# Video: CBSE Class X • Pack 1 • 2018 • Question 13B

CBSE Class X • Pack 1 • 2018 • Question 13B

04:48

### Video Transcript

If 𝐴 negative five, seven; 𝐵 negative four, negative five; 𝐶 negative one, negative six; and 𝐷 four, five are the vertices of a quadrilateral, find its area.

Here, we have plotted the vertices of the quadrilateral: 𝐴, 𝐵, 𝐶, and 𝐷. However, now, we don’t need the axis. So now, we’ve created our quadrilateral. And the way that we can find the area of this quadrilateral is to create a rectangle and find the area of the rectangle and then subtract all of the outside triangles that it makes.

Let’s actually plot this picture. So here’s a quick sketch of a rectangle, where the quadrilateral is completely inside of it. So if we would find the area of the rectangle and then subtract the area that’s not inside of the quadrilateral, we would have the area of the quadrilateral itself.

So first, let’s get some dimensions. How long would this be? Well, it would be left and right. So it’s 𝑥-values. So it would be from negative five to positive four. The difference between them is nine, which means this distance will be nine. Now, these heights would be on the 𝑦-axis. And they go from seven to negative six. And the distance between those values would be 13.

So this means that we have a rectangle that is nine by 13. This means the area of the rectangle will be nine times 13, which is 117. So now, let’s break up the area that’s not inside of the quadrilateral, but is inside of the rectangle and break it up into triangles.

We have this triangle, this triangle, this triangle, this triangle, and this small rectangle. So let’s label these pieces: one, two, three, four, and five. Therefore, to find the area of quadrilateral 𝐴𝐵𝐶𝐷, we need to take the area of the rectangle and then subtract the area of triangle one, area of triangle two, area of triangle three minus area of triangle four minus the area of rectangle five.

So we already know the area of the rectangle: it’s 117. So now, let’s find the dimensions of these triangles and the one rectangle. For triangle number one, we know this side will be nine. And then, from five to seven, found here and here, their difference is two.

So the area of triangle one would be one-half times nine times two. This length would be 13 minus two. So we have 11. This length will go from four to negative one on the 𝑥-axis. So that will give us a length of five. So the area of triangle number two will be one-half times 11 times five.

For triangle number three, this length will go from negative one to negative four. So it’s three. And this length will go from negative five to negative six, so one. And now for triangle number four, this length would be 13 minus one, so 12. And now, this side of that triangle would be nine — the whole side — minus five minus three. So it’s one. So its area would be one-half times 12 times one.

And now, this little rectangle is a one by one because the sides of it are one and one. So the area of a rectangle is length times width. So it’s just one times one. So evaluating, we have 117 minus nine because the twos cancel minus fifty-five halves minus three-halves minus six because 12 divided by two is six minus one because one times one is one. So 117 minus nine minus six minus one is equal to 101. And then negative fifty-five halves minus three-halves would be negative fifty-eight halves.

So to combine these, we need to make 101 have a denominator of two. So we need to multiply 101 times two over two. So we have two hundred and two halves minus fifty-eight halves, which is equal to one hundred and forty-four halves, which is then equal to 72. We also could have reduced fifty-eight halves to be 29. And 101 minus 29 would still get us 72.

So the area of this quadrilateral will be 72. Now, they don’t give us any units. So we simply say the area of this quadrilateral would be 72 units squared or 72 square units.