Question Video: Finding the Area of a Quadrilateral Using the Relation between Quadrilaterals Sharing the Same Base That Lie between Parallel Lines | Nagwa Question Video: Finding the Area of a Quadrilateral Using the Relation between Quadrilaterals Sharing the Same Base That Lie between Parallel Lines | Nagwa

Question Video: Finding the Area of a Quadrilateral Using the Relation between Quadrilaterals Sharing the Same Base That Lie between Parallel Lines Mathematics • Second Year of Preparatory School

Given that the area of 𝐷𝑀𝑌𝐶 = 68 cm² and 𝐵𝑋 = 𝐶𝑌, find the area of 𝐴𝑀𝑋𝐵.

02:35

Video Transcript

Given that the area of 𝐷𝑀𝑌𝐶 is equal to 68 square centimeters and 𝐵𝑋 is equal to 𝐶𝑌, find the area of 𝐴𝑀𝑋𝐵.

We’re given information about the area of a quadrilateral in our shape. So let’s begin by highlighting that quadrilateral. It’s this one. And we’re being asked to find the area of 𝐴𝑀𝑋𝐵, which is this one. And so let’s begin by looking at our shape. It’s a quadrilateral with one pair of parallel sides. So that tells us it’s a trapezium. We might also notice that the line segments 𝐴𝐶 and 𝐵𝐷 are the diagonals of this trapezium or trapezoid.

Now, what happens is that when we divide our trapezoid into four triangles by using its diagonals, the area of the two triangles opposite one another whose bases are not the parallel sides are equal. In other words, the area of triangle 𝐷𝑀𝐶 must be equal to the area of triangle 𝐴𝑀𝐵. But that’s not good enough. We still have a little bit of our shape 𝐷𝑀𝑌𝐶 left over. And so we recall that if two triangles share the same vertex and then have bases of equal lengths on the same line, then their areas must be equal.

If we consider vertex 𝑀 in the middle of our triangle, we can then see that there are two triangles coming off of vertex 𝑀 who have bases of equal length on the same line. Those bases are made up of line segments 𝐶𝑌 and 𝑋𝐵. And so this means that the area of triangle 𝐶𝑀𝑌 has to be equal to the area of 𝑋𝑀𝐵. And we can now consider the fact that the quadrilateral 𝐷𝑀𝑌𝐶 is made up of the triangles 𝐷𝑀𝐶 and 𝐶𝑀𝑌. The area of 𝐷𝑀𝑌𝐶 is 68 square centimeters. So the sum of the area of triangle 𝐷𝑀𝐶 and 𝐶𝑀𝑌 is also 68 square centimeters.

If we replace the area of triangle 𝐷𝑀𝐶 with the area of triangle 𝐴𝑀𝐵, because we said that they’re the same, and similarly triangle 𝐶𝑀𝑌 with triangle 𝑋𝑀𝐵, we see that the sum of the areas of triangle 𝐴𝑀𝐵 and 𝑋𝑀𝐵 must also be equal to 68 square centimeters. But we also said that 𝐴𝑀𝑋𝐵, the quadrilateral we highlighted earlier, is made up of these two triangles. And so that the area of 𝐴𝑀𝑋𝐵 is the same as the area of 𝐷𝑀𝑌𝐶. It’s 68 square centimeters.

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