The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is

#### Options

28 cm

^{2}48 cm

^{2}96 cm

^{2}24 cm

^{2}

#### Solution

**Given:** Rhombus with diagonals measuring 16cm and 12 cm.

**To find:** Area of the figure formed by lines joining the midpoints of the adjacent sides.

**Calculation: **We know that, ‘**Area of a rhombus is half the product of their diagonals’. **

H and F are the midpoints of AD and BC respectively.

`AH = 1/2 AD `

`BE = 1/2 BC`

Now ABCD is a parallelogram which means

`AD = BC `

`1/2AD = 1/2BC`

AH = BF ……..(1)

Also , AD || BC

AH || BF ……(2)

From 1 and 2 we get that ABFH is a parallelogram.

Since Parallelogram FHAB and ΔFHE are on the base FH and between the same parallels HF and AB.

∴` ar (Δ FHE ) = 1/2 ar ( "||"^(gm) FHAB )` ……(3)

Similarly ,

`ar (ΔFHG) = 1/2ar ("||"^(gm) FHDC)` ……(4)

Adding 3 and 4 we get,

`ar (Δ FHE ) + ar (ΔFHG) = 1/2 ar ("||"^(gm) FHAB)+1/2ar("||"^(gm)FHDC)`

`ar (EFGH) = 1/2 (ar("||"^(gm) FHAB ) + ar ("||"^(gm) FHDC))`

`ar (EFGH) = 1/2 (ar("||"^(gm) ABCD))`

`ar (EFGH) = 1/2 (1/2 (16xx12))`

`ar (EFGH) = 1/4 (16 xx 12)`

`ar (EFGH) = 48 cm^2`