Question

if A(5, 4), B(4, -1), C(-1, -2), D(0, 3) be four points in a plane, show that ABCD is q rhombus but not a square. find the area of the rhombus​

Answer 1

Before we solve:-

A rhombus is a figure that four sides are equal, and a square is a figure that is a rhombus and a rectangle. To show the figure is a rhombus which is not a square, we can consider 'a rhombus which is not a rectangle'. Let's get to the problem.

• Rhombus

→ All four sides are equal.

• Rectangle

→ All four angles are equal.

Solution:-

Assuming A, B, C, D are in order, let's test if the figure is a rhombus.

• If it is a rhombus, $\overline{AB}=\overline{BC}=\overline{CD}=\overline{DA}$.
• If it is not a rhombus, at least one length is different.

$\overline{AB}=\sqrt{(5-4)^2+(4+1)^2} =\sqrt{1+25} =\sqrt{26}$

$\overline{BC}=\sqrt{(4+1)^2+(-1+2)^2} =\sqrt{25+1} =\sqrt{26}$

$\overline{CD}=\sqrt{(0+1)^2+(3+2)^2} =\sqrt{1+25} =\sqrt{26}$

$\overline{DA}=\sqrt{(0-5)^2+(3-4)^2} =\sqrt{25+1} =\sqrt{26}$

$\therefore \overline{AB}=\overline{BC}=\overline{CD}=\overline{DA}=\sqrt{26}$ ...[Eqn. 1]

Then now let's test if the figure is not a rectangle.

• If it is a rectangle, $\overline{CA}^2=\overline{AB}^2+\overline{BC}^2$.
• If it is not a rectangle, $\overline{CA}^2\neq\overline{AB}^2+\overline{BC}^2$.

$\overline{CA}=\sqrt{(-1-5)^2+(-2-4)^2} =\sqrt{36+36} =6\sqrt{2}$

$\therefore\overline{CA}^2=72$

But,

$\overline{CA}^2=72\neq\overline{AB}^2+\overline{BC}^2=52$ ...[Eqn. 2]

According to [Eqn. 1, 2] it is a rhombus that is not a square.

Area of the rhombus:-

$\dfrac{1}{2} \times \overline{AC}\times \overline{BD}$

$=\dfrac{1}{2} \times \sqrt{(5+1)^2+(4+2)^2} \times \sqrt{(4-0)^2+(-1-3)^2}$

$=\dfrac{1}{2} \times \sqrt{36+36} \times \sqrt{16+16}$

$=\dfrac{1}{2} \times 6\sqrt{2} \times 4\sqrt{2}$

$=24$ units².