Video: Finding the Area of the Part between the Two Tangents and the Smaller Arc of Two Points Lying on the Circumference

𝐴𝐡 and 𝐡𝐢 are two tangents to the circle 𝑀 where 𝐡 and 𝐢 lie on the circumference. 𝑀𝐴 = 14 centimeters and the radius of the circle is seven centimeters. Find the area of the part between the two tangents and the smaller arc 𝐡𝐢 giving the answer to the nearest square centimeter.

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Video Transcript

𝐴𝐡 and 𝐡𝐢 are two tangents to the circle 𝑀 where 𝐡 and 𝐢 lie on the circumference. 𝑀𝐴 is equal to 14 centimeters and the radius of the circle is seven centimeters. Find the area of the part between the two tangents and the smaller arc 𝐡𝐢 giving the answer to the nearest square centimeter.

This might look a little tricky at first. But we can break the problem down into several smaller parts. First, we see that if we add the radii 𝐡𝑀 and 𝑀𝐢 onto the diagram, we can form two right-angled triangles. That’s 𝐴𝐡𝑀 and 𝐴𝐢𝑀. This is because the angle between a tangent and a radius is 90 degrees. So angles 𝐴𝐡𝑀 and 𝐴𝐢𝑀 must both be 90 degrees.

Once we spot this, we can use right angle trigonometry to work out the measure of angle 𝐴𝑀𝐡 and 𝐴𝑀𝐢. Drawing triangle 𝐴𝑀𝐡 out, we were told that 𝑀𝐴 is 14 centimeters. So the hypotenuse of our triangle is 14 centimeters. It’s the side opposite the right angle and it’s always the longest side in the triangle. The radius 𝑀𝐡 is seven centimeters. That’s the adjacent side and it’s the side next to the included angle.

We can now use the cosine ratio to find the measure of the angle πœƒ. Cos of πœƒ is equal to adjacent divided by hypotenuse. In this case, that’s seven divided by 14 which is 0.5. We can solve this equation for πœƒ by finding the inverse of the cosine ratio. That’s inverse cos of 0.5. This is one of those standard results that we should know by heart. But if we don’t, we get πœ‹ over three radians.

Now that we have this information, we can find the area of the triangle 𝐴𝐡𝑀. Once we have that, we can then find the area of the quadrilateral 𝐴𝐡𝑀𝐢. And then, we will be able to subtract the area of the arc to find the area required. There are actually two ways we could find the area of this triangle.

One way would be to find the length of the missing side and then use the formula a half base times height. However, we can use the trigonometric formula: a half 𝘒𝘣 sin 𝘀. The included angle in our triangle 𝐴𝐡𝑀 is πœ‹ over three. And the sides 𝘒 and 𝘣 are seven centimeters and 14 centimeters. So the area is a half multiplied by seven multiplied by 14 multiplied by sine of πœ‹ over three.

Once again, sine of πœ‹ over three is one of those standard results we should know by heart. Its root three over two. And we can simplify somewhat by dividing through by two. Once we do, we can see that the area of our right-angled triangle is 49 root three over two units squared or square units.

We can find the area of the quadrilateral then by multiplying this value by two. That’s two multiplied by 49 root three over two. So the area of π΄π΅π‘€π˜Š is 49 root three units squared.

Remember to find the shaded area, we said we’d need to subtract the area of sector π˜‰π˜Šπ˜”. The formula for area of a sector with radius 𝘳 and angle πœƒ radians is a half 𝘳 squared πœƒ. We know the size of the angle in our sector. It’s two lots of πœ‹ over three, two πœ‹ over three.

So the area is a half multiplied by seven squared multiplied by two πœ‹ over three. Seven squared is 49. And we can simplify by dividing through by two. And the area of our sector is 49 πœ‹ over three.

The area we need is the difference between these. It’s 49 root three minus 49 πœ‹ over three. That’s 33.557. Remember we were asked to give our answer to the nearest square centimeter. And if we do, we can see that the area we require is 34 units squared or 34 centimeters squared.

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