Question

1. Find the height of a parallelogram with 10 sides and 20 inches long, and an included angle of 35 degrees. Also, calculate the area of the figure.
2. A certain city block is in the form of a parallelogram. Two of its sides measures 32 ft and 41 ft. If the area of the land in the block is 656 ft^2, what is the length of its longer diagonal?
3. The area of an isosceles trapezoid is 246 m^2. If the height and the length of one of its congruent sides measures 6m and 10m, respectively, find the length of the two bases.

Answer 1

Answer on Question #57001 – Math – Geometry

Question

Find the height of a parallelogram with 10 sides and 20 inches long, and an included angle of 35 degrees. Also, calculate the area of the figure.

Solution

From the rectangle triangle we can find $h$:

$$\sin(35{}^\circ) = \frac{h}{AB}$$ $$h = AB \cdot \sin(35{}^\circ) = 10 \sin(35{}^\circ)$$

Now, the area of parallelogram is given by

$$S = AD * h = 20 * 10 \sin(35{}^\circ) = 200 * \sin(35{}^\circ) \approx 114.72$$

**Answer**: $10 \sin(35{}^\circ)$ in, $114.72 \text{ in}^2$

Question

A certain city block is in the form of a parallelogram. Two of its sides measures 32 ft and 41 ft. If the area of the land in the block is 656 ft^2, what is the length of its longer diagonal?

Solution

We know the formula of area of parallelogram:

$$S = A B * A D * \sin (\alpha),$$

hence

$$\sin (\alpha) = \frac {S}{A B * A D} = \frac {6 5 6}{3 2 * 4 1} = 0. 5,$$

therefore

$$\alpha = 3 0 {}^ {\circ}$$

Using the cosine rule a larger diagonal is equal to

$$d = \sqrt {A B ^ {2} + B C ^ {2} - 2 \cdot A B \cdot B C \cdot \cos (1 8 0 {}^ {\circ} - \alpha)} = =$$ $$\sqrt {A B ^ {2} + A D ^ {2} + 2 * A B * A D * \cos \alpha} = \sqrt {1 0 2 4 + 1 6 8 1 + 1 3 1 2 * 0 . 8 6 6} \approx 6 2,$$

$BC$ and $AD$ are congruent, $\cos (180{}^{\circ} - \alpha) = -\cos (\alpha)$ .

Answer: 62 ft.

Question

The area of an isosceles trapezoid is $246\mathrm{m}^2$. If the height and the length of one of its congruent sides measures $6\mathrm{m}$ and $10\mathrm{m}$, respectively, find the length of the two bases.

Solution

We can find $x$ using the Pythagorean Theorem from the rectangle triangle:

$$10^2 = x^2 + 6^2$$ $$x = \sqrt{100 - 36} = 8$$

Area of trapezoid is given by

$$S = \frac{1}{2}(BC + AD)h = \frac{1}{2}(2x + 2y)h$$ $$S = (x + y) \cdot 6 = 246,$$

hence

$$x + y = 41$$ $$y = 41 - x = 33$$

So

$$BC = y = 33, \quad AD = y + 2x = 33 + 2 \cdot 8 = 49$$

Answer: $33\mathrm{m}$, $49\mathrm{m}$.

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