Question

A park with flower plants is to be developed within a quadrilateral with points A(0, −1), B(6, 7), C(−2, 3) and D(8, 3) as vertices and AB and CD as diagonals. Show that AB and CD bisect each other and AD2 + DB2 = AB2. Find the area of the park. (All distances are in km) ​

Answer 1

Answer: The area is 40 sq. km.

Step-by-step explanation:  As shown in the attached figure, ABCD is a quadrilateral with vertices A(0, −1), B(6, 7), C(−2, 3) and D(8, 3) and diagonals AB and CD.

Let the diagonals AB and CD intersect at the point 'O'.

The co-ordinates of the mid-point of AB are

$M_{AB}=\left(\dfrac{0+6}{2},\dfrac{-1+7}{2}\right)=(3,3),$

and the co-ordinates of the mid-point of CD are

$M_{CD}=\left(\dfrac{-2+8}{2},\dfrac{3+3}{2}\right)=(3,3).$

Therefore, the co-ordinates of 'O' are (3,3), which is th eintersecting point of AB and CD.

So, AB and CD bisect each other.

Now,

$AD^2+BD^2=(8-0)^2+(3+1)^2+(8-6)^2+(3-7)^2=64+16+4+16=68+32=100,\\\\AB^2=(6-0)^2+(7+1)^2=36+64=100.$

Thus, $AD^2+BD^2=AB^2.$

Also, this condition implies that the triangle ABD is a right-angled triangle, where ∠D=90°.

Now, slope of AC is

$m_1=\dfrac{3+1}{-2-0}=-2,$

and slope of BD is

$m_2=\dfrac{3-7}{8-6}=-2.$

Since the slopes of AC and BD are equal, so the lines are parallel. From here, we can conclude that

∠D=∠A=90°.

Similarly, we can show that

∠B=∠C=90°.

We have

AD² = 80 ⇒ AD = 4√5,

AC² = 20 ⇒ AC = 2√5.

Therefore, area of the quadrilateral ABCD (rectangle) will be

$AD\times AC = 4\sqrt 5\times 2\sqrt 5=40~\textup{sq. km.}$