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

Aug 9, 2018

$\text{Area of parallelogram "ABCD=10 " sq. units}$

#### 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)|........................$\left(1\right)$

Plot the graph as shown below.

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

Let $A \left(2 , 5\right) , B \left(5 , 10\right) , C \left(10 , 15\right) \mathmr{and} D \left(7 , 10\right)$ be the vertices of Parallelogram $A B C D$.

We know that ,

$\text{Each diagonal of a parallelogram separates parallelogram}$

$\text{into congruent triangles.}$

Let $\overline{B D}$ be the diagonal.

So, $\triangle A B D \cong \triangle B D C$

$\therefore \text{Area of parallelogram "ABCD=2xx "area of"triangleABD }$

Using $\left(1\right)$,we get

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

Expanding we get

$\therefore D = 2 \left(10 - 10\right) - 5 \left(5 - 7\right) + 1 \left(50 - 70\right)$

$\therefore D = 0 + 10 - 20 = - 10$

$\therefore \Delta = \frac{1}{2} | | - 10 | | = | | - 5 | |$

$\therefore \Delta = 5$

$\therefore \text{Area of parallelogram "ABCD=2xx "area of"triangleABD }$

$\therefore \text{Area of parallelogram } A B C D = 2 \times \left(5\right) = 10$

$\therefore \text{Area of parallelogram "ABCD=10 " sq. units}$