Step-by-step explanation:
the sum of all angles in any quadrilateral (object with 4 sides) is 360°.
so, as the sum of the base angles is 140°, the sum of the 2 top angles must be
360 - 140 = 220°.
since it is an isoceles trapezoid, both base angles must be the same, and also both of the top angles must be the same.
a single base angle is therefore 140/2 = 70°.
a single top angle is 220/2 = 110°.
consider the height of the trapezoid is cutting off 90° there, so that leaves 110-90 = 20° for the inner angle of the extending triangle (see also further below).
the shorter base must be equal to the longer base minus the 2 extensions to the side under the non-parallel sides.
these extensions to the side are right-angled triangles.
7 ft is the Hypotenuse (baseline opposite of the 90° angle), the baseline extension is one leg, and the height of the trapezoid is the second leg.
the law of sine says
a/sinA = b/sinB = c/sinC
with a, b, c being the sides of a triangle, and A, B, C are the corresponding opposite angles.
for our triangle here :
Hypotenuse/sin(90°) = base extension leg/sin(20°)
sin(90°) = 1
so,
7/1 = base extension leg/sin(20°)
base extension leg = 7×sin(20°) = 2.394141003... ft
that means the
short base = 22 - 2×2.394141003... = 17.21171799 ≈ 17.21 ft
sorry, it got a bit longer, but since I don't know what you don't know I had to explain the background of every step.
Answer:
17.21 ft
Step-by-step explanation:
An isosceles trapezoid is a quadrilateral with one pair of parallel sides (bases), and non-parallel sides (legs) that are equal in length.
In the given isosceles trapezoid (see attached diagram), the sum of the base angles adjacent to the base measuring 22 feet is 140°. Since the sum of the interior angles of any quadrilateral is 360°, the sum of the base angles adjacent to the shorter base must be 360° - 140° = 220°. The base angles of an isosceles trapezoid are congruent, so each of the base angles adjacent to the shorter base measures 220° / 2 = 110°.
To find the length of the shorter base (AB), we can use triangle ABC formed by the diagonal AC and the non-parallel sides AB and BC. In this triangle, diagonal AC is opposite angle B, and sides AB and BC include angle B, so the easiest way to find the length of AB is by using the Law of Cosines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Cosines}}\\\\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:
[tex]AC^2=AB^2+BC^2-2(AB)(BC)\cos B[/tex]
Substitute AC = 20.68, BC = 7, and B = 110° into the equation, and rearrange in the form a quadratic equation:
[tex]20.68^2=AB^2+7^2-2(AB)(7)\cos 110^{\circ} \\\\427.6624=AB^2+49-(AB)14\cos 110^{\circ} \\\\AB^2-(AB)14\cos 110^{\circ}-378.6624=0[/tex]
To solve the quadratic equation, we can use the quadratic formula:
[tex]\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}[/tex]
In this case:
Therefore:
[tex]AB=\dfrac{-(-14\cos 110^{\circ}) \pm \sqrt{(-14\cos 110^{\circ})^2-4(1)(-378.6624)}}{2(1)} \\\\\\ AB=\dfrac{14\cos 110^{\circ} \pm \sqrt{(-14\cos 110^{\circ})^2+1514.6496}}{2} \\\\\\ AB=17.21183541065..., AB=-22.0001174172...[/tex]
As length is positive, we take the positive solution. Therefore:
[tex]AB=17.21183541065...\\\\AB=17.21\; \sf ft\;(nearest\;hundredth)[/tex]
So, the length of the shorter base is:
[tex]\Large\boxed{\boxed{17.21\; \sf ft}}[/tex]
