Question

Consideräm trapezul dreptunghic ortodiagonal ABCD, AB I CD, AD L AB,
AC intersectat BD = (0). Dacà DO = 6 cm si DB = 30 cm, atunci determinati:
a) lungimea diagonalei AC;
b) aria trapezului.

Answer 1

Explicație pas cu pas:

a)

$OB=DB-DO = 30 - 6 = 24 \: cm$

$AO²=DO \times OB = 6 \times 24 = 144 \\ AO = 12 \: cm$

$DO²=AO \times OC \\ OC = \frac{ {6}^{2} }{12} = 3 \: cm$

$AC=AO + OC = 12 + 3 \\ AC= 15 \: cm$

b)

$AB²=AO²+OB² = {12}^{2} + {24}^{2} = 720 \\ AB = 12 \sqrt{5} \: cm$

$AD²=AO² + DO² = {12}^{2}+{6}^{2}= 180 \\ AD = 6 \sqrt{5} \: cm$

$DC²=DO² + OC² = {6}^{2} + {3}^{2} = 45 \\ DC = 3 \sqrt{5}\: cm$

$Aria = \frac{(DC + AB) \times AD}{2} \\ = \frac{ (3 \sqrt{5} + 12 \sqrt{5}) \times 6 \sqrt{5}}{2} \\ = \frac{15 \sqrt{5} \times 6 \sqrt{5} }{2} = 225 \: {cm}^{2}$

Answer 2

a)

$\it OB=BD-DO=30-6=24\ cm\\ \\ \Delta DAB-\ dreptunghic,\ \hat A=90^o, \ AO\ - \ \hat\imath n\breve al\c{\it t}imea\ corespunz\breve a toare\ ipotenuzei\\ \\ Teorema\ \hat\imath n\breve al\c{\it t}imii\ \Rightarrow AO^2=DO\cdot OB\ \Rightarrow AO^2=6\cdot24=144=12^2\ \Rightarrow \\ \\ \ \Rightarrow AO=12\ cm$

$\it \Delta CDA-\ dreptunghic,\ \hat D=90^o, \ DO\ - \ \hat\imath n\breve al\c{\it t}imea\ corespunz\breve a toare\ ipotenuzei\\ \\ Teorema\ \hat\imath n\breve al\c{\it t}imii\ \Rightarrow DO^2=AO\cdot OC\ \Rightarrow 36=12\cdot OC|_{:12} \Rightarrow OC=3\ cm\\ \\ AC=AO+OC=12+3=15\ cm$

b)

$\it AC\perp BD \Rightarrow \mathcal{A}=\dfrac{AC\cdot BD}{2}=\dfrac{15\cdot30}{2}=225\ cm^2$