# CLASS-9AREA OF A QUADRILATERAL WHERE DIAGONALS INTERSECTS AT RIGHT ANGLES

AREA OF A QUADRILATERAL WHERE DIAGONALS INTERSECTS AT RIGHT ANGLES -

When diagonals of a quadrilateral interest intersect at right angles.

Let the diagonals AC & BD of quad ABCD intersect at O at right angle.

Then area of quad ABCD =

= Area (ABC) + Area (ACD)

1                         1

= (------ X AC X BO) + (------ X AC X OD)

2                         2

1

= ------- X AC X (BO + OD)

2

1

= ------- X AC X BD

2

1

= ------- X (Product of the Diagonals)

2

Example.) Find the area of the quadrilateral whose sides are 10 m, 50 m, 30 m, and 25 m respectively and the angle between first two sides is a right angle.

Ans.) Let ABCD be the given quadrilateral in which AB = 10 m, BC = 50 m, CD = 25 m, DA = 20 m, and ABC = 90⁰

By Pythagoras theorem, we have –

AC = √AB² + BC²

= √10² + 50² = √100 + 2500

= √2600 = 50.99 = 51 m

1

Area of ABC = ------- X AB X BC

2

1

= ------- X 10 X 50 m² = 250 m²

2

So, In ACD,

AC = a = 51 m

CD = b = 30 m, and

DA = c = 25 m

1                        1

Then, s = ------- (a + b + c) = ------- (51 + 30 + 25)

2                        2

106

= ------- = 53 m

2

So, now =>  (s – a) = (53 – 51) = 2 m

(s – b) = (53 – 30) = 23 m

(s – c) =  (53 – 25) = 28 m

So, the area ACD = √s(s – a) (s – b) (s – c)

= √53 X 2 X 23 X 28 m²

= √68264 m² = 261.27 m² = 261 m²

So, the area of ABCD = Area (ABC) + Area (ACD)

= (250 + 261) m²= 511 m²

so, the area of quadrilateral is 511 m² (Ans.)