Area of a region

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October 8th 2010, 05:53 AM
Punch

Area of a region

The diagram shows a logo $ABCD$ in the shape of a trapezium. $AB=13x cm, CD=12x cm, AD=2y cm$ and Angle $BCD = 90^o$ . The shaded region PQDR is a parallelogram such that $RD=\frac{1}{3}AD$ .

Before reaching the part i have problems with,

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

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

Quote:

Originally Posted by Punch

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

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

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 $\overline{AB}$ 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:

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

with $\overline{AD} = 2y$

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

$\overline{DC} = 12x$
October 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
October 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?
October 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
October 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:

$area = base \cdot height$

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

3. Thus the area of the parallelogram is:

$a = \dfrac23 y \cdot 12x = 8xy$
October 10th 2010, 12:50 AM
Punch

Quote:

Originally Posted by earboth

1. The area of a parallelogram is calculated by:

$area = base \cdot height$

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

3. Thus the area of the parallelogram is:

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

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

Thanks