1. Apr 4, 2016

### zak100

• Originally posted in a non-homework forum section
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.

Zulfi.

2. Apr 4, 2016

### ProfuselyQuarky

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

And is triangle DBC a right triangle?

Last edited: Apr 4, 2016
3. Apr 4, 2016

### ProfuselyQuarky

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

4. Apr 4, 2016

### 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 $6$) ...

5. Apr 4, 2016

### ProfuselyQuarky

distance formula :)

6. Apr 4, 2016

### 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 hw 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 cant 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.

7. Apr 4, 2016

### SammyS

Staff Emeritus
How do you know that ΔDBC is a right triangle ?

8. Apr 4, 2016

### 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 cant we take sides DB & BC??

Zulfi.

9. Apr 4, 2016

### SammyS

Staff Emeritus
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.

Last edited: Apr 4, 2016
10. Apr 4, 2016

### Staff: Mentor

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

11. Apr 4, 2016

### ProfuselyQuarky

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.

12. Apr 4, 2016

### SammyS

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

Heron's theorem is not needed here.

13. Apr 4, 2016

### ProfuselyQuarky

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

14. Apr 5, 2016

### 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:

15. Apr 5, 2016

### LCKurtz

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

16. Apr 6, 2016

### Samy_A

Geogebra

17. Apr 10, 2016

### micromass

Staff Emeritus