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 4:
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 5:
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

Leave a Comment