Respuesta :
Answer:
(a) a = 22 or 2
(b) The equations of AD are
y = x/2 + 7
or
y = x/2 + 17
(c) The equation of CD is y = -2·x - 3
(d) The coordinate of the point D is either (-8, 13) or (-4, 5)
(e) the possible areas are;
250 square units or 270 square units
Step-by-step explanation:
With only the details of the trapezium, without the drawing, we have as follows;
(a) The given points are;
A(a, 18), B(12, -2), and C(2, -7)
The length of BC is given from the formula for finding the length, l, of a line with the coordinates of the end points as follows;
[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex]
[tex]l_{BC} = \sqrt{\left ((-7)-(-2) \right )^{2}+\left ((2)-(12) \right )^{2}} = \sqrt{\left ((-5) \right )^{2}+\left (-(10) \right )^{2}} = 5\cdot \sqrt{5}[/tex]
∴ From [tex]l_{AB} = l_{BC}[/tex], we have;
[tex]l_{AB}[/tex] = 2 × 5·√5 = 10·√5
Which gives;
[tex]l_{AB} = \sqrt{\left ((18)-(-2) \right )^{2}+\left (a-12 \right )^{2}} = \sqrt{\left 20 \right ^{2}+\left (a-12 \right )^{2}}= 10 \cdot \sqrt{5}[/tex]
20² + (a - 12)² = 500
(a - 12)² = 500 - 20² = 500 - 400 = 100
(a - 12)² = 100
a - 12 = ±√100 = ±10
a = 10 + 12 or -10 + 12
a = 22 or 2
(b) The equation of BC is given as follows;
The slope, m, of BC = (-7 -(-2)/(2 - 12) = -5/-10 = 1/2
The equation of BC is therefore;
y - (-7) = 1/2×(x - 2)
y + 7 = x/2 - 1
y = x/2 - 1 - 7 = x/2 - 8
y = x/2 - 8
Therefore, the slope of AD = m = 1/2
The equation of AD can be
y - 18 = 1/2×(x - 22)
y = x/2 -11 + 18 = x/2 + 7
y = x/2 + 7
or
y - 18 = 1/2×(x - 2)
y = x/2 -1+ 18 = x/2 + 17
y = x/2 + 17
(c) The equation of CD is given as follows;
CD is perpendicular to BC, therefore, the slope of CD = -1/m = -2
The equation of CD is therefore;
y - (-7) = -2×(x - 2)
y = -2·x + 4 - 7 = -2·x - 3
y = -2·x - 3
(d) The coordinate of the point D is found as follows;
At point D,
At
x/2 + 17=-2·x - 3
2.5·x = -20
x = -8
y = -8/2 + 17 = 13
or
x/2 + 7 =-2·x - 3
2.5·x = -10
x = -4
y = -4/2 + 7 = 5
The possible coordinates of the point D are (-8, 13) or (-4, 5)
(e) The area of the trapezium is found as follows;
The vertices points are;
(2, 18) or (22, 18), (12, -2), (2, -7) and (-8, 13) or (-4, 5)
The formula for the area of a trapezium = (a + b)/2×h
Length of a = [tex]l_{BC}[/tex] = 5·√5
h = [tex]l_{CD} = \sqrt{\left ((13)-(-7) \right )^{2}+\left ((-8)-2 \right )^{2}} = \sqrt{\left 20 \right ^{2}+10^{2}}= 10 \cdot \sqrt{5}[/tex]
or
[tex]l_{CD} = \sqrt{\left ((5)-(-7) \right )^{2}+\left ((-4)-2 \right )^{2}} = \sqrt{\left 12 \right ^{2}+6^{2}}= 6 \cdot \sqrt{5}[/tex]
b = [tex]l_{AD} = \sqrt{\left (13-18 \right )^{2}+\left ((-8)-2 \right )^{2}} = \sqrt{\left (-5 \right )^{2}+(-10)^{2}}= 5 \cdot \sqrt{5}[/tex]
[tex]l_{AD} = \sqrt{\left (5-18 \right )^{2}+\left ((-4)-22 \right )^{2}} = \sqrt{\left (-13 \right )^{2}+(-26)^{2}}= 13 \cdot \sqrt{5}[/tex]
Therefore, the possible areas are;
(5×√5 + 5×√5)/2 × 10×√5 = 250 square units
(5×√5 + 13×√5)/2 × 6×√5 = 270 square units
The value of 'a' is 22 or 2, the equation of AD is (y = 0.5x + 17) or (y = 0.5x + 7) and the point D is (-8,13) or (-4,5) and this can be determine by using the point slope form.
Given :
- The diagram shows a trapezium in which AD is parallel to BC and angle ADC = angle BCD = 90°.
- The points A, B, and C are (a, 18), (12, -2) and (2, -7) respectively.
- AB = 2 BC
a) To determine the value of 'a' use the relation (AB = 2 BC).
[tex]\sqrt{(12-a)^2+(-2-18)^2}=2\times \sqrt{(2-12)^2+ (-7+2)^2}[/tex]
[tex]\sqrt{(12-a)^2+400}=2\times \sqrt{125}[/tex]
Squaring both sides in the above expression.
[tex](12-a)^2+400=4\times 125[/tex]
[tex]144+a^2-24a=100[/tex]
[tex]a^2-24a+44=0[/tex]
[tex]a^2-22a-2a+44=0[/tex]
[tex]a(a-22)-2(a-22) = 0[/tex]
a = 2 or 22
b) The equation of BC is given by:
[tex]\dfrac{y+2}{x-12}=\dfrac{-7+2}{2-12}[/tex]
[tex]2(y+2)=(x-12)[/tex]
2y + 4 = x - 12
2y - x + 16 = 0
y = 0.5x - 8
Given that AD is parallel to BC so, the slope of 0.5.
First, take a = 2. The equation of line AD is given by:
[tex]y-18 =0.5(x-2)[/tex]
Now, take a = 22. The equation of line AD is given by:
[tex]y-18 =0.5(x-22)[/tex]
c) The line CD is perpendicular to line BC. So, the slope of line CD is -2. The equation of the line CD is given by:
y - (-7) = -2(x - 2)
y + 7 = -2x + 4
y + 2x + 3 = 0
d) The point D is given by:
0.5x + 17 = -2x - 3
2.5x = -20
x = -8
y = -4 + 17 = 13
or
0.5x + 7 = -2x - 3
x = -4
Now, y = - 2 + 7 = 5
e) Area of the trapezium is given by:
[tex]\rm A = L_{CD} \times L_{AD}[/tex]
So, the possible area of the trapezium is:
[tex]\dfrac{(5\times \sqrt{5} +5\times \sqrt{5} )}{2}\times 10 \times \sqrt{5} = 250[/tex]
[tex]\dfrac{(5\times \sqrt{5} +13\times \sqrt{5} )}{2}\times 6 \times \sqrt{5} = 270[/tex]
For more information, refer to the link given below:
https://brainly.com/question/18666670
