2
$\begingroup$

What is the area in square units, of a quadrilateral whose vertices are $$(5,3), (6,-4), (-3,-2), (-4,7)?$$

I have tried creating the triangles, but didn't know how to find the diagonal. I wanted to try the shoelace method but I thought it only worked for triangles. The answer that was provided is $69$.

$\endgroup$
  • $\begingroup$ I drew the quadrilateral but I wasn't able to get the area of the 2 triangles it makes. I am not even sure if making 2 triangles is the best way to get the area of the quadrilateral. $\endgroup$ – Rubab Apr 30 '15 at 13:06
  • $\begingroup$ You can definitely break it into two triangles. If you want to do it that way I'd suggest using Heron's formula to find the area of each triangle. $\endgroup$ – user137731 Apr 30 '15 at 13:30
  • $\begingroup$ But I don't know how to find the length of the diagonal. $\endgroup$ – Rubab Apr 30 '15 at 14:23
  • $\begingroup$ The distance between two points $(a_1, a_2)$ and $(b_1, b_2)$ is given by $\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2}$. $\endgroup$ – user137731 Apr 30 '15 at 14:27
  • $\begingroup$ I thought the distance between two points $(a_1,a_2)$ and $(b_1,b_2)$ is given by $\sqrt{(a_2−a_1)^2 + (b_2−b_1)^2.}$ – ? Am I right? $\endgroup$ – Rubab Apr 30 '15 at 15:19
3
$\begingroup$

Let $A(5,3), B(6,-4), C(-3,-2), D(-4,7)$.

The area of $\triangle ABC$ is given by $$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}.$$ One can get the area of $\triangle{ACD}$ in the same way as above.

Added : Since $$\vec{AB}=(6-5,-4-3)=(1,-7)$$$$\vec{AC}=(-3-5,-2-3)=(-8-5)$$$$\vec{AB}\cdot \vec{AC}=1\cdot(-8)+(-7)\cdot (-5)=27,$$one has $$\frac{1}{2}\sqrt{\left|\vec{AB}\right|^2\left|\vec{AC}\right|^2-\left(\vec{AB}\cdot \vec{AC}\right)^2}=\frac{1}{2}\sqrt{(1+(-7)^2)\cdot((-8)^2+(-5)^2)-27^2}.$$

$\endgroup$
  • $\begingroup$ How can i get the value of AC? AB=7, BC=9, CD=9,DA=7, AC=? $\endgroup$ – Rubab Apr 30 '15 at 14:19
  • $\begingroup$ @Rubab: What does $AB=7$ mean? If you mean the distance between $A$ and $B$, then $AB=5\sqrt 2$, not $7$. $\endgroup$ – mathlove Apr 30 '15 at 22:08
  • $\begingroup$ After I drew the shape, I saw that AB has |3|+|-4| as the length. How did you get the $5\sqrt2$? $\endgroup$ – Rubab May 1 '15 at 3:53
  • $\begingroup$ @Rubab: $AB=\sqrt{(6-5)^2+(-4-3)^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt 2$. By the way, my answer is about vectors. Do you know vectors? $\endgroup$ – mathlove May 1 '15 at 9:49
2
$\begingroup$

There's a neat method we can use in this case called the shoelace formula (which is derived from just adding triangles together, and is applicable in the case where we just have list of the coordinate points).

Application of this formula gives us that the area is

$$\begin{align}A&=\frac{1}{2}\left|x_1y_2+x_2y_3+x_3y_4+x_4y_1-x_2y_1-x_3y_2-x_4y_3-x_1y_4\right| \\&=\frac{1}{2}\left|5\cdot(-4)+6\cdot(-2)+(-3)\cdot7+(-4)\cdot3-6\cdot3-(-3)\cdot(-4)-(-4)\cdot(-2)-5\cdot7\right| \\&=\frac{1}{2}\left|-20-12-21-12-18-12-8-35\right| \\&=\frac{1}{2}\left|-138\right| \\&=69 \end{align}$$

$\endgroup$
  • $\begingroup$ Thanks a lot for telling me about this method. I knew this method but I thought it only worked for triangles, not quadrilaterals. $\endgroup$ – Rubab Apr 30 '15 at 14:21
2
$\begingroup$

From the diagram:

$$11\times 10-\frac12 \times 9 \times 4 - 4 \times 1 - \frac12 \times 1 \times 7 -\frac12 \times 9 \times 2 - 2 \times 1 - \frac12 \times 1 \times 9$$ $$= 110-18-4 -3.5-9-2-4.5 =69$$

enter image description here

$\endgroup$
  • $\begingroup$ Thanks a lot for this solution. :) 1 Large Rectangle - 4 Triangle-2 Small Rectangle $\endgroup$ – Rubab May 1 '15 at 3:39
0
$\begingroup$

You want a really easy way to do this?? Check this out!

Given the four vertices: (5,3),(6,−4),(−3,−2),(−4,7), find the Area:

(1) Determine the diagonal "vectors" (by subtracting the opposite vertices):

d1 = (5, 3)  - (-3,-2)  =   (8, 5)

d2 = (6,-4) -  (-4, 7)  =   (10,-11)

(2) The area equals 1/2 of the absolute value of the "cross-product" of the diagonals:

X = 1/2 * | (8 * -11) - (10 * 5) | = 1/2 * 138 = 69.

This works for any irregular quadrilateral.

$\endgroup$
0
$\begingroup$

Use the formula
Area of Triangle = 1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2)
Join the diagonal (any one) then find the area of two triangles formed by above formula

$\endgroup$
  • $\begingroup$ Please use mathjax $\endgroup$ – Shailesh Mar 1 '18 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.