Question

ABCD is a trapezium in which AD is parallel to BC and AB=6cm,BC=7cm,CD=8cm,AD=17cm.If sides AB and DC are extended to meet at E ,then the resulting angle BEC

Answer 1

Answer:

ABCD is a trapezium in which AD is parallel to BC and AB=6cm,BC=7cm,CD=8cm,AD=17cm.If sides AB and DC are extended to meet at E ,then the resulting angle BEC.

Answer 2

The measure of ∠BEC is 90°.

Given:

A trapezium with sides AB=6cm, BC=7cm, CD=8cm, AD=17cm.

Sides AB and DC are extended to meet at E.

To Find:

The ∠BEC

Solution:

We have to apply the similarity property in the triangle ΔEAD and ΔEBC

We have ∠E common in both the triangles.

Then we have ∠EAD and ∠EBC equal, since they are corresponding angles.

Similarly, ∠EDA and ∠ECB are corresponding angles and hence equal.

Therefore, the two triangles hold the ASA (Angle-Side-Angle) property of similarities and hence are similar.

Now, the corresponding sides of the similar triangle are in ratio,

$\implies \frac{AD}{BC} = \frac{EA}{EB} \\\\\implies \frac{17}{7} = \frac{6+x}{x} \\\\\implies 17x = 42 + 7x\\\\\implies 10x = 42\\\\\implies x = 4.2$

Now,

$\implies \frac{AD}{BC} = \frac{ED}{EC} \\\\\implies \frac{17}{7} = \frac{y+8}{y}\\\\\implies 17y = 7y +56\\\\\implies 10y = 56\\\\\implies y = 5.6$

Now, if we apply Pythagoras theorem in ΔEBC,

$\implies EB^{2} + EC^{2} = BC^2\\\\\implies x^{2} + y^{2} = 7^2\\\\\implies 4.2^{2} + 5.6^{2} = 49\\\\\implies 17.64 + 31.36 = 49\\\\\implies 49 = 49$

Since, LHS = RHS, therefore we can say ∠BEC is 90°.

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