PSLEThe figure is made up of two rectangles, BCJG and GHEF, and two right-angled isosceles triangles, CDJ and EDH. CB = 12 cm, EF = 6 cm and CJ =
k cm. BGF and DJHG are straight lines.
- Find the length of BF in terms of k. Give your answer in the simplest form.
- Find the total area of the figure when k = 14.
(a)
Length JH
= 12 - 6
= 6 cm
Length DJ =
k cm (Isosceles triangle)
Length HE
= Length CG
= (
k + 6) cm (Isosceles triangle)
Length BF
=
k +
k + 6
= (2
k + 6) cm
(b)
Area of Rectangle BCGJ
= 14 x 12
= 168 cm
2 Length GF
=
k + 6
= 14 + 6
= 20 cm
Area of Rectangle EFGH
= 20 x 6
= 120 cm
2 Area of Triangle CDJ
=
12 x 14 x 14
= 98 cm
2 Area of Triangle DEH
=
12 x 20 x 20
= 200 cm
2 Total area of the figure
= 168 + 120 + 98 + 200
= 586 cm
2 Answer(s): (a) (2
k + 6) cm; b) 586 cm
2