Video: Pack 4 β€’ Paper 2 β€’ Question 18

Pack 4 β€’ Paper 2 β€’ Question 18

10:55

Video Transcript

𝐴𝐡𝐢𝐷𝑃 is a regular solid pyramid. 𝐴𝐡𝐢𝐷 is a square, where 𝐡𝐢 is equal to eight centimetres. The angle between any sloping edge and the base 𝐴𝐡𝐢𝐷 is 65 degrees. Work out the surface area of the pyramid, giving your answer to two significant figures.

Now, before we delve into this question, let us define two more points to make the different components of our pyramid easier to refer to. We shall let π‘Œ be the midpoint of the square 𝐴𝐡𝐢𝐷, which forms the base of the pyramid. And we shall let 𝑋 be the midpoint of the line 𝐡𝐢.

Next, when we examine this question, it’s really important to pay attention to the wording. The question tells us that the angle between any sloping edge and the base 𝐴𝐡𝐢𝐷 is equal to 65 degrees. Now, it may be tempting to immediately draw on the right-angled triangle, here represented as π‘ƒπ‘‹π‘Œ.

We would then assume that the question is telling us that the angle marked π‘ƒπ‘‹π‘Œ here is equal to 65 degrees. However, doing this would actually be incorrect. To see why, let us recall that in a 3D shape, an edge is the line segment where two faces meet. And the face is one of the flat surfaces on the solid object.

We should, therefore, be able to see that the sloping edges go from any of the four corners of the square 𝐴, 𝐡, 𝐢, or 𝐷 to the top of the pyramid 𝑃. Here, we have marked on the sloping edge 𝐡𝑃.

Now that we have done this, we can draw on a different right-angled triangle π‘ƒπ΅π‘Œ. And we are now correct in saying that the angle marked here π‘ƒπ΅π‘Œ is indeed equal to 65 degrees. And now that we understand what the question is telling us, we begin to wonder how this helps us. Unfortunately, we do not have a lot of information about this triangle π‘ƒπ‘Œπ΅ since the question has neither given us the side length π‘ƒπ‘Œ which is the height of the pyramid or the side length π΅π‘Œ.

Now, the solution to this problem has a lot of different components. So we’re gonna briefly spell out the different steps will be taken before continuing.

The first thing we’re going to do is to find length π΅π‘Œ using a different right-angled triangle, which is triangle π‘Œπ‘‹π΅. We should be able to do this using only the side length of the square, which is given in the question as eight centimetres.

Next, we’ll use triangle π‘ƒπ‘Œπ΅ to find the length π‘ƒπ‘Œ or the height of the pyramid. We will be able to do this using the newly found length π΅π‘Œ and the angle of 65 degrees given in the question.

As step three, we’ll find length 𝑃𝑋, using the triangle π‘ƒπ‘Œπ‘‹ and the information that we have already found at this point. Once we have done this, we will know both the base and the height of the triangle which forms the four slanted triangular faces of the pyramid. This information will allow us to calculate the area of these faces. Here, we have marked the area 𝑃𝐡𝐢 on the diagram.

Finally, we’ll add the area of our four triangular faces to the area of the square base to find the total surface area of the pyramid.

Okay, now that we understand the process that we will be going through, let us get to the solution. Step one, find π΅π‘Œ using triangle π‘Œπ‘‹π΅.

First, we remember that we defined point 𝑋 as the midpoint between 𝐡 and 𝐢. The question gives us that the length 𝐡𝐢 is equal to eight centimetres. And we can, therefore, say that 𝐡𝑋 is half of this length or four centimetres. We should also be able to see that the line of π‘Œπ‘‹ is also equal to four centimetres since the distance from the midpoint of any of the sides of the square to the centre of the square is equal to half of the side lengths.

We’re therefore able to say that both 𝐡𝑋 and π‘Œπ‘‹ are equal to four centimetres. Now, it should be easy to see that triangle π‘Œπ‘‹π΅ is a right-angled triangle. And because of this, we are able to use Pythagoras’s theorem to find the length π΅π‘Œ.

Pythagoras’s theorem tells us that the square of the hypotenuse, which is π΅π‘Œ, is equal to the sum of the square of the other two sides, 𝐡𝑋 squared plus π‘Œπ‘‹ squared. We substitute in our value of four for both of these sides. We then square our fours to get 16 plus 16 and we find that π΅π‘Œ squared is equal to 32. By taking the square root of both sides, we see that π΅π‘Œ is equal to the square root of 32. And we perform a simplification to find that π΅π‘Œ is equal to four times root two centimetres. Of course, since π΅π‘Œ is a length, we ignore any negative solutions to the square roots in our answers.

Great! We have now found π΅π‘Œ using the triangle π‘Œπ‘‹π΅. We now move on to finding the length π‘ƒπ‘Œ, which is the height of the pyramid. And we do so using triangle π‘ƒπ‘Œπ΅. We can label the side length that we have just found π΅π‘Œ which is four times the square root of two centimetres. We can also draw on the angle that the question gives us between the sloping edge 𝑃𝐡 and the base which is 65 degrees.

This situation with a right-angled triangle should be familiar to us. We have an angle, unknown adjacent side, and an unknown opposite side that we want to find. In this situation, we can use trigonometry to help us.

Looking at SOHCAHTOA, we can see that we need to use the tan function given by TOA. The identity here is that tan of an angle πœƒ is equal to the opposite side divided by the adjacent side. Let’s use this identity on our triangle.

We have that tan of 65 degrees is equal to the opposite side or π‘ƒπ‘Œ divided by the adjacent side, which we know is four times the square root of two. To find π‘ƒπ‘Œ on its own, we can multiply both sides of our equation by four times the square root of two and we find that π‘ƒπ‘Œ is equal to four times the square root of two times tan of 65 degrees.

When working with trig functions, it is always worth checking that your calculator is set to the right units. And in this case, we want degrees instead of radians. At this stage, we could type the calculation into our calculator to find the length of π‘ƒπ‘Œ is approximately equal to 12.131 centimetres.

This may be good as a sense check to make sure that the value is roughly within the range that you’re expecting. However, since we’ll be using π‘ƒπ‘Œ at later stages in our calculations, but here we’re going to choose not to approximate and to keep using the exact value for π‘ƒπ‘Œ. We do this since we’ll be using π‘ƒπ‘Œ in later calculations. And this helps preserve a higher degree of accuracy.

We have now found the value of π‘ƒπ‘Œ. And we can move on to step three, which is finding 𝑃𝑋 using triangle π‘ƒπ‘Œπ‘‹. For triangle π‘ƒπ‘Œπ‘‹, we can label the side π‘‹π‘Œ, which is four centimetres, as shown earlier. And we can label the side π‘ƒπ‘Œ, which we have just found of four times the square root of two times tan 65 degrees in centimetres.

We’re now in the situation again where we can use Pythagoras’s theorem to find the desired length, which is the hypotenuse of this right-angled triangle 𝑃𝑋. Here, we have that the hypotenuse squared is equal to the sum of the square of the other two sides. So 𝑃𝑋 squared is equal to π‘‹π‘Œ squared plus π‘ƒπ‘Œ squared.

We can substitute in four for π‘‹π‘Œ and four times the square root of two times tan of 65 degrees for π‘ƒπ‘Œ. Taking the square root of both sides, we find that 𝑃𝑋 is equal to the square root of 16 plus four times the square root of two times tan of 65 degrees squared. Although we want to keep our answers as accurate as possible, here our line of working is getting a little bit big. So let’s find a value for 𝑃𝑋. Typing this into our calculator, we find that side 𝑃𝑋 is equal to 12.774 centimetres to three decimal places.

Now that we’ve found side length 𝑃𝑋, we can move on to our next step. For this step, we want to find the area of the triangle 𝑃𝐡𝐢, which is one of the triangular slanted faces of the pyramid. We know the base of this triangle is eight centimetres as it is given in the question. Since we define point 𝑋 as the midpoint of the line 𝐡𝐢, we also know the height of this triangle, which is side length 𝑃𝑋 that we just found.

The area of a triangle we know is half times the base times the height. In our case, the area is half times 𝐡𝐢 times 𝑃𝑋. Here, we substitute in our values to find half times eight times 12.774 is the area of 𝑃𝐡𝐢. Performing this calculation, we find that we have a value of 51.094 centimetres squared.

We now come to our final stage where we calculate the surface area of the pyramid. The surface of the pyramid consists of five different faces. One of these is the square which forms the base, which is the shape 𝐴𝐡𝐢𝐷. The other four faces are triangles. These four slanted triangular faces are all identical to the triangle 𝑃𝐡𝐢, for which we have just found the area.

The final and simple piece of the puzzle is, therefore, the area of the square 𝐴𝐡𝐢𝐷. The area of a square is equal to its side lengths squared. We can, therefore, say the area of 𝐴𝐡𝐢𝐷 is equal to eight squared centimetres squared or 64 centimetres squared. We are now ready to perform our final calculation and find the total surface area of the pyramid.

The total surface area is equal to the area of the square 𝐴𝐡𝐢𝐷 plus four times the area of the triangle 𝑃𝐡𝐢. We’re able to substitute in the values we have found for the area of these shapes. Performing this calculation, we find that the surface area of the pyramid is approximately equal to 268.376 centimetres squared.

Finally, we recall that the question asked for our answer to two significant figures. Rounding our answer, we say that the total surface area of the pyramid is equal to 270 centimetres squared to two significant figures.

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