The figures are not drawn to scale. Figure 1 shows a rectangular piece of paper EGHK that measures 20 cm by 15 cm. EF = JH = 5 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 72°. Find ∠FJH in Figure 2.
(a)
Area of Rectangle EGHK
= 20 x 15
= 300
Area of Triangle EFK
=
12 x 5 x 15
= 37.5 cm
2 Area of Triangle FKJ
= (300 - 37.5 - 37.5) ÷ 2
= 225 ÷ 2
= 112.5 cm
2 Area of EFJHK
= 112.5 + 37.5 + 37.5
= 187.5 cm
2 (b)
∠KFJ
= (180° - 72°) ÷ 2
= 108 ÷ 2
= 54° (Angles on a straight line)
∠FJH
= 360° - 90° - 90° - 54°
= 126° (Sum of angles in a quadrilateral)
Answer(s): (a) 187.5 cm
2; (b) 126°