# Please Help me with this..(Areas of Plane Figures)

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• Feb 14th 2008, 02:02 AM
sarayork
Please Help me with this..(Areas of Plane Figures)
Find the area of each figures;
1. a rhombus with shorter diagonal 6cm and a 120° angle.

2. A regular hexagon with side 12.

3. A regular pentagon with side S and apothem 3.

4. The ratio of the areas of two rectangles is 36 : 64. Find the scale factor and the ratio of the perimeters.

5. Two similar polygons have scale factor 3 : 4, the area of the smaller polygon is 108. Find the area of the larger polygon.

*Please explain a brief statement of how to find the answers... ㅠㅠ...
Thank- U so much...
• Feb 14th 2008, 04:29 AM
earboth
Quote:

Originally Posted by sarayork
Find the area of each figures;
1. a rhombus with shorter diagonal 6cm and a 120° angle.

2. A regular hexagon with side 12.

...

to #1:

The shorter diagonal bisect the greater angle. Therefore you are dealing with 4 right triangles with the shorter leg of 3 cm and the longer leg is

$\displaystyle l = 3\ cm \cdot \tan(60^\circ) = 3 \sqrt{3} \ cm$.

The area is then $\displaystyle A = 4 \cdot \underbrace{(3 \cdot 3\sqrt{3})}_{\text{area of one triangle}}\ cm^2 = 36 \sqrt{3}\ cm^2$

to #2:

A regular hexagon consists of 6 equilateral triangles. If the length of one side of the triangle is a then the height of the triangle is: $\displaystyle h = \frac12 \cdot a \cdot \sqrt{3}$

Therefore the area of an equilateral triangle is:

$\displaystyle A = \frac12 \cdot a \cdot \underbrace{\frac12 \cdot a \cdot \sqrt{3} }_{height}= \frac{a^2}{4} \sqrt{3}$

And finally the area of a regular hexagon is:

$\displaystyle A_6=6 \cdot \frac{a^2}{4} \sqrt{3} = \frac32 \cdot a^2 \sqrt{3}$. Plug in the value of a (= 12) to calculate the area.
• Feb 14th 2008, 04:42 AM
earboth
Quote:

Originally Posted by sarayork
...

4. The ratio of the areas of two rectangles is 36 : 64. Find the scale factor and the ratio of the perimeters.

5. Two similar polygons have scale factor 3 : 4, the area of the smaller polygon is 108. Find the area of the larger polygon.

...

to #4:

If the lengthes(?) of corresponding sides of similar plane figures have the ratio

$\displaystyle \frac{l_1}{l_2}=\frac ab$ then the ratio of the areas is $\displaystyle \frac{A_1}{A_2}=\frac{a^2}{b^2}$

If the areas of similar plane figures have the ratio

$\displaystyle \frac{A_1}{A_2}=\frac pq$ then the ratio of the lengthes(?) of corresponding sides is $\displaystyle \frac{l_1}{l_2}=\frac{\sqrt{p}}{\sqrt{q}}$

Because $\displaystyle \frac{A_1}{A_2} = \frac{36}{64}$ the ration of the perimeters is: $\displaystyle \frac{p_1}{p_2} = \frac{6}{8}$

to #5:

Calculate first the ratio of the areas and then the actual value of the greater area.

(For confirmation only: A_2 = 192)

EDIT: I would have helped you with #3 but I couldn't find the word "apothem" in my dictionaries. Is it maybe the radius of the inscribed circle of the pentagon?
• Feb 14th 2008, 05:00 AM
topsquark
Quote:

Originally Posted by earboth
EDIT: I would have helped you with #3 but I couldn't find the word "apothem" in my dictionaries. Is it maybe the radius of the inscribed circle of the pentagon?

I don't have time to answer questions right now, but see here: apothem.

-Dan
• Feb 14th 2008, 05:18 AM
earboth
Quote:

Originally Posted by sarayork
...
3. A regular pentagon with side S and apothem 3.

...

A regular 5-gon consists of 5 isosceles triangles with the base s.
The apothem is the height h perpendicular to the base s.
The height bisect the angle at the top and the base s.
This angle at the top is $\displaystyle \tau = \frac{360^\circ}{5} = 72^\circ$.

To calculate the length of the base you use half of the isoscele triangle which is a right triangle:

$\displaystyle \tan(36^\circ) = \frac{\frac12 s}{h}~\implies~ s = 2h\cdot \tan(36^\circ) \approx 4.359...$
The area of one triangle is calculated by:

$\displaystyle A_{triangle} = \frac12 \cdot s \cdot h= \frac12 \cdot 2h\cdot \tan(36^\circ) \cdot h$

And therefore the area of the 5-gon is:

$\displaystyle A_5 = 5h^2 \cdot \tan(36^\circ)$ which means here: $\displaystyle A_5=45 \cdot \tan(36^\circ) \approx 32.6944$