Question

In trapezul ABCD,AB paralel cu CD ,AC=DB, unghiul B =60 grade si BC=12 cm,AC AC perpendicular pe BC. A)Calculati lungimile bazelor trapezului.

b)Daca CM si BN sunt mediane in triunghiul ABC, M apartine AB ,N apartine AC ,BN intersectat cu AD={P},iar BN intersectat cu CM ={G},calculati BG supra BP.

c) Daca DQ paralel cu BP ,Q apartine NC ,aratati ca AN =2NQ

VA ROG!DAU COROANA.

Answer 1

a. In ΔACB dr in C, avem ∡ABC=60°⇒

∡CAB=180-(90+60)=30°⇒Teoremei ∡30°⇒ 2BC=AB

2×12=AB

AB=24 cm

• Daca AC=BD (diagonalele trapezului)⇒ABCD trapez isoscel⇒ BC=AD

AD=12 cm

m(∡DAB)=60° (trapez isoscel)

m(∡CAB)=30°⇒ m(∡DAC)=30°

m(∡ADC)=[360-(60+60)]:2=120°

⇒ m(∡DCA)=180-(120+30)=30°

Deci ∡DCA=∡DAC∡⇒ΔDAC isoscel⇒ AD=DC

DC=12 cm

• In ΔABC , aplicam Pitgora

AB²=AC²+BC²

AC²=576-144=432

AC=√432=12√3 cm

b. CM=12cm (jumatate din ipotenuza AB, fiind mediana in triunghi dreptunghic)

CD=12 cm

M mijlocul AB⇒ AM=MB=12 cm

Deci CD=AM , CD║AM (CD║AB)

CM=CD⇒CD=AD=CD=AM⇒ CDAM romb

CM║AD⇒ CG║PA

ΔCNG~ΔANP⇒

$\frac{CN}{NA}=\frac{CG}{AP} =\frac{GN}{NP} \\\\\frac{CN}{NA}=\frac{GN}{NP}\\\\1=\frac{GN}{NP}\\\\GN=NP$

• Putem afla BN in ΔCNB dr in C, aplicam Pitagora

BN²=CN²+BC²

BN²=108+144

BN²=252

BN=6√7

BG se afla la doua treimi de varf

$BG=\frac{2}{3} \times BN=\frac{2}{3}\times 6\sqrt{7} =4\sqrt{7}$

GN=6√7-4√7=2√7 cm=NP

BP=PN+BN=2√7+6√7=8√7 cm

Deci raportul dintr BG si BP este:

$\frac{BG}{BP}=\frac{4\sqrt{7} }{8\sqrt{7} } =\frac{1}{2}$

c. Stim de la punctul anterior ca NP=NG, CN=AN si AP=CG

Aflam CG, implicit AP

CM=12cm

CG se afla la doua treimi de varf

$CG=\frac{2}{3}\times 12=8\ cm$

Deci AP=8 cm, AD=12 cm⇒ PD=12-8=4 cm

• NP║DQ⇒ T.Thales

$\frac{AP}{PD}=\frac{AN}{NQ}$

$\frac{8}{4} =\frac{6\sqrt{3} }{NQ} \\\\NQ=3\sqrt{3}$

CN=6√3 ⇒ Q mijlocul lui CN⇒ AN=2NQ