Point P is on the line y = 2x + a. The x-coordinate of P is 4. The a

How come the sum of parallel sides is a + a + 8? I understand a+8 but how come a?

One of the parallel side is along the Y-axis where the side is from the origin to the Y-intercept (x=0) of the line y=2x+a and so the length of this side is calculated from 0 to a which is a.

And as you know the length of the other parallel side is a+8

So the sum of the lengths of parallel sides is a+a+8

y coordinate of P when x=4 is
y=2(4)+a
y=8+a

So, P=(4,8+a)

Split the total area into

1. A rectangle with length \(4\) and breadth \(a\)

2. A right triangle with base \(4\) and height \((8+a)-a = 8\)

Area of the rectangle = \(4a\)
Area of the triangle = \(\frac{1}{2}*4*8\)=\(16\)

Total Area = \(4a+16\) = \(24\)

\(4a=24-16\)
\(4a=8\)
\(a=2\)

Answer is (C)

We know P(4,y). Based on the figure, we can construct a line from point, lets call it Q, to get a complete rectangle, the coordinates of that point , Q, will be (0,y).
We know, slope is 2. So, for 1 unit increase in x, we have 2 units increase in y. Therefore, y coordinate of Point P will be 2 X 4(x-coordinate) = 8.
We can find the area of newly formed rectangle. 8 units(vertical distance) x 4 units (horizontal) = 32 units^2.
We have a right-angled triangle from point Q (0,8).
We have the area of the quadrilateral = 24 units^2. Area of the triangle will be= Area of rectangle - Area of quadrilateral(given in the question) = 32 - 24 = 8 units^2.
distance from point Q and a can be calculated by:
area of triangle = bh/2.
8 = 4b/2, base =6.
6 units down from y axis (0,8) we have (0,2).
As you have already noticed, a is y-intercept value of x will be 0 and y will be 2.

The following sum can be solved by making two equations
Let us first plot the point P as (4,b) and the positive point on the Y axis as (0,a)

1. We have been given that the area of the trapezium (quadrilateral) is 24 = ((l1 +l2) /2) * h (4)
((a + b) / 2) * 4 = 24
a+b = 12 ......Eq.1

2. We know the slope of the line from the equation of the straight line is 2
From the slope we can get another equation
(y2-y1) / (x2-x1) = 2
((x - a) / 4 - 0 ) = 2
b - a = 8......Eq.2
Solving both Equation we get
2a = 4
a = 2

Answer is C

Let y coordinate of P be b
Triangle part of the quadrilateral is (4*(b-a) / 2, and the rest is 4a
so the entire part is
[(4*(b-a)) / 2 ] + 4a = 24
we get b+a=12 or b=12-a

Now we have P x coordinate 4 and y coordinate 12-a = P(4,12-a)
Insert these two figures in the given equation of y=2x+a
12-a = 4(2) + a
a = 2

Answer C

Area of a trapezium = 1/2 h(a+b), where h is the height of the trapezium, and a & b are the 2 parallel sides, of differing lengths.

From the figure, you should be able to figure that
1. the the Point P on the line y=2x +a,
2. the point at which it cuts the Y axis(a, which is the y intercept as mentioned in the equation)
3. the point at which the perpendicular from Point P meets the X axis(4,0)
4. and the origin
> form a trapezium

Fitting the relevant points into the 1/2h(a+b) formula...

here, a = a (the length of the side of the trapezium on the Y axis)
b= y(length of the perpendicular from Point P on the X axis)

h= 4 (the length from the origin the point (4,0)

Solving this equation should give you: a+y = 12----- eq. 1

Now, from the equation of the line(y= 2x +a), by moving the terms around and by substituting x=4[given in the question, the x coordinate of P], you should get an equation of the form a-y = -8 ------ eq. 2


Solving eq.'s 1 and 2 should given you[b] 2a = 4 --> a = 2


Thanks!

Bunuel
Can you please explain this question??

I plan to answer the question in the same way as the solutions already given. Can you let me know what specific parts you want more information about?