Question

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

Answer 1

Hello!

Answer:

$\huge\boxed{59.04 units}$

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 2

Answer:

$\large \boxed{\mathrm{59.05 \ units}}$

Step-by-step explanation:

Take one small triangle, solve for hypotenuse.

$\frac{b}{2} =\frac{28}{2} =14$

Use Pythagorean theorem.

$c=\sqrt{(3\sqrt{5})^2 +14^2 }$

$c= 15.524175...$

Add the hypotenuse twice because there are two triangles, then add to the length of b to find the perimeter.

$15.524175...+15.524175...+28$

$59.048349...$