Answer:
Let's solve each part of the problem:
a) To find how far apart the 18-inch sides are, we use the formula for the area of a parallelogram: area = base × height. Rearranging the formula to solve for height, we get height = area/base. Therefore, the height of the parallelogram corresponding to the 18-inch side is [tex]\( \frac{144 \text{ square inches}}{18 \text{ inches}} = 8 \text{ inches} \).[/tex]
b) Similarly, to find how far apart the 10-inch sides are, we divide the area by the length of the 10-inch side: [tex]\( \frac{144 \text{ square inches}}{10 \text{ inches}} = 14.4 \text{ inches} \).[/tex]
c) Since opposite sides of a parallelogram are equal in length and parallel, the angles opposite these sides are also equal. Therefore, we can use the properties of parallelograms to determine that the angles are supplementary to each other. We can use the cosine rule to find one of the angles. Let's denote the length of the 10-inch side as a, the length of the 18-inch side as b, and the length between the 18-inch sides as h. The cosine of the angle [tex]\( \theta \)[/tex] between the 10-inch and 18-inch sides can be calculated using the formula:
[tex]\[ \cos(\theta) = \frac{a^2 + b^2 - h^2}{2ab} \][/tex]
Plugging in the values, we get:
[tex]\[ \cos(\theta) = \frac{10^2 + 18^2 - 14.4^2}{2 \times 10 \times 18} \]\[ \cos(\theta) = \frac{100 + 324 - 207.36}{360} \]\[ \cos(\theta) = \frac{216.64}{360} \]\[ \cos(\theta) \approx 0.601 \][/tex]
Using the inverse cosine function, we find that \( \theta \) is approximately 53.13 degrees. Since opposite angles of a parallelogram are equal, the other angle is also approximately 53.13 degrees.
d) The diagonals of a parallelogram bisect each other, so each diagonal is split into two equal parts. We can use the Pythagorean theorem to find the length of the diagonals. Let's denote the length of the diagonals as \( d \). Using one of the right triangles formed by the diagonal and two sides of the parallelogram, we have:
[tex]\[ d^2 = (10 \text{ inches})^2 + (8 \text{ inches})^2 \]\[ d^2 = 100 + 64 \]\[ d^2 = 164 \][/tex]
So, the length of the diagonal is approximately [tex]\( \sqrt{164} \)\frac{x}{y}[/tex] inches.
Answer:
[tex]\textsf{a)}\quad 8 \;\textsf{inches}[/tex]
[tex]\textsf{b)}\quad 18\; \textsf{inches}[/tex]
[tex]\textsf{c)}\quad 53.13^{\circ}\;\textsf{and}\;126.87^{\circ}[/tex]
[tex]\textsf{d)}\quad \textsf{Diagonal 1}=4\sqrt{13}\;\textsf{inches}\approx 14.42\;\textsf{inches}\\\\\phantom{bcw.}\textsf{Diagonal 2}=8\sqrt{10}\;\textsf{inches}\approx 25.30\;\textsf{inches}[/tex]
Step-by-step explanation:
A parallelogram is a quadrilateral with opposite sides that are both equal in length and parallel.
A parallelogram has 10-inch and 18-inch sides and an area of 144 square inches.
[tex]\hrulefill[/tex]
The area of a parallelogram is the product of its base (b) and perpendicular height (h). In this case, the area is 144 inches and the base is the 18-inch side. Therefore:
[tex]\textsf{Area}=b \times h\\\\\\144=18 \times h\\\\\\h=\dfrac{144}{18}\\\\\\h=8\; \rm inches[/tex]
So, the 18-inch sides are 8 inches apart.
[tex]\hrulefill[/tex]
The endpoints of the parallel 10-inch sides are connected by the 18-inch sides. Therefore, the distance between the 10-inch sides is equal to the length of the 18-inch sides. So, the 10-inch sides are 18 inches apart.
[tex]\hrulefill[/tex]
As we have the sides and height of a parallelogram, we can use trigonometry to find the interior angles.
If we draw the height from the top left vertex to the base, we create a right triangle with a height of 8 inches and a hypotenuse of 10 inches. To find the angle between the base and and the 10-inch side, we can use the sine trigonometric ratio.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Sine trigonometric ratio}}\\\\\sf \sin(\theta)=\dfrac{O}{H}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{O is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{H is the hypotenuse (the side opposite the right angle).}\end{array}}[/tex]
In this case, the side opposite the angle is the height of the parallelogram, so h = 8, and the hypotenuse is h = 10. Therefore:
[tex]\sin \theta=\dfrac{8}{10}\\\\\\\theta=\sin^{-1}\left(\dfrac{8}{10}\right)\\\\\\\theta=53.130102354...^{\circ}\\\\\\\theta=53.13^{\circ}\; \sf (2\;d.p.)[/tex]
As consecutive angles in a parallelogram are supplementary, to find the measure of the other angle, we can subtract 53.13° from 180°:
[tex]\alpha=180^{\circ}-53.13^{\circ}\\\\\\\alpha=126.87^{\circ}[/tex]
Therefore, the angles of the parallelogram are 53.13° and 126.87°.
[tex]\hrulefill[/tex]
To find the length of the diagonals, we can use the Cosine Rule.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Cosine Rule}}\\\\c^2=a^2+b^2-2ab \cos C\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides.}\\\phantom{ww}\bullet\;\textsf{$C$ is the angle opposite side $c$.}\end{array}}[/tex]
In this case, C is angle, c is the diagonal, and a and b are the lengths of the adjacent sides.
Substitute angle 53.13°:
[tex]c^2=18^2+10^2-2(18)(10)\cos (53.13...^{\circ})\\\\c^2=324+100-360(0.6)\\\\c^2=424-216\\\\c^2=208\\\\c=\sqrt{208}\\\\c=4\sqrt{13}[/tex]
Substitute angle 126.87°:
[tex]c^2=10^2+18^2-2(10)(18)\cos (126.87...^{\circ})\\\\c^2=100+324-360(-0.6)\\\\c^2=424+216\\\\c^2=640\\\\c=\sqrt{640}\\\\c=8\sqrt{10}[/tex]
Therefore, the exact lengths of the diagonals are [tex]4\sqrt{13}[/tex] and [tex]8\sqrt{10}[/tex].
