Question

If perimeter of rhombus is 40 and whose one diagonal is 16 cm find out the length of other diagonal and area of the Rhombus

Answer 1

Answer:

Let the sides of the rhombus = x

OB = 16/2 =8

perimeter = 40

=> 4x = 40

=> x= 40/4 =10

InΔ AOB , by Pythagoras theorem

(AD)$^{2}$ + (OB)$^{2}$= AO$^{2}$

(10)$^{2}$ +(8)$^{2}$ =AO$^{2}$

AO = $\sqrt{100-64} \\=\sqrt{36} =6$

AO= 6+6= 12cm

A  rhombus has 4 edges of equal length if it has perimeter of 40cm

then each length would be = 40cm

=> 40 /4 =10cm

Divide the rhombus into 4 right - angle triangle by cutting along 2 diagonals.

each right angle triangle will be have a hypotenuse 10cm and one other edge with length.

12 /2 = 6

$\sqrt{(10)^{2}-(6)^{2}  } =8cm$

now the find area of triangle

$\frac{1}{2}$ ×6×8

=> area of triangle of 24cm$^{2}$

the area of rhombus be

24×4 = 96cm$^{2}$

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@GauravSaxena01