Page 48 - ACE YR IGCSE A TOP APPR' TO MATHS

13 Midpoint of AB = ( 1 – 3 , 5 – 3 ) EC = 2.5 × 8
10
2
2
= (–1, 1) EC = 2 cm [2]
ED
– 3 – 5 Area of ΔEDC ( ) 2
Gradient of line AB = (c) =
–3 – 1 Area of ΔABC AB
2.5
–8 ( ) 2
= Area of ΔEDC = × 100
–4 10
= 2 = 6.25 cm 2 [2]
2 (a) ∠A = ∠E and ∠B = ∠D
1 (Alternate angles are equal.)
Gradient of line CD = –
2 ∠ACB = ∠DCE
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1 = – 1 (–1) + c (Vertically opposite)
2 ΔABC ΔEDC [1]
∼
c = 1 (b) 6 = 7x + 4
2 7 11x – 1
Equation of line CD, 66x – 6 = 49x + 28
y = – 1 x + 1 [6] 17x – 34 = 0
2 2 x – 2 = 0
–5 – 3 x = 2 [2]
14 Gradient of line X =
–8 – 10 (c) CE = 11(2) – 1
4 = 21 [1]
=
7
9 (d) Area of ΔABC = ( ) 2
–15 – (–5) Area of ΔEDC 21
Gradient of line Y =
1
2 – 8 = ( ) 2
5 3
=
3 = 1
4 5 20 9
× = (≠ – 1)
9 3 27 Area of ΔABC : area of ΔEDC = 1:9 [2]
X is not perpendicular to Y. [3] 3 (a) (i) 3 = 4x – 5
x 4x – 5 + 2x
9 – (–3) 2
15 Gradient of line X = 18x – 15 = 4x – 5x
–2 – (–10) 4x – 23x + 15 = 0 (shown) [2]
2
3 (ii) (x – 5)(4x – 3) = 0
=
2 3
–6 – (–9) x = 5 or x = 4 (invalid) [1]
Gradient of line Y =
2 – 0 Area of ΔRQS ( ) 2
3
3 (b) (i) Area of ΔRPT = 5
=
2 9
X and Y are parallel lines. [3] = 25
Area of ΔRQS : Area of ΔRPT = 9:25 [2]
4 4 Geometry Area of ΔRQS
(ii)
Area of QPTS
Similar shapes = Area of ΔRQS
Area of ΔRPT – Area of ΔRQS
1 (a) ∠A = ∠E and ∠B = ∠D = 9
(Alternate angles are equal.) 25 – 9
∠ACB = ∠ECD = 9
(Vertically opposite) 16
ΔBAC ΔDEC [1] Area of ΔRQS : Area of QPTS = 9:16 [2]
∼
(b) BC = AB Volume of A 1 3 1
DC ED 4 Volume of A + B = ( 1 + 1 ) = 8
BC = 10 × 3 Volume of A 1 3 1
2.5 = ( ) =
BC = 12 cm Volume of A + B + C 1 + 1 + 2 64
EC ED Volume of B = 8 – 1 = 7 = 1
= Volume of C 64 – 8 56 8
AC AB

Cambridge IGCSE
TM
168 Ace Your Mathematics

Answers.indd 168 15/03/2022 11:08 AM