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
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+(25−5)+3+4=58 cm
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+(25−5)+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=252−72 =625−49 =576AB=√576 =24 cmBC=24 cm÷2 =12 cmCE2=52+122 =25+144 =169CE=√169 =13 cm
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=252−72 =625−49 =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=52−32=16BE=4 cmBC2+(5+4)2=172 BC2=172−92=208 BC=√208 BC=14.42 cm
Diagram below shows two right-angled triangles, ABE and CBD. BED is a straight line.

Solution:
32+BE2=52 BE2=52−32=16BE=4 cmBC2+(5+4)2=172 BC2=172−92=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
Diagram below shows two right-angled triangles, ABC and ADE. ACD is a straight line.

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