ZYXW is a trapezium with ZY || WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of .

Question

Homi · Accepted Answer

The key idea is to show that two small triangles, $\triangle ZAY$ and $\triangle BAX$ , have the same area, and then use this to relate the areas of the trapezium and the larger triangle.

Step 1 — Identify congruent triangles

Let us look at the two triangles, $\triangle ZAY$ and $\triangle BAX$ .

We are given that A is the midpoint of the line segment XY. This means that the length of AY is equal to the length of AX. $AY = AX$

The lines ZY and WX are parallel because ZYXW is a trapezium. Since WX is part of the line WB, this means that ZY is parallel to XB. When two parallel lines (ZY and XB) are cut by a transversal line (XY), the alternate interior angles are equal. So, $\angle ZYA$ is equal to $\angle BXA$ . $\angle ZYA = \angle BXA$

Also, when two straight lines (ZB and XY) intersect, the angles opposite to each other at the intersection point are called vertically opposite angles. These angles are always equal. So, $\angle ZAY$ is equal to $\angle BAX$ . $\angle ZAY = \angle BAX$

Now we have two angles and the included side of $\triangle ZAY$ equal to two angles and the included side of $\triangle BAX$ . This means that $\triangle ZAY$ is congruent (exactly the same shape and size) to $\triangle BAX$ by the Angle-Side-Angle (ASA) congruence criterion.

$\boxed{\triangle ZAY \cong \triangle BAX \text{ (by ASA congruence)}}$

Diagram 1

Step 2 — Relate the areas of the trapezium and the triangle

Since $\triangle ZAY$ and $\triangle BAX$ are congruent triangles, their areas must be equal. $\text{Area}(\triangle ZAY) = \text{Area}(\triangle BAX)$

Now, let us consider the area of the trapezium ZYXW. The trapezium ZYXW can be seen as the sum of the area of the quadrilateral ZWXA and the area of the triangle ZAY. $\text{Area}(\text{trapezium ZYXW}) = \text{Area}(\text{quadrilateral ZWXA}) + \text{Area}(\triangle ZAY)$

Next, let us consider the area of the large triangle $\triangle ZWB$ . The triangle $\triangle ZWB$ can also be seen as the sum of the area of the quadrilateral ZWXA and the area of the triangle BAX. $\text{Area}(\triangle ZWB) = \text{Area}(\text{quadrilateral ZWXA}) + \text{Area}(\triangle BAX)$

From the previous step, we know that $\text{Area}(\triangle ZAY) = \text{Area}(\triangle BAX)$ . Let's substitute $\text{Area}(\triangle BAX)$ in place of $\text{Area}(\triangle ZAY)$ in the equation for the trapezium's area. $\text{Area}(\text{trapezium ZYXW}) = \text{Area}(\text{quadrilateral ZWXA}) + \text{Area}(\triangle BAX)$

Now, we can see that the expression for the area of the trapezium is exactly the same as the expression for the area of $\triangle ZWB$ .

$\boxed{\text{Area}(\text{trapezium ZYXW}) = \text{Area}(\triangle ZWB)}$

Answer

The area of the trapezium ZYXW is equal to the area of $\triangle\text{ZWB}$ .

Question 36

Answer

More questions in FIO