Answer 1

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`