F5 Maths !!!

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 < 15, -3 < p < 1.67

(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 < p < 1.x)