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


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

Solution:
D E 2 = 3 2 + 4 2    =9+16    =25 DE= 25  =5 cm A C 2 = 7 2 + 24 2    =49+576    =625 AC= 625  =25 cm Perimeter 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:
A B 2 = 25 2 7 2    =62549    =576 AB= 576  =24 cm BC=24 cm÷2  =12 cm C E 2 = 5 2 + 12 2    =25+144    =169 CE= 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:
3 2 +B E 2 = 5 2    B E 2 = 5 2 3 2 =16 BE=4 cm B C 2 + ( 5+4 ) 2 = 17 2    B C 2 = 17 2 9 2 =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:
A C 2 = 12 2 + 9 2    =225 AC= 225  =15 cm A E 2 = ( 15+9 ) 2 + 11.5 2    =576+132.25    =708.25 AE= 708.25  =26.6 cm

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