Video: CBSE Class X • Pack 1 • 2018 • Question 17A

CBSE Class X • Pack 1 • 2018 • Question 17A

05:55

Video Transcript

Prove that the area of an equilateral triangle sharing one of its sides with a side of a square is equal to half the area of an equilateral triangle showing one of its sides with a diagonal of the square.

So let’s begin this question with a sketch of what we’re being asked to show. So here, we have a square. And the first equilateral triangle we’re told shares one of its sides with one side of the square. So here, we have that triangle, which we’ll refer to as triangle one. We’re told that the second equilateral triangle shares one of its sides with the diagonal of the square. So I’ve drawn in one of the diagonals of the square. And the second equilateral triangle looks something like this.

Now, the question is asking us to prove a statement about the areas of these equilateral triangles. So we need to think about how to find the area of a triangle. Now, there are two ways to do this. If we know the base and the perpendicular height of a triangle, we can use the formula a half multiplied by base multiplied by height. Or if we know two sides of the triangle and the included angle — so that’s the angle between those two sides — we can use the formula a half 𝑎𝑏 sin 𝑐, where 𝑎 and 𝑏 are two sides of the triangle and 𝑐 is the included angle.

Now, as both of these triangles are equilateral, this means that all of the internal angles are 60 degrees. So we know that the value of 𝑐 in our second formula will be 60 degrees. In order to use the second formula then, we just need expressions for the side length of each triangle.

As our triangles are equilateral, this means that all three sides are the same length. So in fact, our formula becomes a half multiplied by 𝑎 squared multiplied by sin of 60 degrees, where 𝑎 is the side length of each triangle.

So let’s consider triangle one first of all. And we’ll start by denoting the side length of the square as 𝑥. As this first triangle shares one of its sides with a side of the square, this means that all of the side lengths of this equilateral triangle are also 𝑥. So using our formula for the area of a triangle and substituting 𝑥 for the side length of this triangle, we had that the area of triangle one is equal to a half multiplied by 𝑥 squared multiplied by sin of 60 degrees.

Now, 60 degrees is one of those special angles whose sine, cosine, and tangent ratios can be expressed exactly in terms of surds. Remember sin of 60 degrees is equal to root three over two. So substituting this into our area formula, we have a half 𝑥 squared multiplied by root three over two. And remembering that to multiply fractions together, we multiply the numerators and multiply the denominators. This simplifies to root three 𝑥 squared over four.

So we have our expression for the area of the first equilateral triangle in terms of the side length of the square. And now, we need to consider the second.

Now, this triangle doesn’t have a side in common with the side of a square. Instead, it has a side in common with the diagonal of the square. So we need an expression for the length of the square’s diagonal.

Let’s consider the triangle formed by two sides of the square and the diagonal of the square. As this side is a right-angled triangle, we can apply Pythagoras’s theorem in order to find an expression for the length of the diagonal.

Remember Pythagoras’s theorem tells us that in a right-angled triangle the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. In this triangle, the two shorter sides are both 𝑥 as they’re the side length of the original square and the hypotenuse I’ve labelled as 𝐿. So we have 𝐿 squared is equal to 𝑥 squared plus 𝑥 squared. This just simplifies to 𝐿 squared is equal to two 𝑥 squared.

Now, we could solve this equation by square rooting to find 𝐿, exactly in terms of 𝑥. However, let’s look back at the area formula that we’re going to use. This uses the side length of the triangle squared. So we don’t actually need to know what 𝐿 is because we’re going to be using 𝐿 squared anyway.

So we can substitute the value of 𝐿 squared which is two 𝑥 squared into our area formula. And it gives the area of triangle two is a half multiplied by two 𝑥 squared multiplied by sin of 60 degrees. Again, remember that sin of 60 degrees is exactly equal to root three over two. So the area of the second equilateral triangle becomes a half multiplied by two 𝑥 squared multiplied by root three over two. This simplifies to two root three 𝑥 squared over four.

Now, we could cancel a factor of two from the two in the numerator and the four in the denominator. But remember what we’re being asked to prove is that the area of the first equilateral triangle is half the area of the second equilateral triangle.

If instead we take this factor of two out of our expression for the area of triangle two, we have that the area of triangle one is root three 𝑥 squared over four and the area of triangle two is two root three 𝑥 squared over four.

So we can say that the area of triangle two is twice the area of triangle one or phrased in other way the area of triangle one is a half of the area of triangle two, which is what we were asked to prove in the question.

Now, just a quick note, it is of course best if you can remember the exact value of sin of 60 degrees and substitute this into your two areas. However, if you couldn’t, then you’d still have a factor of sin of 60 degrees in each of the areas. And as they both have this factor, you would still have been able to prove that the area of triangle one is half the area of triangle two.

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