Respuesta :

DᴀʀᴋPᴀʀᴀᴅᴏx

[tex]▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ [/tex]

Here's the solution :

1. The given figure is of a Rectangle, and we know that diagnonals of a Rectangle bisects each other.

so, PR = 2 × PT

that is :

  • PR = 2 × 10 = 20 units

and both the diagnonals of a Rectangle are equal,

  • PR = QS = 20 units

2. let's take the length of PQ = x and QR = y

Angle RPS = Angle PRQ = 60°

(because they form pair of Alternate Interior angles, which are equal)

By using trigonometry :

  • [tex] \sin(60) = \dfrac{x}{20} [/tex]

  • [tex] \dfrac{ \sqrt{3} }{2} = \dfrac{x}{20} [/tex]

  • [tex]x = \dfrac{20 \times \sqrt{3} }{ {2} } [/tex]

  • [tex]x = 10 \sqrt{3} \: \: units[/tex]

similarly,

  • [tex] \cos(60) = \dfrac{y}{20} [/tex]

  • [tex] \dfrac{1}{2} = \dfrac{y}{20} [/tex]

  • [tex]y = \dfrac{20}{2} [/tex]

  • [tex]y = 10 \: \: units[/tex]

And now, we know that area of triangle PQR is equal to :

  • [tex] \dfrac{1}{2} \times base \times height[/tex]

  • [tex] \dfrac{1}{2} \times x \times y[/tex]

  • [tex] \dfrac{1}{2} \times 10 \sqrt{3} \times 10[/tex]

  • [tex]50 \sqrt{3} \: \: unit {}^{2}[/tex]

3. Area of Rectangle PQRS :

  • [tex]length \times width[/tex]

  • [tex]x \times y[/tex]

  • [tex]10 \times 10 \sqrt{ 3} [/tex]

  • [tex]100 \sqrt{3} \: \: unit {}^{2} [/tex]