Answer :
Answer:
13.4m
Step-by-step explanation:
Please find the attached.
Answer:
13.4 m
Step-by-step explanation:
A parallelogram is a quadrilateral with opposite sides that are parallel and equal in length.
The diagonals of a parallelogram bisect each other and form two pairs of congruent triangles. Each triangle has side lengths that are half the length of each diagonal. Given that the diagonals of the parallelogram are 26.5 meters and 39.4 meters, the corresponding side lengths of each triangle are 13.25 meters and 19.7 meters.
When the diagonals bisect each other, they form two pairs of vertically opposite angles at their intersection. These angles are the apex angles of each pair of congruent triangles. Given that the diagonals intersect at an angle of 137.5°, the apex angle for one pair of congruent triangles is 137.5°, while the apex angle for the other pair is 180° - 137.5° = 42.5°.
The unknown sides of the pair of congruent triangles with an apex angle of 137.5° will be the longer sides of the parallelogram. The unknown sides of the pair of congruent triangles with an apex angle of 42.5° will be the shorter sides of the parallelogram.
Since we have the lengths of two sides and the included angle, we can use the Law of Cosines to find the length of the side opposite the angle.
[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:
- a = 13.25 m
- b = 19.7 m
- c = shorter side length of parallelogram
- C = 42.5°
Substitute the values into the equation and solve for c.
[tex]c^2=13.25^2+19.7^2-2(13.25)(19.7)\cos 42.5^{\circ} \\\\ c^2=175.5625+388.09-522.05\cos 42.5^{\circ} \\\\ c^2=563.6525-522.05\cos 42.5^{\circ} \\\\ c=\sqrt{563.6525-522.05\cos 42.5^{\circ}} \\\\ c=13.369998740... \\\\c=13.4\; \sf (1\;d.p.)[/tex]
Therefore, the length of the shorter side of the parallelogram rounded to one decimal place is:
[tex]\LARGE\boxed{\boxed{\sf 13.4\; m}}[/tex]