Answers Study Guide Ch 11 Geom

Unit 11 Test Study Guide: Volume & Surface Area

Topic 1: Area of Plane Figures

  • General Instructions: Find the area of each figure, rounding to the nearest hundredth when necessary.

  • Problem 1: Triangle

    • Base b=10.8b = 10.8 in, Height h=3.7h = 3.7 in
    • Area A=bh=(10.8)(3.7)=39.96A = bh = (10.8)(3.7) = 39.96 in\textsuperscript{2}

  • Problem 2: Circle

    • Radius r=11r = 11 m
    • Area A=πr2=π(11)2=121π≈380.13A = \pi r^2 = \pi (11)^2 = 121\pi \approx 380.13 m\textsuperscript{2}

  • Problem 3: Triangle (Right Triangle)

    • Hypotenuse = 17.3 ft, One leg = 16.5 ft
    • Using the Pythagorean theorem: x2+16.52=17.32x^2 + 16.5^2 = 17.3^2
    • x2=17.32−16.52=299.29−272.25=27.04x^2 = 17.3^2 - 16.5^2 = 299.29 - 272.25 = 27.04
    • x=27.04=5.2x = \sqrt{27.04} = 5.2 ft (This is the base of the triangle)
    • Area A=12bh=12(5.2)(16.5)=42.9A = \frac{1}{2}bh = \frac{1}{2}(5.2)(16.5) = 42.9 ft\textsuperscript{2}

  • Problem 4: Trapezoid

    • Parallel sides: 7 mm and 15 mm, One of the non-parallel sides = 8.5 mm, Height = x
    • Using the Pythagorean theorem to find height: x2+42=8.52x^2 + 4^2 = 8.5^2
    • x2=8.52−42=72.25−16=56.25x^2 = 8.5^2 - 4^2 = 72.25 - 16 = 56.25
    • x=56.25=7.5x = \sqrt{56.25} = 7.5 mm
    • Area A=12(b<em>1+b</em>2)h=12(7+15)(7.5)=12(22)(7.5)=11(7.5)=82.5A = \frac{1}{2}(b<em>1 + b</em>2)h = \frac{1}{2}(7 + 15)(7.5) = \frac{1}{2}(22)(7.5) = 11(7.5) = 82.5 mm\textsuperscript{2}

  • Problem 5: Semicircle

    • Area of semicircle = 76.97 square yards
    • Full circle area: A=πr2A = \pi r^2; Semicircle area: A=12πr2A = \frac{1}{2} \pi r^2
    • 76.97=12πr276.97 = \frac{1}{2} \pi r^2
    • 153.94=πr2153.94 = \pi r^2
    • r2=153.94π≈49r^2 = \frac{153.94}{\pi} \approx 49
    • r=49=7r = \sqrt{49} = 7 yards
    • Diameter d=2r=2(7)=14d = 2r = 2(7) = 14 yards

  • Problem 6: Rectangle

    • Area = 104 cm\textsuperscript{2}, Width = 8 cm
    • Length L=AreaWidth=1048=13L = \frac{Area}{Width} = \frac{104}{8} = 13 cm
    • Perimeter P=2L+2W=2(13)+2(8)=26+16=42P = 2L + 2W = 2(13) + 2(8) = 26 + 16 = 42 cm

Topic 2: Area of Sectors

  • General Formula: Area of a sector A=θ360πr2A = \frac{\theta}{360} \pi r^2, where θ\theta is the angle in degrees.

  • Problem 7:

    • Angle = 114°, Radius = 8 cm
    • Area A=114360π(8)2=114360π(64)≈43.67A = \frac{114}{360} \pi (8)^2 = \frac{114}{360} \pi (64) \approx 43.67 cm\textsuperscript{2}

  • Problem 8:

    • Angle = 37°, Radius = 15 m
    • Area A=37360π(15)2=37360π(225)≈634.21A = \frac{37}{360} \pi (15)^2 = \frac{37}{360} \pi (225) \approx 634.21 m\textsuperscript{2}

Topic 3: Area of Composite Figures

  • Problem 9:

    • Figure consists of a trapezoid and a triangle.
    • Trapezoid: bases 7 m and 16 m, height 9 m
    • Area of trapezoid: A1=12(9)(7+16)=4.5(23)=103.5A_1 = \frac{1}{2}(9)(7+16) = 4.5(23) = 103.5 m\textsuperscript{2}
    • Triangle: base 16 m, height 838\sqrt{3} m (using 30-60-90 triangle ratios or Pythagorean theorem)
    • Area of triangle: A2=12(16)(83)=8(83)=643≈110.85A_2 = \frac{1}{2}(16)(8\sqrt{3}) = 8(8\sqrt{3}) = 64\sqrt{3} \approx 110.85 m\textsuperscript{2}
    • Total Area: A=103.5+110.85=214.35A = 103.5 + 110.85 = 214.35 m\textsuperscript{2}

  • Problem 10:

    • Figure consists of a semicircle and a triangle.
    • Semicircle: radius = 11.9 ft
    • Area of semicircle: A1=12π(11.9)2≈71.4A_1 = \frac{1}{2} \pi (11.9)^2 \approx 71.4 ft