Focus Practice


Question 6:
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:
GH2=DH2+DG2   =242+72   =576+49   =625GH=25 cmPerimeter of the whole diagram=12+12+12+26+7+14+25=108 cm

Question 7:
In the diagram, ABC and EFD are right-angled triangles.


Calculate the perimeter, in cm, of the shaded region.

Solution:
DE2=32+42   =9+16   =25DE=25 =5 cmAC2=72+242   =49+576   =625AC=625 =25 cmPerimeter of the shaded region=24+7+(255)+3+4=58 cm

Question 8:
In the diagram, ABD and BCE are right-angled triangles. ABC and DBE are straight lines. The length of AB is twice the length of BC.


Calculate the length, in cm, of CE.

Solution:
AB2=25272   =62549   =576AB=576 =24 cmBC=24 cm÷2 =12 cmCE2=52+122   =25+144   =169CE=169 =13 cm

Question 9:
Diagram below shows two right-angled triangles, ABE and CBD. BED is a straight line.
Find the length, in cm, of BC. Round off the answer to two decimal places.

Solution:
32+BE2=52   BE2=5232=16BE=4 cmBC2+(5+4)2=172   BC2=17292=208 BC=208 BC=14.42 cm


Question 10:
Diagram below shows two right-angled triangles, ABC and ADE. ACD is a straight line.
Find the length, in cm, of AE. Round off the answer to one decimal places.

Solution:
AC2=122+92   =225AC=225 =15 cmAE2=(15+9)2+11.52   =576+132.25   =708.25AE=708.25 =26.6 cm

6.2.1 Pythagoras’ Theorem, PT3 Focus Practice


6.2.1 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,
BC= 52 – 42
= 25 – 16
= 9
BC = √9
  = 3 cm

In ∆ ABD,
BD = BC + CD
  = 3 + 6
  = 9
AD= 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,
AC= 62 + 82
= 36 + 64
= 100
AC = √100
  = 10 cm
AD = 5 cm

In ∆ EDC,
EC= 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,
BC= 252 – 242
= 625 – 576
= 49
BC = √49
  = 7 cm
AB = 17 – 7 = 10 cm

In ∆ DAB,
AD= 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
BC= 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:
AD= 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

6.1 Pythagoras’ Theorem


6.1 Pythagoras’ Theorem

6.1.1 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:
x2=52+122=25+144x=169=13  

Example 2:
 
Solution:
x2=15292=22581x=144=12

3. Pythagorean triples
are three whole numbers that form the sides of a right-angled triangle.

Example:
(a) 3, 4, 5
(b) 6, 8, 10
(c) 5, 12, 13
(d) 8, 15, 17
(e) 9, 12, 15


6.1.2 The Converse of the Pythagoras’ Theorem
 
 
  In a triangle, if the sum of the squares of the two sides
  is equal to the square of the longest side, then the angle
  opposite the longest side is a right angle.