Mathematics, 07.09.2020 15:01 ovaldez5237

20. 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.
Given that AB 2BC, find
(a) the value of a,
(b) the equation of AD,
(c) the equation of CD,
(d) the coordinates of D,
(e) the area of the trapezium.

Answers: 2

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Answers

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).

$\sqrt{(12-a)^2+(-2-18)^2}=2\times \sqrt{(2-12)^2+ (-7+2)^2}$

$\sqrt{(12-a)^2+400}=2\times \sqrt{125}$

Squaring both sides in the above expression.

$(12-a)^2+400=4\times 125$

144+a^2-24a=100

a^2-24a+44=0

a^2-22a-2a+44=0

a(a-22)-2(a-22) = 0

a = 2 or 22

b) The equation of BC is given by:

$\dfrac{y+2}{x-12}=\dfrac{-7+2}{2-12}$

2(y+2)=(x-12)

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:

y-18 =0.5(x-2)

Now, take a = 22. The equation of line AD is given by:

y-18 =0.5(x-22)

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

0.5x + 7 = -2x - 3

x = -4

Now, y = - 2 + 7 = 5

e) Area of the trapezium is given by:

$\rm A = L_{CD} \times L_{AD}$

So, the possible area of the trapezium is:

$\dfrac{(5\times \sqrt{5} +5\times \sqrt{5} )}{2}\times 10 \times \sqrt{5} = 250$

$\dfrac{(5\times \sqrt{5} +13\times \sqrt{5} )}{2}\times 6 \times \sqrt{5} = 270$

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Answer from: musa79

(a) a = 22 or 2

(b) The equations of AD are

y = x/2 + 7

y = x/2 + 17

(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;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

$l_{BC} = \sqrt{\left ((-7)-(-2) \right )^{2}+\left ((2)-(12) \right )^{2}} = \sqrt{\left ((-5) \right )^{2}+\left (-(10) \right )^{2}} = 5\cdot \sqrt{5}$

∴ From $l_{AB} = l_{BC}$ , we have;

$l_{AB}$ = 2 × 5·√5 = 10·√5

Which gives;

$l_{AB} = \sqrt{\left ((18)-(-2) \right )^{2}+\left (a-12 \right )^{2}} = \sqrt{\left 20 \right ^{2}+\left (a-12 \right )^{2}}= 10 \cdot \sqrt{5}$

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

y - 18 = 1/2×(x - 2)

y = x/2 -1+ 18 = x/2 + 17

y = x/2 + 17

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,

x/2 + 17=-2·x - 3

2.5·x = -20

x = -8

y = -8/2 + 17 = 13

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 = $l_{BC}$ = 5·√5

h = $l_{CD} = \sqrt{\left ((13)-(-7) \right )^{2}+\left ((-8)-2 \right )^{2}} = \sqrt{\left 20 \right ^{2}+10^{2}}= 10 \cdot \sqrt{5}$

$l_{CD} = \sqrt{\left ((5)-(-7) \right )^{2}+\left ((-4)-2 \right )^{2}} = \sqrt{\left 12 \right ^{2}+6^{2}}= 6 \cdot \sqrt{5}$

b = $l_{AD} = \sqrt{\left (13-18 \right )^{2}+\left ((-8)-2 \right )^{2}} = \sqrt{\left (-5 \right )^{2}+(-10)^{2}}= 5 \cdot \sqrt{5}$

$l_{AD} = \sqrt{\left (5-18 \right )^{2}+\left ((-4)-22 \right )^{2}} = \sqrt{\left (-13 \right )^{2}+(-26)^{2}}= 13 \cdot \sqrt{5}$

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

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20. The diagram shows a trapezium in which AD is parallel to BC and angle ADC = angle BCD = 90°.

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