Step-by-step explanation:
distance QR :
Pythagoras, as the distance between 2 points is the Hypotenuse (baseline) of a right-angled triangle of the differences of the x and y coordinates.
QR² = (12 - 2)² + (-2 - -7)² = 10² + 5² = 125
QR = sqrt(125) = sqrt(25×5) = 5×sqrt(5)
PQ = 2×QR = 10×sqrt(5) = sqrt(100×5) = sqrt(500)
PQ² = (h - 12)² + (18 - -2)² = h² - 24h + 144 + 20²
500 = h² - 24h + 144 + 400
0 = h² - 24h + 44
since a quadratic equation :
h = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -24
c = 44
h = (24 ± sqrt(576 - 176))/2 = (24 ± sqrt(400))/2 =
= (24 ± 20)/2 = 12 ± 10
h1 = 22
h2 = 2
since the trapezium opens up to the right at the angle Q, it must be the value further to the right : h = 22.
the line QR has the slope (ratio y difference / x difference) :
x changes by +10 (from 2 to 12).
y changes by +5 (from -7 to -2).
the slope is 5/10 = 1/2
SP has the same slope (parallel line).
SR is perpendicular to these 2 lines. its slope is x and y differences upside-down in the ratio, and the sign is flipped.
the SR slope is therefore -2/1 = -2
the general line equation is
y = ax + b
a is the slope, b is the y- intercept (y value when x = 0).
we get b by using the known point coordinates in the half finished equation with the slope.
the SP equation is
y = (1/2)x + b
we know P (22, 18). so,
18 = (1/2)×22 + b = 11 + b
7 = b
the full SP line equation is
y = (1/2)x + 7
the RS equation :
y = -2x + b
we know R (2, -7). so,
-7 = -2×2 + b = -4 + b
-3 = b
the full RS equation is
y = -2x - 3
the coordinates of S we get by crossing RS and SP :
(1/2)x + 7 = -2x - 3
x + 14 = -4x - 6
5x = -20
x = -4
now using that in any of the 2 line equations
y = -2×-4 - 3 = 8 - 3 = 5
so, S = (-4, 5)
the distance SP is
SP² = (22 - -4)² + (18 - 5)² = 26² + 13² = 676 + 169 =
= 845
SP = sqrt(845) = sqrt(169×5) = 13×sqrt(5)
the distance RS is
RS² = (-4 - 2)² + (5 - -7)² = (-6)² + 12² = 36 + 144 = 180
RS = sqrt(180) = sqrt(36×5) = 6×sqrt(5)
the area of the trapezium is the sum of the rectangle of SRQ (and the 4th point being around the middle of SP) and the triangle of PQ and this 4th rectangle point.
the area of the rectangle is
6×sqrt(5)×5×sqrt(5) = 30×5 = 150
the area of the triangle is
base×height/2
height is the same as RS = 6×sqrt(5).
base is the part of SP from the 4th rectangle point to P, and that is the same as
SP - QR = 13×sqrt(5) - 5×sqrt(5) = 8×sqrt(5)
so, we have for the area of the triangle
6×sqrt(5)×8×sqrt(5)/2 = 3×sqrt(5)×8×sqrt(5) = 24×5 = 120
the total area of the trapezium is
150 = 120 = 270
