Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 2 β€’ Question 27

𝐿𝑀𝑁𝑂𝑃 is a rectangular pyramid. The horizontal base 𝐿𝑀𝑁𝑂 has side lengths 8 cm and 12 cm and centre 𝐢. Angle 𝑃𝐢𝐿 is 90Β°. Angle 𝐿𝑃𝐢 is 34Β°. Remember that the volume of a pyramid = (1/3) Γ— area of base Γ— perpendicular height. Work out the volume of the pyramid.

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Video Transcript

𝐿𝑀𝑁𝑂𝑃 is a rectangular pyramid. The horizontal base 𝐿𝑀𝑁𝑂 has side lengths eight centimeters and 12 centimeters and centre 𝐢. Angle 𝑃𝐢𝐿 is 90 degrees. Angle 𝐿𝑃𝐢 is 34 degrees. Remember that the volume of a pyramid is equal to one-third times the area of the base times the perpendicular height. Work out the volume of the pyramid.

Let’s begin by going through the given information. We’re told this is a rectangular pyramid. So the horizontal base 𝐿𝑀𝑁𝑂 would be a rectangle with side lengths eight centimeters, which means over here would be eight centimeters and since it’s a rectangle and opposite sides of a rectangle are equal, and 12 centimeters, which means this had to be 12 centimeters.

And we’re also told that the centre of this rectangle is 𝐢. And this is important because when combined with the next given piece of information, the fact the angle 𝑃𝐢𝐿 is 90 degrees and 𝐢 is the centre of the rectangle, then 𝑃𝐢 will be the perpendicular height of this pyramid. And we don’t know this length and we need it for the volume of the pyramid. So let’s go ahead and call it π‘₯. And lastly, angle 𝐿𝑃𝐢 is 34 degrees.

So for the volume of the pyramid, we need to take one-third times the area of the base times the perpendicular height. Since our base is a rectangle, the area of our base will be length times width: eight centimeters times 12 centimeters giving us 96 centimeters squared. And then last, we’ll multiply by the perpendicular height, which we don’t know. So we’ll call it π‘₯. So this perpendicular height is the only piece of information that we’re missing to find this volume.

And here, we have a right-angled triangle and π‘₯ is the length of one of its sides. And we know an angle of this triangle: angle 𝐿𝑃𝐢 is equal to 34 degrees. So if we could find one of these side lengths, we would be able to solve for the perpendicular height 𝑃𝐢, which we’re calling π‘₯ using trigonometry. Well, since the base is a rectangle, triangle 𝐿𝑁𝑂 would be a right triangle because angles in a rectangle are 90 degrees.

So if we found the length of the hypotenuse β€” the longest side of this triangle β€” and then found half of that, that would be a length of the side of the pink triangle that we could use to help us find π‘₯ and this length would be half of 𝐿𝑁. That’s what we need to find. So let’s begin by first finding the length of 𝐿𝑁. We can find the length of 𝐿𝑁 by using Pythagoras’s theorem. The length of the hypotenuse squared is equal to, so 𝐿𝑁 squared is equal to the sum of the other two side lengths squared. Eight squared is 64 and 12 squared is 144. Adding these together, we get 208.

So if we have that 𝐿𝑁 squared is equal to 208, to solve for 𝐿𝑁, we simply square root both sides. 208 is 16 times 13. And we’re doing this to simplify the square root. 16 is a perfect square. It’s four times four. So this means we find that 𝐿𝑁 is equal to four square root 13. So if we would like to find half of 𝐿𝑁 for the pink triangle, we need to take one-half times four square root of 13. Two goes into four twice. So we are left with two square root of 13.

Notice we only took one-half times four. That’s because both of these numbers are not inside of a square root. So we cannot take one-half times the square root of 13 because the square root of 13 is underneath the square root, where one-half is not. So now we can use this length to help us find the perpendicular height π‘₯.

So here is our pink triangle. From our angle of 34 degrees, two square root of 13 will be considered our opposite side and π‘₯ would be considered the adjacent side. So out of sine, cosine, and tangent, it’s the tangent function that uses the opposite and adjacent sides. The tangent of an angle is equal to the opposite side divided by the adjacent side. So we have that the tangent of 34 degrees is equal to two square root of 13 divided by π‘₯.

And since we want to solve for π‘₯, we can multiply both sides of the equation by π‘₯ β€” this way it’s moved up to the numerator β€” and then divide both sides of the equation by tangent 34. So we are left with π‘₯ is equal to two square root of 13 divided by the tangent of four. So we’ll plug this into our calculator. So this means π‘₯ is about 10.69089 and so on. Now to avoid rounding error, we want to keep this entire number on our calculator and then multiply by 96 and then multiply by one-third or dividing by three. And remember this perpendicular height is in centimeters. So when we have centimeters squared times centimeters, our volume should be in centimeters cubed.

And after multiplying these together and we get 342.108773822 cubic centimeters. Now this is not specify how many decimal places to round. Normally, it’s one or two. Let’s go ahead and round one. But either way would be fine. So we need to decide whether to round the one up or round it down. So we look at the number to right, zero. Since zero is less than five, we round down. So we’ll keep the one a one.

Therefore, the volume of this rectangular pyramid would be 342.1 cubic centimeters.

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