**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:Solution:

In ∆

*ABC*,*BC*

^{2 }= 5

^{2}– 4

^{2}

= 25 – 16

= 9

*BC*= √9

= 3 cm

In ∆

*ABD*,

*BD*=

*BC*+

*CD*

= 3 + 6

= 9

*AD*

^{2 }= 4

^{2}+ 9

^{2}

= 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:Solution:

In ∆

*ABC*,*AC*

^{2 }= 6

^{2}+ 8

^{2}

= 36 + 64

= 100

*AC*= √100

= 10 cm

*AD*= 5 cm

In ∆

*EDC*,

*EC*

^{2 }= 12

^{2}+ 5

^{2}

= 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:Solution:

In ∆

*DBC*,*BC*

^{2 }= 25

^{2}– 24

^{2}

= 625 – 576

= 49

*BC*= √49

= 7 cm

*AB*= 17 – 7 = 10 cm

In ∆

*DAB*,

*AD*

^{2 }= 10

^{2}+ 24

^{2}

= 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 cm^{2}.Calculate the length, in cm, of

*BC.*

Solution:Solution:

*BD*= √144

= 12 cm

*BC*

^{2 }= 12

^{2}+ 9

^{2}

= 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 cm

^{2}, of the shaded region.

Solution:Solution:

*AD*

^{2 }= 5

^{2}+ 12

^{2}

= 25 + 144

= 169

*AD*= √169

= 13 cm

Area of trapezium

*ABDC*= ½ (13 + 15) × 9

= 126 cm

^{2}Area of triangle

*AED*

= ½ × 5 × 12

= 30 cm

^{2}Area of the shaded region

= 126 – 30

=

**96 cm**^{2 }