Video: CBSE Class X • Pack 3 • 2016 • Question 28

CBSE Class X • Pack 3 • 2016 • Question 28


Video Transcript

In the given figure, the vertices of triangle 𝐴𝐵𝐶 are 𝐴 four, six; 𝐵 one, five; and 𝐶 seven, two. A line segment 𝐷𝐸 is drawn to intersect 𝐴𝐵 and 𝐴𝐶 at 𝐷 and 𝐸, respectively, such that 𝐴𝐷 over 𝐴𝐵 is equal to 𝐴𝐸 over 𝐴𝐶, which is equal to one over three. Calculate the area of triangle 𝐴𝐷𝐸 and compare it with the area of triangle 𝐴𝐵𝐶.

Let’s look at two different methods to solve this problem. The first one we’ll call the box method. This will be helpful if we don’t have a right triangle and we’re only given the vertices of a triangle. The first thing we do is sketch a coordinate plane and then grab these points accurately. 𝐵 is located at point one, five and 𝐴 at four, six, 𝐶 at seven, two.

We then notice that the figure we were given is not drawn to scale or to the correct shape of our triangle. But back to the box method, we wanna create a rectangle around this triangle. To find the area of triangle 𝐴𝐵𝐶, we take the area of the rectangle and then we find the area of these three right triangles. And we subtract the areas of the three right triangles from the area of the rectangle. But how do we do this?

We see that the distance from point 𝐵 to point 𝐶 on the 𝑥-axis is six units and the distance from point 𝐶 to point 𝐵 on the 𝑦-axis is three units. Let’s just move this formula down to give us a little bit more room. Back to the area of the red right triangle, the area of this triangle will be equal to one-half six times three: one-half its height times its base.

And for the yellow triangle, the distance from 𝐶 to 𝐴 on the 𝑦-axis is four units and the distance from 𝐴 to 𝐶 on the 𝑥-axis is three units, which means the area of the yellow triangle equals one-half times three times four. We’ll rewrite that with multiplication just to make it clear.

And finally, we repeat the same process for the pink triangle. The distance from 𝐴 to 𝐵 on the 𝑦-axis is one unit and the distance from 𝐴 to 𝐵 on the 𝑥-axis is three units. The area of the pink triangle is one-half times three times one.

We can’t forget the area of the rectangle. This rectangle has a length of six and a width of four. The area of the rectangle is found by multiplying six times four. Now, we’re ready to solve. Six times four is 24 minus one-half times six times three which equals nine minus one-half times three times four which equals six. One-half times three times one equals three over two or one and a half. 24 minus nine minus six, this part equals nine.

To subtract three-halves from nine, we need to write nine with the denominator of two. The whole number nine written as a fraction with the denominator of two is 18 over two. 18 over two minus three over two equals 15 over two. The area of triangle 𝐴𝐵𝐶 equals 15 over two units squared.

We’re still interested in the area of the smaller triangle, triangle 𝐴𝐷𝐸. And we want to set up a relationship between the area of the small interior triangle, triangle 𝐴𝐷𝐸, and the large outside triangle, triangle 𝐴𝐵𝐶. These two triangles share an angle. On either side of this shared angle, both of their sides are in the same proportion.

And that means triangle 𝐴𝐷𝐸 is similar to triangle 𝐴𝐵𝐶. The ratio of the areas of two similar triangles is their ratio squared. And that means the area of 𝐴𝐷𝐸 over the area of 𝐴𝐵𝐶 will be one over nine. We can use this to solve for the area of 𝐴𝐷𝐸. We know that the area of our larger triangle is 15 over two. And we’re missing the area of our smaller triangle.

By cross multiplying 15 over two times one, we see that 15 over two equals nine times the area of the smaller triangle. After that, I wanna get rid of the nine on the right side. And I can do that by multiplying both sides of the equation by one over nine. On the right side of the equation, we’re left with 𝑆. The numerator of the other side is one times 15 and the denominator is nine times two. The area of the smaller triangle is 15 over 18. Both the numerator and the denominator are divisible by three. 15 divided by three is five. 18 divided by three equals six.

The area of the smaller triangle, triangle 𝐴𝐷𝐸, is five-sixths units squared.

Our question was asking for two things: the area of triangle 𝐴𝐷𝐸, which we found five-sixths units squared, and it wanted us to compare this with the area of triangle 𝐴𝐵𝐶. The comparison looks like this: the area of the smaller triangle over the area of the larger triangle is in the ratio one to nine.

But if you remember all the way at the beginning of the problem, I said I was going to show you two different methods. We looked at the box method. I wanna show you a formula for finding the area of a triangle if you’re given coordinates.

The area 𝐴 𝑥 times 𝐵 𝑦 minus 𝐶 𝑦 plus 𝐵 𝑥 times 𝐶 𝑦 minus 𝐴 𝑦 plus 𝐶 𝑥 times 𝐴 𝑦 minus 𝐵 𝑦 over two then take the absolute value. When it says 𝐴 sub 𝑥, it means the 𝑥-coordinate of the 𝐴 vertex; for us that’s four. And then, we need to multiply it by 𝐵 𝑦; that’s the 𝑦-coordinate of the 𝐵 value and the 𝑦-coordinate of the 𝐶 vertex. We’ll continue with this pattern. 𝐵 𝑥 equals one 𝐶 𝑦 equals two minus six plus seven times six minus five all over two.

If we solve what’s in the parentheses, five minus two equals three, two minus six equals negative four, six minus five equals one. Now, we’ll do the multiplication. Four times three equals 12. One times negative four equals negative four. And seven times one equals seven. 12 plus seven equals 19 minus four equals 15. 15 over two take the absolute value which equals 15 over two.

Using a different method, we’ve confirmed that the area of the larger triangle is 15 over two units. Once you found the area of the larger triangle using this formula, you could find the area of the smaller triangle in the same way by setting up a proportion. You would still find that it was one over nine. And then the area of the smaller triangle would be five over six units squared.

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