What is the area of a parallelogram with vertices (2,5), (5, 10), (10, 15), and (7, 10)?

1 Answer
Aug 9, 2018

# "Area of parallelogram "ABCD=10 " sq. units"#

Explanation:

We know that ,

#color(blue)("If "P(x_1,y_1) ,Q(x_2,y_2),R(x_3,y_3)# are the vertices of

#color(blue)(triangle PQR#, then area of triangle:

#color(blue)(Delta=1/2||D||,# where , #color(blue)(D=|(x_1,y_1,1) ,(x_2,y_2,1),(x_3,y_3,1)|#........................#(1)#

Plot the graph as shown below.

Consider the points in order, as shown in the graph.

enter image source here

Let #A(2,5) ,B(5,10) ,C(10,15) and D(7,10)# be the vertices of Parallelogram #ABCD#.

We know that ,

#"Each diagonal of a parallelogram separates parallelogram"#

#"into congruent triangles."#

Let #bar(BD)# be the diagonal.

So, #triangleABD~=triangleBDC#

#:. "Area of parallelogram "ABCD=2xx "area of"triangleABD "#

Using #(1)#,we get

#color(blue)(Delta=1/2||D|| ,where, # #color(blue)(D=|(2,5,1),(5,10,1),(7,10,1)|#

Expanding we get

#:.D=2(10-10)-5(5-7)+1(50-70)#

#:.D=0+10-20=-10#

#:.Delta=1/2||-10||=||-5||#

#:.Delta=5#

#:. "Area of parallelogram "ABCD=2xx "area of"triangleABD "#

#:. "Area of parallelogram "ABCD=2xx(5)=10#

#:. "Area of parallelogram "ABCD=10 " sq. units"#