Question

ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of $\triangle$AEB is:

• A. 6
• B. 7
• C. 8
• D. 9
• E. 10

Hint:

In such geometry problems with a trapezium and intersecting non-parallel sides, use the similarity of the two triangles formed to relate the sides.

Answer 1

The Correct Option is C

Step 1: Understand the Geometry:
In trapezium ABCD, AB $\parallel$ CD. Extending AD and BC meets at E, forming triangle AEB. Since AB $\parallel$ CD, $\triangle$EDC $\sim$ $\triangle$EAB. This is the key similarity.
Step 2: Relate the Sides:
Let the perimeter of $\triangle$AEB be $P$. From similarity, the ratio of corresponding sides is constant. \[ \frac{CD}{AB} = \frac{1}{2} \] This means all sides of $\triangle$EDC are half of the corresponding sides of $\triangle$EAB.
Step 3: Express the Perimeter of ABCD:
Perimeter of trapezium ABCD = AB + BC + CD + DA = $2 + BC + 1 + DA = 6$ (given). Thus, $BC + DA = 3$.
Step 4: Relate to $\triangle$AEB:
Let $AE = x$ and $BE = y$. Then $DA = x - ED$ and $BC = y - EC$. Since $\triangle$EDC has sides proportional, $ED = \frac{1}{2} EA = \frac{x}{2}$ and $EC = \frac{1}{2} EB = \frac{y}{2}$. So, $DA = x - \frac{x}{2} = \frac{x}{2}$ and $BC = y - \frac{y}{2} = \frac{y}{2}$. From Step 3, $BC + DA = \frac{x}{2} + \frac{y}{2} = 3$, which implies $x + y = 6$.
Step 5: Find the Perimeter of $\triangle$AEB:
Perimeter of $\triangle$AEB = $AB + AE + BE = 2 + x + y = 2 + 6 = 8$.
Step 6: Final Answer:
The perimeter of $\triangle$AEB is 8 cm.