Question

Find the area of the quadrilateral ABCD given in Fig . The diagonals AC and BD measure 48 m respectively and are perpendicular to each other​

Answer 1

Step-by-step explanation:

Area of rhombus= 1/2*d¹*d²

= 1/2*48

= 24

Answer 2

Area of ABCD is 1152 m².

Step-by-step explanation:

It is given that,

AC= 48 m

BD = 48 m

Let, ‘O’ is an imagenery point at which AC and BD bisects each other.

Hence,

AO = (48 *1/2)

      = 24 m

OC = (48*1/2)

    = 24 m

OD = (48*1/2)

     = 24 m

OB = (48*1/2) m

    = 24 m

For Δ AOD

AO = 24m

OD = 24 m

∴ Area = 1/2 * (24*24) [By pythagoras theorem]

           = 12*24

          = 288 m²

For Δ DOC

DO = 24m

OC = 24 m

∴ Area = 1/2 * (24*24) [By pythagoras theorem]

           = 12*24

          = 288 m²

For Δ COB

CO = 24m

OB = 24 m

∴ Area = 1/2 * (24*24) [By pythagoras theorem]

           = 12*24

          = 288 m²

For Δ AOB

AO = 24m

OB = 24 m

∴ Area = 1/2 * (24*24) [By pythagoras theorem]

           = 12*24

          = 288 m²

∴ Area of quadrilateral ABCD =area of Δ AOD +area of Δ DOC +area of Δ COB +area of Δ AOB

= (288+288+288+288) m²

= 1152 m²

∴ Area of ABCD is 1152 m².