Activity 1
Solution AP = AQ (PQ ∣∣ BC, prop theorem) PB QC ∴ 5/3 = 4/x ∴ 5x = (3)(4) ∴ x = 12/5 = 2,4 cm 3 [4] |
In ∆ABC, AB ∣∣ FD; AF ∣∣ DE and FE : EC = 3 : 4.
Determine EC : BF (7)
NOTE:
3 : 4 does not mean that
FE = 3 and EC = 4.
For any a, we can say that
FE = 3a and EC = 4a
For every 3 of a in FE, there is
4 of a in EC.
Solution Work with two different triangles: ∆ACF and ABC In ∆ACF: AD = FE (AF ∣∣ DE, prop intercept theorem) DC EC In ∆ABC: AD = BF (AB ∣∣ FD, prop intercept theorem) DC FC ∴ FE = BF(both = AD) EC FC DC FE = 3a and BF = BF EC 4a FC 7a ∴ 3a = BF 4a 7a ∴ BF = 3(7a/4) = 21a____4 ∴ EC = 4a ÷ 21a/4 BF = 4a × 4 1 21a = 16 21 ∴ EC : BF = 16 : 21 [7] |
Determine the value of x if PQ ∣∣ BC. (4)
Solution AP = AQ (prop theorem , PQ ∣∣ BC) PB QC 5/3 = 4/x 5x = (3) (4) x = 12/5 = 2,4cm [4] |
RH = 3 units, RK = 9 units, HF = 2 units. GE: EK = 1:3
Calculate (stating reasons) the lengths of:
4.1 FG
4.2 FD (8)
Solutions 4.1 In ∆HKG FG = 9 S (line ∣∣ one side of a ∆) 3 R or (RF ∣∣ KG) 2 3 FG = 6 units S (3) 4.2 GD = GE = ¼ S (line ∣∣ one side of a ∆) 3 R or (ED ∣∣ KH) GH GK GD = ¼ .GH GD = ¼ .(8) S GD = 2 S ∴ FD = 6 − 2 = 4units 3 R OR In ∆HKG, HK ∣∣ DE GD = EG = 1/3 S DH EK (line ∣∣ one side of a ∆) R or (proportional theorem, HK//DE ) 6 − FD = 1/3 S 2 + FD 18 − 3FD = 2 + FD ∴ FD = 4 units (5) [8] |
Activity 2
Solution a) | Solution b) AE = EF = AF (||| ∆s) AM MD AD AM = ME (radii) ∴ AE = 2AM ∴ 2AM = AF AM AD ∴ AF = 2 (5) AD |
c) In ∆CDB and ∆ADE ^C = ^A2 (∠s in same seg) ^B = ^E2 (∠s in same seg) ^D4 = ^D1 + ^D2 (opp ∠) ∴ ∆CDB ||| ∆ADE (AAA) (4) | Solution d) CD = DB (III ∆s) AD DE ∴ CD.DE = AD.DB But AD = DB (MD ⊥ AB, M is centre) ∴ CD. DE = AD ∴ AD2 = CD.DE (3) [17] |
Solutions b) In ∆BHD and ∆FED c) FE = FD (||| ∆s) |
In the diagram ∆ABC is such that F is on AB and G is on AC. CB is produced to meet GF produced at E .DGFE is a straight line. BFA ∣∣ CD.
AB = 20, BC = 10, EF = 8, EB = 5 and FB = 6.
3.1 Determine the numerical value of EF (3)
ED
3.2 Calculate the length of ED (2)
3.3 Complete, without stating the reasons: ∆EFB III ∆ …. (1)
3.4 Hence, calculate the length of DC (3)
3.5 Prove that: AF = FG(4)
CD DG
[13]
Solutions |
Solutions 4.1 Proof: In ΔAPQ and ΔACB ^A = ^A (common) S/R ^P2 = ^C S (ext ∠ of a cyclic quad ) 3 R ^P2 = ^ B (sum ∠s of ∆) or ( ext ∠ of cyclic quad ) ∆APQ ∣∣∣ ∆ACB ( ∠.∠.∠) R (4) 4.2 AQ = PQ S (∆APQ ∣∣∣ ∆ACB) S AB BC AQ = PQ S ( AQ = BC ) AB AQ AQ2 = AB.PQ (3) [7] |