Question

help me faster please​

Answer 1

The answer to your questions are given below:

1. The length of diameter is 13cm.
2. The area of semicircle is 66.39$cm^{2}$.
3. The area of  ΔPQR is 30$cm^{2}$.

Step-by-step explanation:

We know that, if a right triangle is inscribed in a circle, then its hypotenuse is a diameter of the circle.

⇒ ΔPQR is a right angled triangle right angled at P.

1. As ΔPQR is right angled at P, QR is the hypotenuse. QR is the diameter    too.

    By Pythagoras theorem,

    $h^{2}$ = $a^{2}$ + $b^{2}$

⇒ $QR^{2}$ = $PR^{2}$ + $PQ^{2}$

   $QR^{2}$ = $5^{2}$ + $12^{2}$

   $QR^{2}$ = 25 + 144

   $QR^{2}$ = 169

   QR = $\sqrt{169}$

   QR = 13cm

The length of Diameter is 13cm.

2.  Diameter of the Circle = 13cm

   Radius of the Circle =  $\frac{Diameter}{2}$

                                     = $\frac{13}{2}$

                                     = 6.5cm

   

   Area of Semicircle = Area of Circle/2

                                   = π$r^{2}$ /2

                                   = 3.14 x 6.5 x 6.5 / 2

                                   = 66.39$cm^{2}$.

The area of semicircle is 66.39$cm^{2}$.

3. Area of ΔPQR =  $\frac{1}{2}$ × b × h

                           = $\frac{1}{2}$ × 12 × 5

                           = $\frac{1}{2}$ × 60

                           = 30$cm^{2}$.

The area of ΔPQR is 30$cm^{2}$.

Sorry Mate, it took me more than 20 minutes to answer your question.