Question

Please need help with the second question and if not hard also the third ^) WILL GIVE BRAINLEST

Answer 1

$▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪$

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 :

• $\sin(60) = \dfrac{x}{20}$

• $\dfrac{ \sqrt{3} }{2} = \dfrac{x}{20}$

• $x = \dfrac{20 \times \sqrt{3} }{ {2} }$

• $x = 10 \sqrt{3} \: \: units$

similarly,

• $\cos(60) = \dfrac{y}{20}$

• $\dfrac{1}{2} = \dfrac{y}{20}$

• $y = \dfrac{20}{2}$

• $y = 10 \: \: units$

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

• $\dfrac{1}{2} \times base \times height$

• $\dfrac{1}{2} \times x \times y$

• $\dfrac{1}{2} \times 10 \sqrt{3} \times 10$

• $50 \sqrt{3} \: \: unit {}^{2}$

3. Area of Rectangle PQRS :

• $length \times width$

• $x \times y$

• $10 \times 10 \sqrt{ 3}$

• $100 \sqrt{3} \: \: unit {}^{2}$