Question

please please answer this​

Answer 1

Step-by-step explanation:

gave any further questions

Answer 2

Question:

Find the area of the shaded region in figure, if PQ = 12 cm, PR = 5 cm and O is the centre of the circle.

Answer:

Area of the shaded region is 36.39 cm². (approx.)

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure, O is the centre of circle.

$\sf\triangle\:PQR$ is inscribed in the circle.

$\sf\:In\:\triangle\:PQR\:,\\\\\\\bullet\sf\:PQ\:=\:12\\\\\\\bullet\sf\:PR\:=\:5\:cm\\\\\\\bullet\sf\:\angle\:RPQ\:=\:90^{\circ}\\\\\\\sf\:RQ^2\:=\:PQ^2\:+\:PR^2\:\:\:-\:-\:[\:Pythagors\:theorem\:]\\\\\\\implies\sf\:RQ^2\:=\:(\:12\:)^2\:+\:(\:5\:)^2\\\\\\\implies\sf\:RQ^2\:=\:144\:+\:25\\\\\\\implies\sf\:RQ^2\:=\:169\\\\\\\implies\sf\:RQ\:=\:\sqrt{169}\:\:\:-\:-\:[\:Taking\:square\:roots\:]\\\\\\\implies\boxed{\red{\sf\:RQ\:=\:13\:cm}}$

We have to find the area of the shaded region.

RQ is the diameter of the circle.

From figure, we know that,

$\pink{\sf\:Area\:of\:shaded\:region\:=\:Area\:of\:semicircle\:-\:A\:(\:\triangle\:PQR\:)}\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{\:\pi\:r^2}{2}\:-\:\dfrac{1}{2}\:\times\:PQ\:\times\:PR\\\\\\\implies\sf\:Area\:\:of\:\:shaded\:\:region\:\:=\:\:\dfrac{\dfrac{22}{7}\:\times\:\dfrac{13}{2}\:\times\:\dfrac{13}{2}}{2}\:-\:\dfrac{1}{2}\:\times\:12\:\times\:5\:\:\:-\:-\:[\:\:Radius\:\:is\:\:half\:\:the\:\:diameter\:\:]\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{22}{7}\:\times\:\dfrac{13}{2}\:\times\:\dfrac{13}{2}\:\times\:\dfrac{1}{2}\:-\:\dfrac{1}{2}\:\times\:12\:\times\:5\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{1}{2}\:\bigg[\:\dfrac{\cancel{22}}{7}\:\times\:\dfrac{13}{\cancel2}\:\times\:\dfrac{13}{2}\:-\:60\:\bigg]\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{1}{2}\:\bigg[\:\dfrac{11\:\times\:169}{14}\:-\:60\:\bigg]\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{1}{2}\:\bigg[\:\dfrac{1859}{14}\:-\:60\:\bigg]\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{1}{2}\:\bigg[\:\dfrac{1859\:-\:840}{14}\:\bigg]\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{1}{2}\:\times\:\dfrac{1019}{14}\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:\dfrac{1019}{28}\\\\\\\implies\sf\:Area\:of\:shaded\:region\:=\:36.392\\\\\\\implies\boxed{\red{\sf\:Area\:of\:shaded\:region\:\approx\:36.39\:cm^2}}$

Area of the shaded region is 36.39 cm² (approx.).