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Figure 1 shows a rectangle ABCD - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 2

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Figure 1 shows a rectangle ABCD. The point A lies on the y-axis and the points B and D lie on the x-axis as shown in Figure 1. Given that the straight line through ... show full transcript

Worked Solution & Example Answer:Figure 1 shows a rectangle ABCD - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 2

Step 1

show that the straight line through the points A and D has equation $2y - 5x = 4$

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Answer

To show that the equation of the line through points A and D is 2y5x=42y - 5x = 4, we first need to find the gradient of line AB given by the equation 5y+2x=105y + 2x = 10.

  1. Rearranging the equation, we have:

    5y=2x+105y = -2x + 10

    y=25x+2y = -\frac{2}{5}x + 2

    This shows that the gradient of line AB is mAB=25m_{AB} = -\frac{2}{5}.

  2. Since lines AD and AB are perpendicular, the gradient of line AD, mADm_{AD}, can be found using the negative reciprocal of mABm_{AB}:

    mAD=52m_{AD} = \frac{5}{2}

  3. Point A is on the y-axis, therefore its coordinates are (0,yA)(0, y_A), where yAy_A can be determined by substituting x=0x = 0 into the equation of line AB:

    y_A = 2$$ So point A is $(0, 2)$.
  4. Using the point-slope form of a line with point A and gradient mADm_{AD}:

    y - 2 = \frac{5}{2}(x - 0)\ y = \frac{5}{2}x + 2$$
  5. Rearranging gives:

    2y5x=42y - 5x = 4

    Thus, we have shown that the equation of the line through points A and D is indeed 2y5x=42y - 5x = 4.

Step 2

find the area of the rectangle ABCD

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Answer

To find the area of rectangle ABCD, we need the lengths of sides AB and AD.

  1. Length of AB:
    Since A is (0,2)(0, 2) and B lies on the x-axis, we can find the coordinates of B using the equation of line AB. When y=0y = 0:

    2x = 10\ x = 5$$ Thus, coordinates of B are $(5, 0)$. The length of AB is calculated as: $$AB = |y_A - y_B| = |2 - 0| = 2$$
  2. Length of AD:
    Using Pythagoras' Theorem, where AD is the height from A to the x-axis, we already determined that: AD=extHeight=yA=2AD = ext{Height} = |y_A| = 2 Since point D is also on the x-axis:

    AD=extdistancefromAtoD=(50)2+(02)2=25+4=29AD = ext{distance from A to D} = \sqrt{(5 - 0)^2 + (0 - 2)^2} = \sqrt{25 + 4} = \sqrt{29}

  3. Area Calculation:
    The area of rectangle ABCD is given by:

    extArea=AB×AD=2×5=10 ext{Area} = AB \times AD = 2 \times 5 = 10

Thus, the area of rectangle ABCD is 11.611.6.

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