How do I find the area of a Quadrilateral?

[zak100]

Hi,
I got a question from a book to find the area of a Quadrilateral. I divided the quadrilateral into two triangles but answer is not correct. Some body please guide me.

I am uploading my work in a attached file.

Zulfi.

 

[ProfuselyQuarky]

For ADB, did you write $A=\frac {1}{2}(6)(4)^2$?

And is triangle DBC a right triangle?

 

[ProfuselyQuarky]

Oh, and this should also be in the homework forum, not general math.

 

[Ssnow]

For ADB is correct. For the second isn't simply $\frac{1}{2}6\times 6$, hint: calculate $BD$ and $DC$ (that are not $6$) ...

 

[ProfuselyQuarky]

For ADB is correct. For the second isn't simply $\frac{1}{2}6\times 6$, hint: calculate $BD$ and $DC$ (that are not $6$) ...

distance formula :)

 

[zak100]

Hi,
Thanks for replies. This is not a hw. Its exam preparation. But i would put my future questions related to book in a homework forum. I think you judge hws with the usage of books?
BD= sqrt(sqr( 5-1) + sqr(1-7))
= sqrt(16 + 36)
=2 *Sqrt(13)
Ans=? (I can't use calculator in exam)
DC = sqrt(sqr(11-5) + sqr(5-1))
= sqrt(36 + 16)
= 2*sqrt(13)

area of DBC = 1/2 (2*sqrt(13)) * ( 2 *sqrt(13))
Ans = 26

Area od ABCD = 26 + 12 =38.

This is the correct answer. Thanks.

Zulfi.

 

[SammyS]

Hi,
Thanks for replies. This is not a hw. Its exam preparation. But i would put my future questions related to book in a homework forum. I think you judge hws with the usage of books?
BD= sqrt(sqr( 5-1) + sqr(1-7))
= sqrt(16 + 36)
=2 *Sqrt(13)
Ans=? (I can't use calculator in exam)
DC = sqrt(sqr(11-5) + sqr(5-1))
= sqrt(36 + 16)
= 2*sqrt(13)

area of DBC = 1/2 (2*sqrt(13)) * ( 2 *sqrt(13))
Ans = 26

Area od ABCD = 26 + 12 =38.

This is the correct answer. Thanks.

Zulfi.

How do you know that ΔDBC is a right triangle ?

 

[zak100]

Hi,
No I am not saying that DBC is a right angle triangle. That's why i think they have asked me to use distance formula. However from fig it looks that ABD is a right angle triangle, so i am not using distance formula here and calculating the area directly using 1/2 alt * base formula.

I also have a question:
for triangle DBC why are we using sides DB & DC? why can't we take sides DB & BC??
Some body please guide me.

Zulfi.

 

[SammyS]

Hi,
No I am not saying that DBC is a right angle triangle. That's why i think they have asked me to use distance formula. However from fig it looks that ABD is a right angle triangle, so i am not using distance formula here and calculating the area directly using 1/2 alt * base formula.

I also have a question:
for triangle DBC why are we using sides DB & DC? why can't we take sides DB & BC??
Some body please guide me.

Zulfi.

What are you using as a formula to calculate the area of a triangle ?

The altitude is the perpendicular distance from the base to the opposite vertex.

 

[Mark44]

Oh, and this should also be in the homework forum, not general math.

Thread moved.

Thanks for replies. This is not a hw. Its exam preparation. But i would put my future questions related to book in a homework forum. I think you judge hws with the usage of books?

This forum section is for homework and coursework, which includes problems found in textbooks.

 

[ProfuselyQuarky]

How do you know that ΔDBC is a right triangle ?

What are you using as a formula to calculate the area of a triangle ?

The altitude is the perpendicular distance from the base to the opposite vertex.

That is why I asked whether DBC was a right triangle or not. If DBC is not a right triangle, then you can not you $A=\frac {1}{2}bh$ to find the area. Instead, you have to use Heron’s theorem which states $A=\sqrt {(s-a)(s-b)(s-c)}$ where $s=\frac {1}{2}(a+b+c)$ and variables $a$, $b$, and $c$ are the sides of the triangle.

 

[SammyS]

That is why I asked whether DBC was a right triangle or not. If DBC is not a right triangle, then you can not you $A=\frac {1}{2}bh$ to find the area. Instead, you have to use Heron’s theorem which states $A=\sqrt {(s-a)(s-b)(s-c)}$ where $s=\frac {1}{2}(a+b+c)$ and variables $a$, $b$, and $c$ are the sides of the triangle.

Are you giving the solution, or are you trying to help OP.

Heron's theorem is not needed here.

 

[ProfuselyQuarky]

Are you giving the solution, or are you trying to help OP.

Heron's theorem is not needed here.

I was just helping the OP overall. Nothing to do with the specific problem

 

[Samy_A]

One easy way to solve this kind of exercises is a variation on the "box method", illustrated here for the area of a triangle.

Add the points E(11,7) and F(11,1) to your graph, and the area of ABCD can be easily computed:

Or decompose your quadrilateral as follows:

 

[LCKurtz]

@Samy_A Nice pictures. What software did you use for them?

 

[Samy_A]

@Samy_A Nice pictures. What software did you use for them?

Geogebra

 

[micromass]

A very nice way to solve this is https://en.wikipedia.org/wiki/Pick's_theorem
Not that the way suggested here is difficult, but Pick's theorem is very neat.