ZYXW is a trapezium with . is the midpoint of . Show that the area of the trapezium ZYXW is equal to the area of .

Understand the Question In trapezium , the sides , and is the midpoint of . The line segment is extended to meet the extended line at point . To show that , we: 1. Prove using the ASA congruence criterion. 2. Since congruent triangles have equal areas, adding the common region (quadrilateral ) to both triangles equates the area of trapezium with the area of . Step 1 · Prove Consider and . Since is the midpoint of Since (or ) with transversal , alternate interior angles are equal Vertically opposite angles are equal By the Angle-Side-Angle (ASA) congruence criterion Step 2 · Equate Areas of Trapezium and Triangle Since congruent triangles have equal areas Area of trapezium is the sum of quadrilateral and Area of is the sum of quadrilateral and Substituting Therefore Answer > Common Mistakes Alternate Interior Angles Error: Forgetting that holds because lies on the line extension of side . Congruence Criteria Confusion: Incorrectly citing SAS instead of ASA; the equal side lies strictly between the two equal pairs of angles (alternate interior and vertically opposite angles).

Area of Polygons | FIO

Question 36

ZYXW is a trapezium with $ZY \parallel WX$ . $A$ is the midpoint of $XY$ . Show that the area of the trapezium ZYXW is equal to the area of $\triangle ZWB$ .

Solution

Understand the Question

In trapezium $ZYXW$ , the sides $ZY \parallel WX$ , and $A$ is the midpoint of $XY$ .
The line segment $ZA$ is extended to meet the extended line $WX$ at point $B$ .
To show that $\text{Area}(\text{trapezium } ZYXW) = \text{Area}(\triangle ZWB)$ $Area (trapezium Z Y X W) = Area (△ Z W B)$ , we:
1. Prove $\triangle ZAY \cong \triangle BAX$ using the ASA congruence criterion.
2. Since congruent triangles have equal areas, adding the common region (quadrilateral $ZWXA$ ) to both triangles equates the area of trapezium $ZYXW$ with the area of $\triangle ZWB$ .

Step 1 · Prove $\triangle ZAY \cong \triangle BAX$

Consider $\triangle ZAY$ and $\triangle BAX$ . Question diagram Diagram 1

Since $A$ is the midpoint of $XY$ $AY = AX$

Since $ZY \parallel WX$ (or $ZY \parallel XB$ ) with transversal $XY$ , alternate interior angles are equal $\angle ZYA = \angle BXA$

Vertically opposite angles are equal $\angle ZAY = \angle BAX$

By the Angle-Side-Angle (ASA) congruence criterion $\triangle ZAY \cong \triangle BAX$

Step 2 · Equate Areas of Trapezium and Triangle

Since congruent triangles have equal areas $\text{Area}(\triangle ZAY) = \text{Area}(\triangle BAX)$

Area of trapezium $ZYXW$ is the sum of quadrilateral $ZWXA$ and $\triangle ZAY$ $\text{Area}(\text{trapezium } ZYXW) = \text{Area}(\text{quadrilateral } ZWXA) + \text{Area}(\triangle ZAY)$

Area of $\triangle ZWB$ is the sum of quadrilateral $ZWXA$ and $\triangle BAX$ $\text{Area}(\triangle ZWB) = \text{Area}(\text{quadrilateral } ZWXA) + \text{Area}(\triangle BAX)$

Substituting $\text{Area}(\triangle ZAY) = \text{Area}(\triangle BAX)$ $\text{Area}(\text{trapezium } ZYXW) = \text{Area}(\text{quadrilateral } ZWXA) + \text{Area}(\triangle BAX)$

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

Answer

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

Common Mistakes

Alternate Interior Angles Error: Forgetting that $ZY \parallel XB$ holds because $B$ lies on the line extension of side $WX$ .
Congruence Criteria Confusion: Incorrectly citing SAS instead of ASA; the equal side $AY = AX$ lies strictly between the two equal pairs of angles (alternate interior and vertically opposite angles).

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Q19

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Q36

ZYXW is a trapezium with $ZY \parallel WX$ . $A$ is the midpoint of $XY$ . Show that the area of the trapezium ZYXW is equal to the area of $\triangle ZWB$ .