Video: Finding All Possible Values for a Radius given the Area of a Circular Sector and the Perimeter

The area of a circular sector is 12 cmΒ² and the perimeter is 16 cm. Find all possible values for the radius.

04:16

Video Transcript

The area of a circular sector is 12 centimetres squared and the perimeter is 16 centimetres. Find all possible values for the radius.

Let’s begin by recalling what we know about sector area and arc length when working with a sector with an angle measured in radians. For a sector with a radius π‘Ÿ and angle πœƒ radians, the area is a half π‘Ÿ squared πœƒ. And the arc length, that’s the curve length, is π‘Ÿ multiplied by πœƒ.

If we add the remaining measurements to this diagram, that’s the radius π‘Ÿ and then the curve length, the arc length π‘Ÿπœƒ, we can form an expression for the perimeter. It’s π‘Ÿ plus π‘Ÿ plus π‘Ÿπœƒ. Remember, the perimeters just a distance around the shape. Collecting like terms, that simplifies to two π‘Ÿ plus π‘Ÿπœƒ. This is equal to 16. So we form an equation by letting this expression be equal to 16. And we have 16 is equal to two π‘Ÿ plus π‘Ÿπœƒ.

Now let’s set up an equation for the area. The area is a half π‘Ÿ squared πœƒ. And we’re told this is equal to 12. So we have 12 is equal to a half π‘Ÿ squared πœƒ. We can simplify this expression somewhat by multiplying both sides of the equation by two. And doing so, we see that 24 is equal to π‘Ÿ squared πœƒ.

One way of doing this is to rearrange the second equation. We now have a pair of simultaneous equations with two unknown variables, π‘Ÿ and πœƒ. We need to find a way to eliminate one of these variables. We can divide both sides by π‘Ÿ squared. And we can see that πœƒ is equal to 24 over π‘Ÿ squared.

We now have an equation for πœƒ in terms of π‘Ÿ squared. We can substitute that into the first simultaneous equation, the equation we formed for the perimeter. Substituting 24 over π‘Ÿ squared for πœƒ in equation one gives us 16 is equal to two π‘Ÿ plus π‘Ÿ multiplied by 24 over π‘Ÿ squared. We can simplify the second term by spotting that factor of π‘Ÿ. And that gives us 16 is equal to two π‘Ÿ plus 24 over π‘Ÿ. Next, we’re going to multiply through by π‘Ÿ. That tells us that 16π‘Ÿ is equal to two π‘Ÿ squared plus 24. Notice how each of these terms has a factor of two. Before we do anything else, we’re going to divide through by two. And we get eight π‘Ÿ equals π‘Ÿ squared plus 12. This is a quadratic equation.

To solve a quadratic equation, we need to rearrange so that it’s equal to zero. We subtract eight π‘Ÿ from both sides. And that leaves us with π‘Ÿ squared minus eight π‘Ÿ plus 12 is equal to zero. To solve this quadratic equation, we need to factorise. There are no common factors between these three terms. So we know this factorises into two brackets with π‘Ÿ at the front of each bracket.

Next, we need to find the factors of positive 12 that add to make negative eight. Those are negative six and negative two since negative six multiplied by negative two is positive 12. But the sum of them is negative eight. We spot then that the product of these two brackets is zero. And the only way for this to be the case is if either of the brackets is equal to zero. π‘Ÿ minus six is equal to zero. Or π‘Ÿ minus two is equal to zero.

We solve this first equation by adding six to both sides. And we can see that one of the possible values for our radius is six. And we solve the second equation by adding two to both sides. That tells us that π‘Ÿ is equal to two. The radius could be six centimetres or two centimetres.

Notice at this stage, we haven’t been required to find the value of πœƒ. But we could work this out by substituting both six and two back into either the original equations. Doing so, we would’ve seen that, for a radius of six, the angle πœƒ would be two-thirds. And for a radius of two, the angle πœƒ would be six radians.

The radius we were after though was either six centimetres or two centimetres.

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