Question

in a trapezium ABCD , seg AB ll Seg DC , seg BD perpendicular Seg AD , set AC perpendicular BC . If AD = 15 , BC = 15 and AB = 25 . Find area if quadrilateral ABCD

Answer 1

we can find out the area of triangle ADB
so height for this triangle wud be equal to the height of trapezium.

so we get height = 12cm.
now we can easily get to CD=7cm

and we know area of trapezium =1/2*(AB+CD)*HEIGHT

NOW,1/2*(25+7)*12 = 192 cm.

how to get CD=7 cm

when we will drop two perpendicular lines from D to AB as point E and from C to AB as point F.

So,we can see that EF = CD (as the line drawns were perpendicular)

Now In triangleADE,
We know AD=15 and DE=12
So we can easily find AE = 9

Similarly from triangleBCF,
we will get BF = 9cm

therefore,EF = AB -(AE+BF) = 25 - 18 =7cm.

Hope it Helps.:)

Answer 2

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