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I can use the formula for the area of any triangle.

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Licence

This content is © Oak National Academy Limited (2023), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

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Key learning points

The formula for the area of any triangle can be used if the lengths of 2 sides are known
The size of the angle between the two sides must also be known
The formula can be used to find a missing side length or angle if the area is known

Common misconception

Using any two side lengths for the sine formula.

It must be two side lengths and the angle between them. So any two side lengths of the triangle can be used so long as you also use the angle between them.

Keywords

Area - The area is the size of the surface and states the number of unit squares needed to completely cover that surface.
Sine function - The sine of an angle (sin(θ°)) is the y-coordinate of point P on the triangle formed inside the unit circle.

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Hello, I'm Mrs. Lashley and I'm gonna be talking you through the lesson today.

I really hope you're willing to try your best and even if it gets challenging, remember I'm here to support you.

So our learning outcome today is to be able to use the formula for the area of any triangle.

On the screen there are definitions for area and sine of an angle.

You may wish to pause the video to read through them before we make a start with the lesson.

So in this lesson about calculating the area of any triangle when the height is not known, we are gonna split it into 2 learning cycles.

Our first learning cycle is calculating the area of any triangle and then when we get to the second learning cycle, we're gonna find a side length when the area is known.

So let's make a start at calculating the area of any triangle.

There are many ways to find the area of a triangle.

We need to determine what information we have and then choose the most appropriate method.

So on the screen we've got a right-angled triangle, what looks to be a scalene triangle and an isosceles triangle.

Pause the video and think about which methods or what information you need to be able to work out the area of those types of triangles.

Press play when you're ready to move on.

So Sam says, \"If it's a right-angled triangle, I may have several options.\" So when you were thinking about the right-angled triangle, did you consider that maybe you have more than 1 option of how to find the area of the triangle? Sam explains that this is the easiest scenario as you already have 2 perpendicular measures.

So the 3 and the 4 are adjacent edges and they are perpendicular, so we could use area equals 1/2 times base times perpendicular heights and calculate that area of 6.

If we were given the third edge of the triangle, then this is slightly more difficult, Sam suggests, as you need to deduce which measures to use.

So when you've got more than you need, sometimes it's a bit challenging to deduce which ones you need.

But if we're gonna use 1/2 times base times perpendicular height, then we're looking for the 2 measures that are perpendicular to each other.

Why might you find this one even more challenging? Pause video and think about that.

Why might Sam find this one even more challenging? Press play when you're ready to discuss that and move on.

When it's orientated a little bit wonky, sometimes we find that challenging.

But if we look at the fact that we do have 2 perpendicular measures because it's a right-angled triangle, those 2 shorter edges are perpendicular, then we are able to still use a 1/2 times base times perpendicular height.

There's no base edge, which is where sometimes we feel a little bit unsure.

So Sam says, \"What makes this more difficult and what strategies are available?\" So pause the video and think about that.

Why is this more challenging than the other right-angled triangle scenarios? Press play when you're ready to discuss that a little bit further.

So with this one, we do not have 2 perpendicular edges, so we can't use base times perpendicular height divided by 2, but we can calculate it by using Pythagoras's theorem.

Alternatively, we could work out the angle between the 6 and the 10.

That would make the 6 the adjacent and the 10 the hypotenuse, so we'd make use of the cosine function to calculate the angle between them and then we'd be able to use a 1/2 times AB sine C.

So now we've got a right-angled triangle, but it's only the hypotenuse that is labelled and we don't know anything further about this right-angled triangle.

We don't know that it's isosceles and therefore we know the other 2 angles of 45 degrees and we'd be able to calculate the edge length using Pythagoras's theorem.

So Sam says, \"How about this? Can I find more measures? Can I use sine?\" Pause the video once again and think about those 2 questions.

We do not have enough information to be able to find out any more measures and we cannot use sine because we need 2 adjacent edges and the angle between them.

Sam's added the angle of 35 degrees.

Can we find more measures now? Pause the video and think about that for a second.

Press play when you're ready to move on.

So Sam has added the third angle, so we can calculate that too using the 90 degree and the 35 degree angle.

We know that the 3 angles need to sum to 180 degrees.

Does that help us? Sam then says, \"I can find missing side length using sine, cosine or tangent.\" So we can make use of trigonometric ratios to work out missing side length.

So here's a check, which of these calculations will find the area of the triangle? So pause the video and when you're ready to check your answer, press play.

So in fact, all 3 of these calculations will find the area of that triangle that's on the screen and that's because it's a right-angled triangle and we had all 3 edges, so we could select the 2 perpendicular edges, base times perpendicular height divided by 2, that's a calculation part A.

On part B and C we're using the sine formula.

The 6 root 2 times 6 sine of 45 is using the hypotenuse and 1 of the shorter edges with the angle of 45 degrees between them.

Whereas on part C, it's using the 2 perpendicular edges of 6 and the angle between them is 90 degrees.

So it's still using the sine formula for the area, but it's using 6 and 6 and 90.

Sine of 90 is 1 and that's why the part A is actually equivalent to part C.

So here on the screen we've got a non right-angled triangle or we don't have enough information to be able to suggest that this is a right-angled triangle.

Izzy says, \"Why are non right-angled triangles more difficult and is this enough information?\" Pause the video and think about this.

Why is non right-angled triangles more difficult to calculate the area? And do we have enough information currently on this diagram to work out the area for this triangle? Press play when you're ready to move on.

So the answer to do we have enough information is no.

So if we're looking to use the sine formula for area, then we need 2 adjacent edges and the angle between them.

We do have 2 adjacent edges, but we don't have the angle between them and nor can we calculate that.

So knowing just 2 sides and no angles isn't enough information and we can look at the constructions using circles to see how the third length changes.

So if we've got a circle set up with a radius of 9, that would give us our length of one of the edges.

And then at the end of that line segment, the end of that radius, we draw another circle with a radius of 8 and then we choose a point on the circumference of that circle and join it back to the centre of the first one.

And you can see that's happened in both of the diagrams, but we end up with 2 very different looking triangles.

We could have chose any point on the circumference of the circle with a radius of 8 and joined it back to the centre of the radius 9 and we would've got a different triangle.

So here we can see that having 2 adjacent edges is not enough information to be able to calculate the area as the area will be different depending on the shape of the triangle.

And similarly, knowing one angle and a side isn't enough either.

So here we've got an edge of 9 and an angle at one end of that line segment of 47 degrees and there's many places along the ray line that creates that 47 degrees that we could choose to be our third vertex and join the side of 9 units.

And so then we'd get many different looking triangles and therefore their areas would all be different.

So 2 adjacent sides is not enough information.

1 side, 1 angle is also not enough information.

To be sure we can find missing measures, we want 2 side lengths and the angle between them.

So we want 2 adjacent sides, so like the 8 and that 9, but we'd also want to know the angle between the 2 edges or we want to be able to find the angle between them.

So if we were given the angles at the end of those 2 line segments, we can sum them and subtract them from 180 degrees to work out the angle between them.

So Izzy says, \"Now I can use my knowledge of trigonometric functions!\" So we have enough information now to be able to use our knowledge of trigonometric functions.

So here is a check, what is the area of this triangle to 1 decimal place? Pause the video, hopefully your answer is going to be a, b or c, and when you have that answer, press play to check it.

The answer is b.

Using the sine formula for the area of a triangle, you'd do 1/2 times 8 times 9 times sine of 82 degrees and that would be 35.

So onto the first task of the lesson where you need to find the area of each of these triangles.

So there are 6 triangles on the screen.

1 triangle is inside of a regular hexagon, so you need to calculate the shaded triangle for that part.

Press pause whilst you're working through those, think about which strategies you can use, and then when you're ready for the answers, press play.

All the answers are on the screen here.

So first triangle, top left, we had 2 perpendicular measures, therefore we could do 1/2 times base times perpendicular height to get the area.

On the second triangle, it's right-angled.

We had 2 of the 3 edges.

They were not perpendicular edges, but we could use Pythagoras's theorem to calculate the third edge and then use 1/2 times base times perpendicular height.

Alternatively, we could work out the angle between the 5 and the 13 using the cosine ratio and then use 1/2 AB sine C, either way, 30 square centimetres.

The third triangle was isosceles.

So we need to get the perpendicular height of that triangle or the angle between the 2 given side lengths.

We do have all 3 side lengths because it is isosceles, so you would know that the third unmarked edge would also be 9 centimetres.

We could split this isosceles into 2 congruent right-angled triangles and use Pythagoras's theorem to get the perpendicular height, or you could still split it into 2 right-angled triangles and use the trigonometric ratios to get the angle between them.

So quite a few options here, but the area was 25.

46 square centimetres to 2 decimal places.

The 4th triangle is the equilateral triangle with an edge of 1.

Because it's equilateral, then the angles are 60 degrees.

So I would've used a 1/2 times, 1 times 1 times sine of 60 degrees and then rounded that to 2 decimal places.

If you left that as root 3/4, then that is the exact form of the area.

On the fifth triangle, we've got all of the relevant information to use the sine formula for an area.

So 1/2 times 3 times 2 times sine of 105 degrees.

You've got 2 adjacent edges and the angle between them.

And lastly, we've got that grey triangle within the regular hexagon.

Again, there's many methods that you could use, but the answer is 0.

87 square centimetres to 2 decimal places.

So we're now up to the second learning cycle of today's lesson where we're going to look at finding a side length when the area is known.

So making use of the formula to find the side length rather than work out the area.

So like you'll have done with other shapes, we can work backwards from the area of a triangle to find measures in the triangles that are not stated.

So if we look at this right-angled triangle here, we know that one of the edges is 8 centimetres and there is an unknown edge which is labelled as h.

However, Sam says, \"If the area is 20 square centimetres, then we know by our formula that a 1/2 times 8 times h is equal to 20\" because h is perpendicular to the 8, so we can say 1/2 times base times perpendicular height is equal to the area and the area is known to be 20.

So now we have an equation that we can solve.

So pause the video and work out h and then we'll move on.

Okay, so h should have been 5, so well done if you've got 5.

Here we've got another triangle, it's not a right-angled triangle, but the h is the perpendicular height.

, the 9 is the base if you like and there is an angle of 80 degrees.

So if the area of this triangle is 36 square centimetres, then a 1/2 times 9 times h equals 36.

So the 80 degrees is not necessary.

9 and the perpendicular height are perpendicular measures and so we can make use of base times perpendicular height divided by 2.

Once again, pause the video and work out what h is for this scenario.

Press play when you're ready to check.

So h would be 8.

Sam has highlighted that the angle is not necessary to find out h, but I could use the angle, the area, and the side length to find a.

So in the same triangle, if a different, if the height was not, the perpendicular height was not given or wasn't the thing that you were trying to calculate, if it was this length of the side here labelled as a, then we might use a different formula.

So we would then look at using the sine formula for the area, so a 1/2 ab sine C is equal to 36.

So if we were told that the area of this triangle is 36, then we could substitute the known stuff into it to solve and find a.

So we know that the edge b is going to be 9, and our angle between them, so that'll be our angle C is 80 degrees, and then we can solve this to find a.

So a would be 8.

12 centimetres to 2 decimal places.

So here's a check for you.

If the area of this triangle is 100 square centimetres, what is the value of a to 1 decimal place? So pause the video and then when you're ready to check press play.

The answer is b.

So you should have set up an equation very similar to the previous one.

So a 1/2 times a times 12 times sine of 70 degrees is equal to 100, and then go forward to solve that by rearranging to make a the subject and calculating, so 17.

Here is another check.

Press pause whilst you work it out, and then when you're ready to check, press play.

This time the answer is c, 26.

So a 1/2 times a times 8 times sine of 50 equals 80 and then solve it for a.

So up to the last task of the lesson.

On this question, you're gonna find the value of b to 2 decimal places for each of these triangles and they all have an area of 10 square centimetres.

So pause the video whilst you work through those 6 questions.

So there is a edge that's labelled b in all of them and you need to work it out to 2 decimal places.

Press play when you're ready to check the answers.

So the answers are on the screen now.

So on question 1, you needed to work out that the edge was 10.

On the second triangle, it's 6.

40.

On the third, it'd be 5.

39.

On the fourth, it's 23.

09.

On the fifth, it's 10.

35 and the last one is 5.

08.

If you've written the first one as 10.

00, then that yes is to 2 decimal places.

So let's now look at how we could find the value of b in each of these triangles.

So from the first triangle we had 2, b was the perpendicular length to that length 2.

So we can set up the equation 1/2 times 2 times b equals 10 and solve it to find the length b.

On the second triangle, we were looking to find the hypotenuse, so we would need to work out the perpendicular height to the 4 using 1/2 times 4 times, let's call it X equals 10.

And then once you had X, you could use the 4 and the value of X and Pythagoras's theorem to get the hypotenuse.

A similar method was necessary on the third triangle.

We can use the formula for the area 1/2 times base times perpendicular height to get the perpendicular height of the isosceles triangle.

And then once we have the perpendicular height of the triangle, we can use 1/2 of that base edge, which is 5, and Pythagoras's theorem to calculate the hypotenuse, which is the edge marked b.

When we get to the equilateral triangle, so triangle number 4, we can use 1/2 ab sine C.

So 1/2 times root b times root b times sine of 60 degrees equals 10 'cause we were told that all of these had an area of 10, and then we can rearrange that to make b the subject.

That was the same for triangle 5, that we can use the sine formula for the area, so a 1/2 times 2 times b times sine of 105 equals 10 and rearrange that to work out the value of b.

And then lastly, the 2 angles that we were given, we can use to work out the third angle of the triangle, so that would be 80 degrees.

And then we can say 1 1/2 times 4 times b times sine of 80 equals 10 and rearrange it to get the value of b.

Well done if you manage to get majority of those correct.

To summarise today's lesson on calculating the area of any triangle when the height is not known, the formula for the area of any triangle can be used if the lengths of 2 sides are known.

The size of the angle between the 2 sides must also be known and the formula can be used to find a missing side length or the angle if the area is known.

So well done today and I look forward to working with you again in the future.

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^2

" }, "right": { "text": "A triangle with perpendicular height = 5 cm and base = 6 cm" } }, { "left": { "text": "Area = 22 cm

^2

" }, "right": { "text": "A triangle with perpendicular height = 11 cm and base = 4 cm" } }, { "left": { "text": "Area = 16 cm

^2

" }, "right": { "text": "A triangle with perpendicular height = 4 cm and base = 8 cm" } }, { "left": { "text": "Area = 4.5 cm

^2

" }, "right": { "text": "A triangle with perpendicular height = 3 cm and base = 3 cm" } } ], "id": "07c36d26-aeb2-dcfe-e945-2f8ca83122b4", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "Which of the following calculate the area of this triangle?", "image": "https://www.thenational.academy/_next/image?url=https%3A%2F%2Foaknationalacademy-res.cloudinary.com%2Fimage%2Fupload%2Fv1723048212%2Ffascmo6w7ybl0jl0rl8g.png&w=640&q=75", "imageKey": "7f92c3ff-03a6-332f-0954-2b7c47d496fd/question/219664642c49f44c5d9a9a760ca5cdfc.png" }, "questionType": "Multiple choice", "correct": [ { "text": "

\\frac{1}{2}\\times 6.3\\times5\\times\\text{sin}(55.3°)

" }, { "text": "

\\frac{1}{2}\\times 6.3\\times5.4\\times\\text{sin}(49.8°)

" }, { "text": "

\\frac{1}{2}\\times 5.4\\times5\\times\\text{sin}(74.9°)

" } ], "choices": [ { "text": "

\\frac{1}{2}\\times 6.3\\times5\\times\\text{sin}(55.3°)

" }, { "text": "

\\frac{1}{2}\\times 6.3\\times5.4\\times\\text{sin}(49.8°)

" }, { "text": "

\\frac{1}{2}\\times 5.4\\times5\\times\\text{sin}(74.9°)

" }, { "text": "

\\frac{1}{2}\\times 6.3\\times5.4\\times\\text{sin}(55.3°)

" }, { "text": "

\\frac{1}{2}\\times 5.4\\times5\\times\\text{sin}(55.3°)

" } ], "id": "0226f337-380a-183f-85b2-b6e45c7b3042", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "The area of this triangle, given to 2 significant figures, is ______cm

^2

.", "image": "https://www.thenational.academy/_next/image?url=https%3A%2F%2Foaknationalacademy-res.cloudinary.com%2Fimage%2Fupload%2Fv1723048212%2Ffascmo6w7ybl0jl0rl8g.png&w=640&q=75", "imageKey": "7f92c3ff-03a6-332f-0954-2b7c47d496fd/question/219664642c49f44c5d9a9a760ca5cdfc.png" }, "questionType": "Short answer", "correct": [ { "text": "13 " } ], "id": "4f2ab661-93cb-949d-1995-f9ed0412b192", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "A right angled triangle has a height of 12 cm and a hypotenuse of 13 cm. Work out the area." }, "questionType": "Multiple choice", "correct": [ { "text": "30 cm

^2

" } ], "choices": [ { "text": "30 cm

^2

" }, { "text": "60 cm

^2

" }, { "text": "15 cm

^2

" }, { "text": "120 cm

^2

" } ], "id": "fe006798-8a75-ffe3-1f8a-5fca55dd16a4", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "Six equilateral triangles are put together to make a regular hexagon. The length of each triangle is 3 cm. Work out the exact area of the regular hexagon." }, "questionType": "Multiple choice", "correct": [ { "text": "

13.5\\sqrt{3}

^2

" } ], "choices": [ { "text": "

13.5\\sqrt{3}

^2

" }, { "text": "

27\\sqrt{3}

^2

" }, { "text": "

\\sqrt{3}

^2

" }, { "text": "

27\\sqrt{3}

^2

" } ], "id": "07daf3dc-225f-74c6-1e07-a3c167005b41", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "An isosceles triangle has a base of 4 cm and base angles of 68.2°. Select the correct area given to 2 significant figures." }, "questionType": "Multiple choice", "correct": [ { "text": "10 cm

^2

" } ], "choices": [ { "text": "5.0 cm

^2

" }, { "text": "10 cm

^2

" }, { "text": "20 cm

^2

" }, { "text": "40 cm

^2

" } ], "id": "cf435168-e023-f3b9-383d-b4a4acf9b5d5", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" } ], "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51", "meta": { "gradeStageId": "ks4", "subject": "Maths", "tier": "Higher", "grade": "Year 11", "topic": null, "unit": "Non right-angled trigonometry", "lessonCount": "11", "age": [ 15, 16 ], "gradeId": "age_15_16", "gradeStage": "GCSE", "lesson": "Calculating the area of any triangle when the height is not known" } }, { "id": "e5ce643e-d39d-b26c-70d0-e05c6e758987", "type": "Resource", "resourceType": "Quiz", "title": "Calculating the area of any triangle when the height is not known", "quizType": "Exit quiz", "items": [ { "type": "Question", "question": { "text": "A right-angled triangle has a height of 12 cm and a hypotenuse of 15 cm. The area of the triangle is ______cm

^2

." }, "questionType": "Short answer", "correct": [ { "text": "54" } ], "id": "9f07cdbe-150f-7518-378d-0cabe8d8b1b3", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "An isosceles triangle has lengths 26 cm, 26 cm and 20 cm. The area of the isosceles triangle. is ______cm

^2

." }, "questionType": "Short answer", "correct": [ { "text": "240" } ], "id": "41b87da7-fab3-4ce1-3c33-64beacb7b118", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "A triangle is drawn. The adjacent lengths are 7 cm and 9 cm with a 47° angle in between. The area of the triangle, to 1 decimal place, is ______cm

^2

." }, "questionType": "Short answer", "correct": [ { "text": "23.0" } ], "id": "72fdf98a-b947-ef37-c34c-b33fec0bd623", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "An equilateral triangle has lengths of 6 cm. The area of the triangle, to 1 decimal place, is ______cm

^2

." }, "questionType": "Short answer", "correct": [ { "text": "15.6" } ], "id": "26730b71-218e-2bef-29a4-0187fe6232ba", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "Two adjacent lengths of a triangle are

x

cm and 6 cm with a 47° angle in between. The area is 26.3 cm

^2

. Work out the value of

x

, giving your answer rounded to the nearest integer." }, "questionType": "Short answer", "correct": [ { "text": "12" }, { "text": "12 cm" } ], "id": "7784b128-dcdc-e9b2-f197-89a7cc5beece", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" }, { "type": "Question", "question": { "text": "An equilateral triangle with lengths of

x

cm is drawn. It has an area of 3.9 cm

^2

. Work out the length of the equilateral triangle, giving your answer as a whole number." }, "questionType": "Short answer", "correct": [ { "text": "3 cm" }, { "text": "3 " } ], "id": "3411aed7-8b09-9c72-92db-950cdd89d485", "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51" } ], "lessonId": "f72f05f0-949c-5ef6-ccd8-2117eba24a51", "meta": { "gradeStageId": "ks4", "subject": "Maths", "tier": "Higher", "grade": "Year 11", "topic": null, "unit": "Non right-angled trigonometry", "lessonCount": "11", "age": [ 15, 16 ], "gradeId": "age_15_16", "gradeStage": "GCSE", "lesson": "Calculating the area of any triangle when the height is not known" } } ] }