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The diagram shows two angles on a straight line. a) Find the value of a. b) Find the perimeter of the following triangle. c) The area of this trapezium is 6000 cm², find the second parallel length.

Mathematics

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2 parts
b

36 cm36 \text{ cm}

c

−15 cm-15 \text{ cm}

Step-by-step solution

10 steps

Step 1: Set up the equation for angles on a straight line.
Angles on a straight line sum to 180∘180^\circ.

(3a+10)∘+(4a−30)∘=180∘(3a + 10)^\circ + (4a - 30)^\circ = 180^\circ

Step 2: Combine like terms.

3a+4a+10−30=1803a + 4a + 10 - 30 = 180 7a−20=1807a - 20 = 180

Step 3: Isolate the term with aa.

7a=180+207a = 180 + 20 7a=2007a = 200

Step 4: Solve for aa.

a=2007a = \frac{200}{7}

The value of aa is 2007\boxed{\frac{200}{7}}.

Part b· 4 steps

Find the perimeter of the following triangle.

  1. Identify the properties of the triangle.

    The triangle is isosceles, with a base of 16 cm16 \text{ cm} and a height of 6 cm6 \text{ cm}. The height bisects the base, forming two right-angled triangles.

  2. Calculate the base of one right-angled triangle.

    The base of each right-angled triangle is 16 cm2=8 cm\frac{16 \text{ cm}}{2} = 8 \text{ cm}.

  3. Use the Pythagorean theorem to find the length of the equal sides.

    Let ss be the length of the equal sides.

    s2=(8 cm)2+(6 cm)2s^2 = (8 \text{ cm})^2 + (6 \text{ cm})^2 s2=64 cm2+36 cm2s^2 = 64 \text{ cm}^2 + 36 \text{ cm}^2 s2=100 cm2s^2 = 100 \text{ cm}^2 s=100 cm2s = \sqrt{100 \text{ cm}^2} s=10 cms = 10 \text{ cm}

  4. Calculate the perimeter of the triangle.

    Perimeter = base + side + side
    Perimeter = 16 cm+10 cm+10 cm16 \text{ cm} + 10 \text{ cm} + 10 \text{ cm}
    Perimeter = 36 cm36 \text{ cm}
    The perimeter of the triangle is 36 cm\boxed{36 \text{ cm}}.

Part c

−15 cm-15 \text{ cm}

The area of this trapezium is 6000 cm26000 \text{ cm}^2, find the second parallel length.

  1. Identify the given values for the trapezium.

    Area (AA) = 6000 cm26000 \text{ cm}^2
    One parallel length (b1b_1) = 90 cm90 \text{ cm}
    Height (hh) = 160 cm160 \text{ cm}
    Let the second parallel length be b2b_2.

  2. Write the formula for the area of a trapezium.

    A=12(b1+b2)hA = \frac{1}{2} (b_1 + b_2) h

  3. Substitute the known values into the formula.

    6000 cm2=12(90 cm+b2)(160 cm)6000 \text{ cm}^2 = \frac{1}{2} (90 \text{ cm} + b_2) (160 \text{ cm})

  4. Simplify the equation.

    6000=(90+b2)16026000 = (90 + b_2) \frac{160}{2} 6000=(90+b2)806000 = (90 + b_2) 80

  5. Isolate the term with b2b_2.

    600080=90+b2\frac{6000}{80} = 90 + b_2 75=90+b275 = 90 + b_2

  6. Solve for b2b_2.

    b2=75−90b_2 = 75 - 90 b2=−15 cmb_2 = -15 \text{ cm}

    The calculated second parallel length is −15 cm\boxed{-15 \text{ cm}}.
    Note: A physical length cannot be negative. This indicates a potential inconsistency in the problem's given values.

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