Question

ABCD patrat
l=AB+36 cm
M mij. (AB)
N *apartine* AD
ND/NA=1/2

a)Aria MNC;Perimetrul MNC
b)sin CMN
c)d (C;MM)

Answer 1

a)$\frac{ND}{NA}= \frac{1}{2}$⇒$\frac{AD}{NA}= \frac{3}{2}$⇒$\frac{36}{NA}= \frac{3}{2}$⇒NA=36*2/3=24 cm
AM=MB=AB/2
AM=MB=18cm
MN²=AN²+AM²
MN²=576+324
MN²=900⇒MN=30cm⊃
NC²=DN²+DC²
NC²=144+1296
NC²=1440⇒NC=12√10cm
MC²=MB²+BC²
MC²=324+1296
MC²=1620⇒MC=18√5 cm
AΔMNC=A ABCD-AΔAMN-AΔMBC-AΔDNC
AΔMNC=1296-18*24/2-18*36/2-12*36/2
AΔMNC=1296-216-324-216
AΔMNC=540 cm²
b)AΔMNC=CM*MN*sin M/2
540=18√5*30*sin M/2
1080=540√5*sin M
sin M=1080/540√5⇒sin CMN=3√5/5
c)P∈(MN),CP⊥MN
AΔCMN=CP*MN/2
540=30*CP/2
1080=30CP
CP=36 cm