Răspunsuri

2014-05-22T21:34:31+03:00
2014-05-22T22:32:21+03:00
Fie E si F proiectiile punctelor B si A pe dreapta CD.

ABEF - dreptunghi⇒EF=8 cm

ΔBEC≡ΔAFD, sunt si isoscele, deci BE=CE=x⇒x=10√2 cm (cu teorema lui Pitagora sau cu sin45)

Deci CD=10√2+8+10√2=4(2+5√2) cm

P_{ABCD}=48+8+20\sqrt2=4(14+5\sqrt2)\ cm

A_{ABCD}=\dfrac{(AB+CD)\cdot BE}{2}=\dfrac{(16+20\sqrt2)\cdot10\sqrt2}{2}= 40(5+2\sqrt2)\ cm^2