width of the track. 10. REFLECTION ANSWERS: TERM 2 GRADE 8 WEEK 6 LESSON 7 MENTAL MATHS CLASSWORK P= 2,75 + 2.75 + 10 + 10 = 297.75m A= lxB= 2.75 x 10 = 27.5m2 Using Pythagoras’ Theorem, using Pythagoras Find missing side BD2 = DC2 + BC2 (4)2 +(4)2= C2 172 = 22 + 82 ⇒ 32= C2 289 - 64 = DC2 ⇒ 225 = DC2 ⇒ C= 5,6 m 15 = DC ⇒ P= 5.6 + 3 + 4 +7=19.6 m Therefore, length of rectangle = 15 cm Area So, area of rectangle = l × b = x b x h = 15 × 8 cm2 = 120 cm2 = = x 4 x 4 =8 m2 Also, perimeter of rectangle = 2 (15 + 8) cm Area of rectangle = 2 × 23 cm A= lb= 4 x3= 12m2 = 46 cm A= 8 + 12= 20 m2 Semi-perimeter of the triangle = (a + b + c)/2 = (41 + 28 + 15)/2 = 84/2 = 42 cm
width of the track. 10. REFLECTION ANSWERS: TERM 2 GRADE 8 WEEK 6 LESSON 7 MENTAL MATHS CLASSWORK P= 2,75 + 2.75 + 10 + 10 = 297.75m A= lxB= 2.75 x 10 = 27.5m2 using Pythagoras Find missing side (4)2 +(4)2= C2 32= C2 C= 5,6 m P= 5.6 + 3 + 4 +7=19.6 m Area = x b x h = = x 4 x 4 =8 m2 Area of rectangle A= lb= 4 x3= 12m2 A= 8 + 12= 20 m2 Using Pythagoras’ Theorem, BD2 = DC2 + BC2 ⇒ 172 = 22 + 82 ⇒ 289 - 64 = DC2 ⇒ 225 = DC2 ⇒ 15 = DC Therefore, length of rectangle = 15 cm So, area of rectangle = l × b = 15 × 8 cm2 = 120 cm2 Also, perimeter of rectangle = 2 (15 + 8) cm = 2 × 23 cm = 46 cm Semi-perimeter of the triangle = (a + b + c)/2 = (41 + 28 + 15)/2 = 84/2 = 42 cm