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# find the area of the rhombus whose side is 6 cm and whose altitude is 4 cm .if one one of its diagnols is 8 cm long , find the length of other diagnol.

Faith
11 Points
4 years ago
Let abcd be a rhombus.
Side of rhombus=6 cm
Altitude of rhombus=4 cm.
Area of rhombus abcd=area of triangle ABD+area of triangle BCD
=half into b×h+ half into b×h
=half ×6×4+half×6×4
=12 +12
=24 cm2
Half into d1 ×d2 =24
d1×d2=24×2
8× d2=48
d2 =48 upon 8
d2=6cm
Deepraj
11 Points
3 years ago
Divide the rhombus into two triangles:-
Area of triangle= ½ * b * h
½ * b * h + ½ * b * h
½ * 6 * 4 + ½ *6 * 4
= 12 +12
24 sq. cm
area of rhombus = ½ * product of its diagonals
24 = ½ * 8 * d2
24= 4 * d2
d2 = 6 cm
amisha
31 Points
3 years ago
first we will divide the rhombus into two triangle
as rhombus abcd in triangle abc and triangle cda
area of rhombus = area of triangle abc and area of triangle cda
area of triangle abc= ½ * 6 * 4
= 12 cm
area of triangle cda = ½ * 6 * 4
= 12 cm
12cm + 12cm = 24cm
hence the area of rhombus is 24cm
so now by appliying the formula we can find the diagonal
d1 = 8 cm
area = 24cm
area of rhombus = ½ * d1 * d2
24  = ½ * 8 * d2
24*2/8   = d2
6 cm = d2
hence the other diagonal is of 6 cm
Ankit singh
15 Points
3 years ago
Area of rhombus=altitude*side
=6*4sq.m
=24sq.m
Area of rhombus=half into (product of diagonal)
24     =1\2*8*x
X      =3m