Parallelogram ABCD lies in the xy-plane, as shown in the figure above. The coordinates of point C are (-3, 4) and the coordinates of point B are (-7, 7). What is the area of the parallelogram ?

A. 1

B. \(2 \sqrt{7}\)

C. 7

D. 8

E. \(7 \sqrt{2}\)

can u explain the solution?

The easiest way to solve this problem is to draw a rectangle around the parallelogram, find its area, and substract area of the triangles that emerge around the parallelogram, within the rectangle (but that are not part of the parallelogram).
Since ABCD is a parallelogram, line segments AB and CD have the same length and the same slope. Therefore, in the diagram above, point A is at (-4,3), The square has an area of 7*7=49. By drawing carefully and exploiting similar triangles created by various parallel lines, you can label the height of each triangle 3, and each base 7. Each triangles has area 1/2hb=1/2*3*7=21/2. Therefore, the area of the parallelogram ABCD equals 49-4*(21/2)=49-42=7.

First find the co-ordinates of point A. For now, lets assume the co-ordinates are (x,y)
Since it is mentioned that ABCD is a parallelogram. So slope of BC must be equal to slope of AD since BC||AD.
Hence we can get the equation 3x=4y=0 -- equation(1) [by equating slope of line BC and AD]

Similarly by equating the lines AB and CD since (AB||CD), we get the equation 4x+3y= -7 -- (2)

Solving for equations (1) and (2) we get x = -4 and y= 3. Thus A is (-4,3).

If we observe now, all sides are equal in length i.e. each side AB=BC=CD=AD=5. Thus ABCD is a rhombus. The area of a rhombus is (product of lengths of diagonals)/2 i.e. in our case (BD*AC)/2.

BD is 7*(2)^(1/2) and AC is 2^(1/2). Thus area of ABCD is 7{sqrt2} * {sqrt2} = 7

option C

can you please show in figure please for better understanding ?
regards,

There is an easier solution, where we need to connect the diagonal AC and find it's length, which is relatively easy with the elaborate coordinate system given. We can use the distance formula between A,C points, or use an easier technique, forming a small right triangle with A,C by extending lines from the points parallel to both X and Y axes. Let's assume the point of intersection is X. So AX = 1, CX = 1 from the graph, so AC is the hypotenuse which is root(2).

We can calculate the height of the 2 triangles BDC and BDA of the parallelogram by taking half of AC, which is root(2) / 2.

The base BD of the 2 triangles of the parallelogram can also be found out either by the distance formula, or the right angle triangle method. It's 7root(2).

Area of triangle BDC = 1/2 * base * height = 1/2 * BD * root(2)/2 = 1/2 * 7root(2) * root(2)/2 which is 7/2. So the area of the parallelogram is twice the area of the triangle, so 7 * (7/2) = 7.

There is one big quadrant with sides 7 and 7 and the area 49; four congruent triangles with legs 3 and 4 providing for the area of each triangle of 12/2=6; two must-be similar quandrants with sides 3 and 3 and the individual areas of 9. These properties are due to parallelogram having equal opposite sides.

Solving for the parallelogram's area: 49 - 4*6 - 2*9 = 49 - 24 - 18 = 7. Answer is C.

One other way is to connect points BD and divide by 2 to find height of half-triangles or ABC and ACD areas. Height of ABC and ACD=(1/2)*sqroot(2)*7=7/sqroot(2). Base of two congruent triangles ABC and ACD=sqroot(2)*1. Hence, area of ABC or ACD= [7/sqroot(2)*sqroot(2)]/2=7/2. The parallelogram consists of ABC and ACD, and its area is 2*(7/2)=7. Answer is C.

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