Question

What is the area of rhombus ABCD? Enter your answer in the box. Do not round at any steps. put the answer in units²

Answer 1

Answer:

Therefore the Area of Rhombus ABCD is 36 unit².

Step-by-step explanation:

Given:

ABCD is a Rhombus

A = (-1,0)

B = (5,-3)

C = (-1,-6)

D = (-7 ,-3)

To Find:

Area of Rhombus ABCD = ?

Solution:

We know that Area of Rhombus is given as

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times d_{1}\times d_{2}$

Where ,

d₁ and d₂ are the Diagonals.

We have,

Diagonals as AC and BD,

Using Distance Formula we get

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting coordinates A and C we get

$l(AC) = \sqrt{((-1-(-1))^{2}+(-6-0)^{2} )}=\sqrt{36}\\l(AC)=6\ unit$

Similarly for BD we have,

$l(BD) = \sqrt{((5-(-7))^{2}+(-3-(-3))^{2} )}=\sqrt{144}\\l(BD)=12\ unit$

Now Substituting AC and BD in Formula we get

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times AC\times BD$

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times 6\times 12=36\ unit^{2}$

Therefore the Area of Rhombus ABCD is 36 unit².