# Thread: Area of a region

1. ## 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

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

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

$\displaystyle \overline{DC} = 12x$

3. Indeed that was a typo. However, the question hasn't been answered!

The question is Find the area of PQDR

4. 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?

5. 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

6. 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:

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

7. 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