Question

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

Answer 1

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

Explanation:

Take one small triangle, solve for hypotenuse.

`(b)/(2) =(28)/(2) =14`

Use Pythagorean theorem.

`c=\sqrt{(3√(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...`

Answer 2

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! :)