# Area of a region

• Oct 8th 2010, 05:53 AM
Punch
Area of a region
The diagram shows a logo $\displaystyle ABCD$ in the shape of a trapezium. $\displaystyle AB=13x cm, CD=12x cm, AD=2y cm$ and Angle $\displaystyle BCD = 90^o$. The shaded region PQDR is a parallelogram such that $\displaystyle RD=\frac{1}{3}AD$.

Before reaching the part i have problems with,

$\displaystyle BC=2y+5x$ and area of $\displaystyle ABCD=\frac{1}{2}(4y+5x)(6x)$

The question is Find the area of PQDR
http://i952.photobucket.com/albums/a...g/PA050041.jpg
• Oct 8th 2010, 07:59 AM
earboth
Quote:

Originally Posted by Punch
[FONT=Arial]The diagram shows a logo $\displaystyle ABCD$ in the shape of a trapezium. $\displaystyle AB=13x cm, CD=12x cm, AD=2y cm$ and Angle $\displaystyle BCD = 90^o$. The shaded region PQDR is a parallelogram such that $\displaystyle AD=\frac{1}{3}AD$.<=== typo? RD = ...
Before reaching the part i have problems with,

$\displaystyle BC=2y+5x$ and area of $\displaystyle ABCD=\frac{1}{2}(4y+5x)(6x)$

The question is Find the area of PQDR
http://i952.photobucket.com/albums/a...g/PA050041.jpg

1. The right triangle at the left with the hypotenuse $\displaystyle \overline{AB}$ has one leg with the length 12x. Use Pythagorean theorem to determine the length of the missing leg (<--- no pun intended!)

2. The area of a trapezium is calculated by:

$\displaystyle a = \dfrac{\overline{AD} + \overline{BC}}2 \cdot \overline{DC}$

with $\displaystyle \overline{AD} = 2y$

$\displaystyle \overline{BC} = 5x + 2y$

$\displaystyle \overline{DC} = 12x$
• Oct 8th 2010, 05:02 PM
Punch
Indeed that was a typo. However, the question hasn't been answered!

The question is Find the area of PQDR
• Oct 9th 2010, 01:39 AM
earboth
Quote:

Originally Posted by Punch
Indeed that was a typo. However, the question hasn't been answered!

The question is Find the area of PQDR

1. PQDR is a parallelogram.

2. To determine the area of a parallelogram you need the length of the base and the length of the height. All values are known to you.

3. So, where are your difficulties?
• Oct 9th 2010, 01:49 AM
Punch
Quote:

Originally Posted by earboth
1. PQDR is a parallelogram.

2. To determine the area of a parallelogram you need the length of the base and the length of the height. All values are known to you.

3. So, where are your difficulties?

I dont have the value of side PR or QD, note that it is not parallel to side AB
• Oct 9th 2010, 10:53 PM
earboth
Quote:

Originally Posted by Punch
I dont have the value of side PR or QD, note that it is not parallel to side AB

1. The area of a parallelogram is calculated by:

$\displaystyle area = base \cdot height$

2. Take $\displaystyle \overline{PQ} = \dfrac23 y$ as the base then the corresponding height is $\displaystyle \overline{DC} = 12x$

3. Thus the area of the parallelogram is:

$\displaystyle a = \dfrac23 y \cdot 12x = 8xy$
• Oct 10th 2010, 12:50 AM
Punch
Quote:

Originally Posted by earboth
1. The area of a parallelogram is calculated by:

$\displaystyle area = base \cdot height$

2. Take $\displaystyle \overline{PQ} = \dfrac23 y$ as the base then the corresponding height is $\displaystyle \overline{DC} = 12x$

3. Thus the area of the parallelogram is:

$\displaystyle a = \dfrac23 y \cdot 12x = 8xy$

Oh my... Sorry!! I was mistaken that the height was referring to RQ

Thanks