Can someone please help me with this problem?? **It's high-school geometry.

Hello!
Answer:
[tex]\huge\boxed{59.04 units}[/tex]
To solve, we will need to use Right-Triangle Trigonometry:
Begin by solving for angles ∠S and ∠R using tangent (tan = opp/adj)
tan ∠S = a / (1/2b)
tan ∠S = 3√5 / 14
tan ∠S ≈ 0.479
arctan 0.479 = m∠S (inverse)
m∠S and m∠R ≈ 25.6°
Use cosine to solve for the hypotenuse, or the missing side-length:
cos ∠S = 14 / x
x · cos (25.6) = 14
x = 14 / cos(25.6)
x ≈ 15.52
Both triangles are congruent, so we can go ahead and find the perimeter of the figure:
RS + RQ + QS = 28 + 15.52 + 15.52 = 59.04 units.
Hope this helped you! :)
Answer:
[tex]\large \boxed{\mathrm{59.05 \ units}}[/tex]
Step-by-step explanation:
Take one small triangle, solve for hypotenuse.
[tex]\frac{b}{2} =\frac{28}{2} =14[/tex]
Use Pythagorean theorem.
[tex]c=\sqrt{(3\sqrt{5})^2 +14^2 }[/tex]
[tex]c= 15.524175...[/tex]
Add the hypotenuse twice because there are two triangles, then add to the length of b to find the perimeter.
[tex]15.524175...+15.524175...+28[/tex]
[tex]59.048349...[/tex]