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F5 Maths !!!
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Q: http://postimg.org/image/83ynseit9/
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20. (a) Area of trapezium ABCD = (1/2) * [2p + (2p + 8)] * 2p cm2 = (1/2) * (2p + 2p + 8) * 2p cm2 = (4p2 + 8p) cm2 Area of ΔBED = (1/2) * [(2p + 8)/2] * 2p cm2 = (p2 + 4p) cm2 Shaded area = (Area of trapezium ABCD) - (Area of ΔBED) q = (4p2 + 8p) - (p2 + 4p) q = 4p2 + 8p - p2 + 4p q = 3p2 + 4p (b) Shaded area = (7/10) * (Area of trapezium ABCD) 3p2 + 4p = (7/10) * (4p2 + 8p) 30p2 + 40p = 28p2 + 56p 2p2 - 16p = 0 2p(p - 8) = 0 p = 0 (rejected) or p = 8 (c) (i) In the graph, show p as x, and q as y. Draw a straight line y = 5 on the graph. (y = 5 is a horizontal line with y-intercept 5.) The line cut the graph of (y = 3x2 + 4x) at x = 0.79 and x =2.12 Hence, when q = 5, p = 0.79 or p = 2.12 (ii) Draw a straight line y = 15 on the graph. (y = 15 is a horizontal line with y-intercept 15.) The line cut the graph of (y = 3x2 + 4x) at x = 1.67 and x = -3 From the graph, when q
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(a) Area of shaded region = area of trapezium ABCD - area of triangle BED = (AB+CD)*(BC)/2 - (ED*BC)/2[note that ED is half of CD] = (2p+2p+8)(2p)/2 - (p+4)*(2p)/2 = (4p^2+8p) - (p^2+4p) = 3p^2 + 4p (b) q = (7/10)(Area ABCD) 3p^2 + 4p = (7/10)(4p^2+8p) 30p^2 +40p = 28p^2 + 56p 2p^2 - 16p = 0 p = 8 or 0 (rej.)[0 is rejected as BA can't equal to zero.] (c)(i) Function of y = 3x^2+4x is same as q=3p^2 + 4p Add a line of y=5 in the graph of y = 3x^2+4x, the intercepts at y=5 are the values of p. (around -2.x and 0.x) (ii) Add a line of y=15 in the graph of y = 3x^2+4x, the values below y=15 are the possibles values of p. ( -3
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