Question

In trapezium abcd ab is parallel to dc seg bd is perpendicular to ad and seg ac perpendicular to bc given ad =15, bc=15 and ab=25 find area of trapezium

Answer 1

The quadrilateral can be divided into two parts
Sin bac = 15/25
 
<bac = 36.87
 
Cos bad = 15/25
 
<bad = 53.13
 
<cad = <bac - <bad
 
        = 53.13 – 36.87
 
        = 16.26
 
Ac = root (25^2 – 15^2) = 20
 
Area of abc = ½ x 15 x 20 = 150
 
Area of adc = ½ x ac x ad Sin <cad = ½ x 20 x 15 Sin 16.26
                     = 42
 
Total area = 150 + 42
 
                  = 192
The area of abcd = 192 square units

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