Question

Vă rog este urgent dau coroana! Ex. 90​

Answer 1

Explicație pas cu pas:

AB ≡ BC => ΔABC este isoscel

=> ∢A ≡ ∢C => ∢C = 45° => ∢B = 90°

=> ΔABC este dreptunghic isoscel

$Aria_{\triangle ABC} = \frac{AB \cdot BC}{2} = \frac{4 \sqrt{2} \cdot 4 \sqrt{2} }{2} = \bf 16 \: {cm}^{2}$

Answer 2

$\it 90.\\ \\ AB=BC \Rightarrow \Delta ABC-\ isoscel, \ \hat C=\hat A=45^o\\ \\ \hat B=180^o-(45^o+45^o)=180^o-90^o=90^o\\ \\ \mathcal{A}_{ABC}=\dfrac{c_1\cdot c_2}{2}=\dfrac{4\sqrt2\cdot4\sqrt2}{2}=\dfrac{4\cdot4\cdot2}{2}=16\ cm^2$

$\it 91.\\ \\ \Big[2+\Big(\dfrac{1}{2}\Big)^3-\dfrac{1}{4}\Big]:\dfrac{15}{8}=\Big(\ \ ^{8)}2+\dfrac{1}{8}-\dfrac{^{2)}1}{\ 4}\Big):\dfrac{15}{8}=\\ \\ \\ =\dfrac{16+1-2}{8}:\dfrac{15}{8}=\dfrac{15}{8}:\dfrac{15}{8}=1$