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: Understanding the Concept:
When the non-parallel sides of a trapezium are extended to meet, similar triangles are formed. In this case, \( \triangle \text{ECD} \) is similar to \( \triangle \text{EAB} \).
Step 2: Key Formula or Approach:
Similarity ratio: \( \frac{CD}{AB} = \frac{EC}{EB} = \frac{ED}{EA} \).
Step 3: Detailed Explanation:
Given \( AB = 2 \) and \( CD = 1 \).
Ratio of similarity \( = \frac{CD}{AB} = \frac{1}{2} \).
Since \( \frac{ED}{EA} = \frac{1}{2} \), D is the midpoint of EA. Thus, \( ED = DA \).
Similarly, since \( \frac{EC}{EB} = \frac{1}{2} \), C is the midpoint of EB. Thus, \( EC = CB \).
Perimeter of \( ABCD = AB + BC + CD + DA = 6 \).
Substituting known values: \( 2 + BC + 1 + DA = 6 \Rightarrow BC + DA = 3 \).
Perimeter of \( \triangle \text{AEB} = AB + BE + EA \).
\( = AB + (EC + CB) + (ED + DA) \).
Since \( EC = CB \) and \( ED = DA \), we have:
Perimeter \( = AB + 2(CB) + 2(DA) = AB + 2(BC + DA) \).
Perimeter \( = 2 + 2(3) = 8 \) cm.
Step 4: Final Answer:
The perimeter of \( \triangle \text{AEB} \) is 8 cm.