<h2>E</h2> <p>a) \( c^2 = §§V0(1,8,1)§§^2 + §§V1(1,6,0.5)§§^2 \)</p> <p>\( c = \sqrt{§§V0(1,8,1)§§^2 + §§V2(1,6,1)§§^2} \)</p> <p>\( c = §§V3(1,10,0.5)§§ cm \)</p> <p>\( §§V4(1,8,0.5)§§^2 = §§V5(4,10,1)§§^2 + §§V6(1,8,0.5)§§^2 \)</p> <p>\( a = §§V5(4,10,1)§§ - §§V4(1,8,0.5)§§^2 \)</p> <p>\( a = §§V6(1,8,0.5)§§ cm \)</p> <svg width="400" height="400" viewBox="0 0 400 400" xmlns="http://www.w3.org/2000/svg"> <!-- Definicija trokuta --> <polygon points="50,300 350,300 190,100" fill="none" stroke="black" stroke-width="2" /> <!-- Oznake uglova --> <text x="30" y="320" font-size="16" fill="black">120°</text> <text x="340" y="320" font-size="16" fill="black">140°</text> <text x="180" y="80" font-size="16" fill="black">20°</text> </svg> <div class="exercise"> <h2>F</h2> <p>a) \( u = 2 \cdot ( §§V0(1,8,1)§§ + §§V1(1,6,0.5)§§) \)</p> <p> \( u = §§V7(10,50,0.5)§§ cm \)</p> <p>b) \( A = ( §§V2(1,8,0.5)§§ + §§V3(1,4,0.5)§§ ) \cdot §§V4(1,4,0.5)§§ \)</p> <p> \( A = §§V8(10,30,0.5)§§ cm^2 \)</p> <p>c) Zuerst berechnet man die Höhe der Basis mit dem Satz des Pythagoras.</p> <p>\( h^2 = §§V5(1,6,0.5)§§^2 - §§V6(1,6,0.5)§§^2 \)</p> <p>\( h = \sqrt{ §§V5(1,6,0.5)§§^2 - §§V6(1,6,0.5)§§^2 } \)</p> <p>\( h = \sqrt{ §§V9(10,40,1)§§ - §§V10(5,10,0.5)§§ } \)</p> <p>\( h = \sqrt{ §§V11(5,20,1)§§ } \)</p> <p>\( h = §§V12(1,6,0.5)§§ cm \)</p> <p>\( A = \frac{1}{2} \cdot §§V13(1,10,0.5)§§ \cdot §§V12(1,6,0.5)§§ \)</p> <p>\( A = §§V14(10,30,1)§§ cm^2 \)</p> </div> <div class="exercise"> <h2>6 Kreis | Alles klar?, Seite 131</h2> <p>A</p> <p>a) \( u = 2\pi r \)</p> <p> \( u = 2 \cdot \pi \cdot §§V0(1,6,0.5)§§ \)</p> <p> \( u = §§V15(10,30,0.5)§§ cm \)</p> <p>B</p> <p>a) \( u = \pi d \)</p> <p> \( §§V16(10,20,1)§§ = \pi \cdot d \)</p> <p> \( d = \frac{§§V16(10,20,1)§§}{\pi} \)</p> <p> \( d = §§V17(1,6,0.5)§§ cm \)</p> </div> <div class="exercise"> <h2>6 Kreis | Alles klar?, Seite 134</h2> <p>A</p> <p>a) \( A = \pi r^2 \)</p> <p> \( A = \pi \cdot §§V18(1,6,0.5)§§^2 \)</p> <p> \( A = §§V19(200,300,5)§§ cm^2 \)</p> <p>B</p> <p>a) \( A = \pi r^2 \)</p> <p> \( §§V20(20,40,1)§§ = \pi r^2 \)</p> <p> \( r^2 = \frac{§§V20(20,40,1)§§}{\pi} \)</p> <p> \( r = \sqrt{\frac{§§V20(20,40,1)§§}{\pi}} \)</p> <p> \( r = §§V21(1,4,0.5)§§ cm \)</p> </div> <div class="exercise"> <h2>6 Kreis | Alles klar?, Seite 139</h2> <p>A</p> <p>a) \( b = 2\pi r \cdot \frac{\alpha}{360^\circ} \)</p> <p> \( b = 2\pi \cdot §§V0(1,4,0.5)§§ \cdot \frac{§§V22(30,90,5)§§}{360^\circ} \)</p> <p> \( b = §§V23(1,6,0.5)§§ cm \)</p> <p>B</p> <p>a) \( b = 2\pi r \cdot \frac{\alpha}{360^\circ} \)</p> <p> \( §§V24(5,10,0.5)§§ = 2\pi \cdot r \cdot \frac{§§V25(30,90,5)§§}{360^\circ} \)</p> <p> \( r = §§V26(1,12,0.5)§§ cm \)</p> </div> <div class="exercise"> <h2>6 Kreis | Alles klar?, Seite 142</h2> <p>A</p> <p> \( r = §§V0(1,6,0.5)§§ cm \)</p> <p> \( A = \text{Quadrat} - \text{Halbkreis} \)</p> <p> \( A = a^2 - \frac{\pi r^2}{2} \)</p> <p> \( A = §§V27(1,10,0.5)§§^2 - \frac{\pi \cdot §§V0(1,6,0.5)§§^2}{2} \)</p> <p> \( A = §§V28(10,30,0.5)§§ cm^2 \)</p> <p>B</p> <p> \( d^2 = §§V5(1,8,0.5)§§^2 + §§V6(1,8,0.5)§§^2 \)</p> <p> \( d = \sqrt{§§V5(1,8,0.5)§§^2 + §§V6(1,8,0.5)§§^2} \)</p> <p> \( d = §§V29(6,12,0.5)§§ cm \)</p> <p> \( A = \text{Dreieck} + \text{Halbkreis} \)</p> <p> \( A = \frac{1}{2} ab + \frac{\pi r^2}{2} \)</p> <p> \( A = \frac{1}{2} \cdot §§V30(4,8,0.5)§§ \cdot §§V31(4,8,0.5)§§ + \frac{\pi \cdot §§V32(2,6,0.5)§§^2}{2} \)</p> <p> \( A = §§V33(30,50,1)§§ cm^2 \)</p> <p> \( u = a + h + l + 2\pi r \)</p> <p> \( u = §§V34(4, 8,0.5)§§ + §§V35(4,8,0.5)§§ + §§V36(1,4,0.5)§§ + 2\pi \cdot §§V32(2,6,0.5)§§ \)</p> <p> \( u = §§V37(20,30,1)§§ cm \)</p> </div> <div class="exercise"> <h2>6 Kreis | Rückspiegel, Seite 148</h2> <p>A</p> <p>a) \( u = 2\pi r \)</p> <p> \( u = 2 \cdot \pi \cdot §§V0(1,8,0.5)§§ \)</p> <p> \( u = §§V38(20,50,1)§§ cm \)</p> <p>B</p> <p>a) \( A = \pi r^2 \)</p> <p> \( A = \pi \cdot §§V39(4,10,0.5)§§^2 \)</p> <p> \( A = §§V40(100,150,1)§§ cm^2 \)</p> <p>C</p> <p>a) \( A = \pi r^2 \)</p> <p> \( A = \pi \cdot §§V41(10,30,0.5)§§^2 \)</p> <p> \( A = §§V42(400,500,5)§§ m^2 \)</p> </div>