Question

the lengths of two parallel sides of trapezium are 28cm and 40 cm. If the lengths of each of its other two sides be 12 cm, then find the area of the trapezium
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Answer 1

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Given P is equidistant from points A and B

PA=PB .....(1)

and Q is equidistant from points A and B

QA=QB .....(2)

In △PAQ and △PBQ

AP=BP from (1)

AQ=BQ from (2)

PQ=PQ (common)

So, △PAQ≅△PBQ (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In △PAC and △PBC

AP=BP from (1)

∠APC=∠BPC from (3)

PC=PC (common)

△PAC≅△PBC (SAS congruence)

∴AC=BC by CPCT

and ∠ACP=∠BCP by CPCT ....(4)

Since, AB is a line segment,

∠ACP+∠BCP=180

∘

Thus, AC=BC and ∠ACP=∠BCP=90

∘

∴,PQ is perpendicular bisector of AB.

Hence proved.

Tusen Takk ✌️❣️