Question

3. In a trapezium ABCD, seg AB || seg DC, seg BD perpendicular seg AD, seg AC
perpendicular seg BC, if AD = 15, BC = 15 and AB = 25. Find A ( []ABCD).​

Answer 1

Answer:

we have find out height so we construct DE ⊥ AB and CF ⊥ AB

In ΔADB, as BD ⊥ AD,

By Pythagoras theorem...

(AB)2 = (AD)2 + (BD)2

252 = 152 + (BD)2

(BD)2 = 625 - 225 = 400

BD = 20 cm.

Similarly,

AC = 20 cm.

Now, In ΔAED and ΔABD

∠AED = ∠ADB [Both 90°]

ΔAED ~ ΔABD [By Angle-Angle Criteria]

therefore, DE/BD=AD/AB=AE/AD..... ( By property of similar triangles)

As we know (given), AD = 15 cm,

BD = 20 cm

and AB = 25 cm

So, DE/20=15/25

therefore,DE = 12 cm

Also, DE/BD=AE/AD

Therefore, 12/20=AE/15

so, AE=9cm.

Similarly, BF = 9 cm

Now,

DC = EF [By construction]

DC = AB - DE - AE

DC = 25 - 9 - 9 = 7 cm

Also, we know

Area of trapezium= 1/2×(sum of parallel sides)×height.

=1/2×(7+25)×12

=192 cm.sq

Therefore, Area of trapezium = 192cm.sq

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