The figures are not drawn to scale. Figure 1 shows a rectangular piece of paper EGHK that measures 20 cm by 12 cm. EF = JH = 4 cm. The paper is folded along the dotted line FJ such that point G touches point K, as shown in Figure 2.
- Find the area of Figure 2. EFJHK, after the folding.
- In Figure 2, ∠EFK is 78°. Find ∠FJH in Figure 2.
(a)
Area of Rectangle EGHK
= 20 x 12
= 240
Area of Triangle EFK
=
12 x 4 x 12
= 24 cm
2 Area of Triangle FKJ
= (240 - 24 - 24) ÷ 2
= 192 ÷ 2
= 96 cm
2 Area of EFJHK
= 96 + 24 + 24
= 144 cm
2 (b)
∠KFJ
= (180° - 78°) ÷ 2
= 102 ÷ 2
= 51° (Angles on a straight line)
∠FJH
= 360° - 90° - 90° - 51°
= 129° (Sum of angles in a quadrilateral)
Answer(s): (a) 144 cm
2; (b) 129°