Question Video: Finding the Perimeter of a Region That Consists of Half the Circumference of a Circle and the Side of a Square Mathematics

The given figure shows circle 𝑀 inscribed in square 𝐴𝐡𝐢𝐷. The area of the shaded region of the figure is 2 1/3 cmΒ². Using the approximation πœ‹ = 22/7, find the perimeter of the shaded region.

04:51

Video Transcript

The given figure shows circle 𝑀 inscribed in square 𝐴𝐡𝐢𝐷. The area of the shaded region of the figure is two and one-third square centimeters. Using the approximation πœ‹ equals 22 over seven, find the perimeter of the shaded region.

Now, the first thing we notice is that we haven’t been given any measurements at all on the diagram. In fact, the only information we’ve been given is that the area of the shaded region is two and one-third square centimeters. This area will be equivalent to the area of rectangle 𝐴𝐸𝐹𝐷 minus the area of the semicircle.

Let’s see if we can use this to work out some information about the dimensions of either the square or the circle. We’ll begin by letting the radius of our circle be equal to π‘Ÿ. Now, this radius is half the diameter of the circle. And as the diameter of the circle is the same as the side length of the square, then the square’s side length will be equal to two π‘Ÿ. This rectangle, 𝐴𝐷𝐹𝐸, therefore has a width of π‘Ÿ units and has a length of two π‘Ÿ units. Its area, using the formula length multiplied by width for the area of a rectangle, is therefore equal to two π‘Ÿ squared.

The area of the semicircle will be half the area of a full circle of radius π‘Ÿ. So that’s πœ‹π‘Ÿ squared over two. And so we have an equation involving the radius of our circle. Two π‘Ÿ squared minus πœ‹π‘Ÿ squared over two is equal to two and one-third. We can factor π‘Ÿ squared from the terms on the left-hand side, giving two minus πœ‹ over two all multiplied by π‘Ÿ squared equals two and one-third.

Now, at this point, we remember we’ve been asked to use the approximation πœ‹ equals 22 over seven. So πœ‹ divided by two or πœ‹ multiplied by a half is the same as 22 over seven multiplied by a half, which is 11 over seven. At the same time, we can think of the integer two as the fraction 14 over seven. So the terms inside our parentheses become 14 over seven minus 11 over seven, which simplifies to simply three over seven or three-sevenths.

We can then convert the mixed number on the right-hand side of our equation, two and one-third, into an improper fraction. And it’s equal to seven over three. To solve this equation then, we need to divide both sides by three-sevenths in order to leave π‘Ÿ squared on its own on the left-hand side. Giving π‘Ÿ squared equals seven over three divided by three-sevenths.

But we recall that to divide by a fraction, we can instead multiply by its reciprocal. So dividing by three-sevenths is equivalent to multiplying by seven-thirds. And we have that π‘Ÿ squared is equal to seven over three multiplied by seven over three. That’s 49 over nine. Or we can write seven over three multiplied by seven over three as seven over three all squared.

If π‘Ÿ squared is equal to seven over three all squared, then to find the value of π‘Ÿ, we square-root both sides of the equation. And we use only the positive value as π‘Ÿ is a length. So we have that π‘Ÿ is equal to seven over three. And we’ve found the radius of our circle.

So now that we know the radius of the circle, we also know the side length of the square. It’s twice this value, which is 14 over three. And now we’re able to calculate the perimeter of the shaded region. Picking a point on the perimeter β€” so I’ve chosen point 𝐹 β€” and then traveling around the shaded region, we see that its perimeter is composed of 𝐹𝐷, 𝐷𝐴, 𝐴𝐸, and then the semicircular arc. 𝐹𝐷 and 𝐴𝐸 are each seven over three centimeters, and 𝐷𝐴 is 14 over three centimeters.

We then recall that the circumference of a full circle is two πœ‹π‘Ÿ. So the arc length of a semicircle is half this. It’s simply πœ‹π‘Ÿ. The length of the semicircular arc is therefore πœ‹ multiplied by seven over three.

Again though, we recall that we need to use 22 over seven as our approximation for πœ‹. And then we can cancel a factor of seven from the numerator and denominator of these fractions. We’re left with seven over three plus 14 over three plus seven over three plus 22 over three. And as these fractions all have common denominator of three, they sum to 50 over three. We can then convert this to a mixed number. It’s 16 and two-thirds. And the units for this perimeter will be centimeters. So we’ve completed the problem. The perimeter of the shaded region is 16 and two-thirds centimeters.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.