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Define quadrilateral, find parallelogram angles w, x, y given z=98°, and calculate trapezium area given dimensions.

Mathematics

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3 parts
a

32 cm2\text{}32 \text{ cm}^2\text{}

b

64 cm2\text{}64 \text{ cm}^2\text{}

c

303 cm2\text{}30\sqrt{3} \text{ cm}^2\text{}

3407 or approximately 48.57\text{}\frac{340}{7}\text{} \text{ or approximately } 48.57

Step-by-step solution

10 steps

Here are the solutions to the problems:

15. Define the term quadrilateral and state two examples.
A quadrilateral is a polygon with four straight sides and four vertices. Its interior angles sum to 360∘360^\circ.
Examples: Square, Rectangle.

17. If z=98∘z=98^\circ find angles w,xw, x and yy.
Assuming the given quadrilateral is a parallelogram based on the diagram:

Step 1: In a parallelogram, opposite angles are equal.
Given z=98∘z = 98^\circ. Angle ww is opposite to zz.

w=zw = z w=98∘w = \boxed{\text{}98^\circ\text{}}

Step 2: In a parallelogram, consecutive angles are supplementary (add up to 180∘180^\circ).
Angle xx is consecutive to zz.

x+z=180∘x + z = 180^\circ x+98∘=180∘x + 98^\circ = 180^\circ x=180∘−98∘x = 180^\circ - 98^\circ x=82∘x = \boxed{\text{}82^\circ\text{}}

Step 3: Angle yy is opposite to xx.

y=xy = x y=82∘y = \boxed{\text{}82^\circ\text{}}

18. Calculate the area of the following:

Part a· 2 steps

Trapezium

  1. Identify the lengths of the parallel sides (aa and bb) and the height (hh).

    a=6 cma = 6 \text{ cm} b=10 cmb = 10 \text{ cm} h=4 cmh = 4 \text{ cm}
  2. Use the formula for the area of a trapezium.

    Area=12(a+b)h\text{Area} = \frac{1}{2}(a+b)h Area=12(6 cm+10 cm)(4 cm)\text{Area} = \frac{1}{2}(6 \text{ cm} + 10 \text{ cm})(4 \text{ cm}) Area=12(16 cm)(4 cm)\text{Area} = \frac{1}{2}(16 \text{ cm})(4 \text{ cm}) Area=12(64 cm2)\text{Area} = \frac{1}{2}(64 \text{ cm}^2) Area=32 cm2\text{Area} = \boxed{\text{}32 \text{ cm}^2\text{}}

Part b

64 cm2\text{}64 \text{ cm}^2\text{}

Square

  1. Identify the side length (ss).

    s=8 cms = 8 \text{ cm}
  2. Use the formula for the area of a square.

    Area=s2\text{Area} = s^2 Area=(8 cm)2\text{Area} = (8 \text{ cm})^2 Area=64 cm2\text{Area} = \boxed{\text{}64 \text{ cm}^2\text{}}

Part c

303 cm2\text{}30\sqrt{3} \text{ cm}^2\text{}

Parallelogram

  1. Identify the base (bb) and the given angle and side.

    Base=10 cm\text{Base} = 10 \text{ cm} Side=6 cm\text{Side} = 6 \text{ cm} Angle=60∘\text{Angle} = 60^\circ
  2. Calculate the height (hh) using trigonometry.

    h=Side×sin⁡(Angle)h = \text{Side} \times \sin(\text{Angle}) h=6 cm×sin⁡(60∘)h = 6 \text{ cm} \times \sin(60^\circ) h=6 cm×32h = 6 \text{ cm} \times \frac{\sqrt{3}}{2} h=33 cmh = 3\sqrt{3} \text{ cm}
  3. Use the formula for the area of a parallelogram.

    Area=Base×h\text{Area} = \text{Base} \times h Area=10 cm×33 cm\text{Area} = 10 \text{ cm} \times 3\sqrt{3} \text{ cm} Area=303 cm2\text{Area} = \boxed{\text{}30\sqrt{3} \text{ cm}^2\text{}}

    (Approximately 51.96 cm251.96 \text{ cm}^2)

    19. A quadrilateral has angles x∘,(x+20)∘,2x∘,3x∘x^\circ, (x+20)^\circ, 2x^\circ, 3x^\circ. Find the value of xx.

  4. The sum of the interior angles of a quadrilateral is 360∘360^\circ.

    Set up the equation with the given angles.

    x+(x+20)+2x+3x=360x + (x+20) + 2x + 3x = 360
  5. Combine like terms.

    (x+x+2x+3x)+20=360(x+x+2x+3x) + 20 = 360 7x+20=3607x + 20 = 360
  6. Solve for xx.

    7x=360−207x = 360 - 20 7x=3407x = 340 x=3407x = \frac{340}{7} x=3407 or approximately 48.57x = \boxed{\text{}\frac{340}{7}\text{} \text{ or approximately } 48.57}
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