5 sentences for geometric world problems with solution
Answer (1)
1. Triangle Angles The second angle is double the first, and the third angle is 40° less than the first. Find the three angles. Let the first angle = [tex]x[/tex]. Second angle = [tex]2x[/tex]. Third angle = [tex]x - 40[/tex]. Equation: [tex]x + 2x + (x - 40) = 180[/tex] → [tex]4x = 220[/tex] → [tex]x = 55^\circ[/tex]. Angles: [tex]55^\circ, 110^\circ, 15^\circ[/tex].2. Rectangle Dimensions Perimeter is 152 m; width is 22 m less than length. Find length and width. Let length = [tex]L[/tex], width = [tex]L - 22[/tex]. Perimeter: [tex]2(L + L - 22) = 152[/tex] → [tex]4L - 44 = 152[/tex] → [tex]4L = 196[/tex] → [tex]L = 49[/tex]. Width = [tex]49 - 22 = 27[/tex] meters.3. Cutting Tubing A 12 cm tube is cut into two pieces; one piece is twice the length of the other. Find both lengths. Let shorter piece = [tex]x[/tex], longer piece = [tex]2x[/tex]. [tex]x + 2x = 12[/tex] → [tex]3x = 12[/tex] → [tex]x = 4[/tex]. Pieces: 4 cm and 8 cm.4. Right Triangle Hypotenuse Legs are 3 cm and 4 cm; find the hypotenuse. By Pythagoras: [tex]c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5[/tex] cm.5. Area of Triangle Base = 6 m, height = 4 m; find the area. Area = [tex]\frac{1}{2} \times 6 \times 4 = 12[/tex] [tex]m^2[/tex].
