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

We need to find the perimeter of triangle AEB.

Perimeter of ABCD = 6 cm

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Since the perimeter of trapezium ABCD is 6 cm:

$AB + BC + CD + DA = 6$

$2 + BC + 1 + DA = 6$

$BC + DA = 3$ cm

Let's call $BC = y$ and $DA = x$. So: $x + y = 3$ ... (equation 1)

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When we extend the non-parallel sides of any trapezium, we always create similar triangles! The triangle AEB and DEC are similar. Why?

As $AB \parallel CD$, we get:

$\angle AEB = \angle CED$ (they're the same angle at point E)

$\angle EAB = \angle ECD$ (these are corresponding angles formed by parallel lines)

$\angle EBA = \angle EDC$ (these are also corresponding angles)

When two triangles have all three angles equal, they are similar triangles.

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For similar triangles, all corresponding sides are in the same ratio.

$AB = 2$ cm is two times $CD = 1$ cm: This means triangle EAB is exactly twice as large as triangle ECD in all measurements.

Therefore:

$\dfrac{EA}{ED} = 2$, which means $EA = 2 \times ED$

$\dfrac{EB}{EC} = 2$, which means $EB = 2 \times EC$

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Let's work with $EA = 2 \times ED$:

Since point D lies on line segment AE, we can write:

$EA = ED + DA$

Substituting our similarity relationship:

$ED + DA = 2 \times ED$

$2 \times ED - ED = DA$

$ED = DA = x$

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Similarly, for $EB = 2 \times EC$:

$EB = EC + CB$

$2 \times EC = EC + CB$

$EC = CB = y$

C divides BE in the ratio 1:1, so EC = CB = y.

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Now we can find each side of triangle AEB:

$AE = ED + DA = x + x = 2x$

$BE = EC + CB = y + y = 2y$

$AB = 2$ cm (given)

Perimeter of triangle AEB = $AE + BE + AB$

$= 2x + 2y + 2$

$= 2(x + y) + 2$

From equation 1, we know $x + y = 3$:

$= 2(3) + 2 = 6 + 2 = 8$

The perimeter of triangle AEB is 8 cm.