F3 Maths Problems

1.
(a)
Let AH be the height of ΔABC from A to BC.

In ΔAHC :
cosC = HC/AC
cosC = (12/2)/24
cosC = 1/4

In ΔBDC :
cosC = DC/BC
1/4 = DC/12
DC = 3


(b) (1)
BC // AP
Hence, ΔPAD ~ ΔBCD
AP/BC = AD/CD (corr. sides of similar Δs)
AP/12 = (24 - 3)/3
AP = 84

BQ = AP = 84 (opp. sides of //gram)

BQ = BC + CQ
84 = 12 + CQ
CQ = 72


(b) (2)
In ΔBCD :
BC² = CD² + CD²
12² = 3² + BD²
BD² = 135
BD = 3√15

ΔPAD ~ ΔBCD
PD/BD = AD/CD (corr. sides of similar Δs)
PD/BD = (24 - 3)/3
PD = 7BD
PD = 7 × 3√15
PD = 21√15

Area of ΔABP
= (1/2) × AD × BP
= (1/2) × (24 - 3) × (3√15 + 21√15)
= 252√15

The diagram of a //gram bisects the //gram.
Area of ΔBPQ
= 252√15

Area of ΔBCD
= (1/2) × BD × DC
= (1/2) × 3√15 × 3
= 4.5√15

Area of quad. CDPQ
= Area of ΔBPQ - Area of ΔBCD
= 252√15 - 4.5√15
= 247.5√15 (square units)
≈ 958.6 (square units)(to 4 sig. fig.)


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2.
5^x + 5^(x + 1) = 750
5^x + 5(5^x) = 750
6(5^x) = 750
5^x = 125
5^x = 5^3
x = 3