Question

problema 5 dau coronita

Answer 1

1)
A (ΔAMN) = A (ABCD) - A(ΔBAN) - A(ΔCNM) - A(ΔAMD) =
= 20×24 - AB × BN/2 - CN × MC/2 - MD × AD/2
= 480 - 20 × (BC - CN) /2 - 8 × 24/2 =
= 480 - 10 × (24 - 9) - 4 × 24 = 480 - 150 - 96 = 234

2)
A (ΔAMN) = MN × h(distanta de la A la MN) /2 = 234
MN × h = 234 × 2 = 468
In Δ dreptunghic CNM aplicam teorema lui Pitagora:
NM² = CN² + CM² = 9² + 12² = 81 + 144 = 225
NM = 15
15 × h = 468
h = 468/15 = 31,2

3)
In ΔBAN: AN² = AB² + BN² = 400 + 225 = 625
AN = 25
In ΔAMD: AM² = AD² + MD² = 576 + 64 = 640
AM = 8√10
∡NAM = 90 - ∡BAN - ∡MAD
sin (NAM) = sin [90 - (BAN + MAD)] = cos (BAN + MAD) = cos(BAN) × cos(MAD) - sin(BAN) × sin(MAD) = AB/AN × AD/AM - BN/AN × MD/AM =
= 20/25 × 24/8√10 - 15/25 × 8/8√10 =
= 4/5 × 3/√10 - 3/5 × 1/√10 = 12/5√10 - 3/5√10 = 9/5√10