Question

In trapezium ABCD seg BD is perpendicular to seg AD, seg AC is perpendicular to seg BC AD =15 BC=15 CD=7 AB=25 FIND AREA OF QUADRILATERAL A (□ABCD )

Answer 1

Explanation:

Construct DE ⊥ AB and CF ⊥ AB

In ΔADB, as BD ⊥ AD, By Pythagoras theorem i.e.

(Hypotenuse)2 = (base)2 + (Perpendicular)2

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

⇒ 252 = 152 + BD2

⇒ BD2 = 625 - 225 = 400

⇒ BD = 20 cm

Similarly,

AC = 20 cm

Now, In ΔAED and ΔABD

∠AED = ∠ADB [Both 90°]

∠DAE = ∠DAE [Common]

ΔAED ~ ΔABD [By Angle-Angle Criteria]

[Property of similar triangles]

As AD = 15 cm, BD = 20 cm and AB = 25 cm

⇒ DE = 12 cm

Also,

⇒ AE = 9 cm

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

Also, we know

Area of trapezium

$\frac{1}{2} \times (sum \: of \: parallels \: sides) \times height$

= 1/2 × (DC + AB ) × DE

= 1/2 ×(7 + 25) × 12

= 192 cm^2

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