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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.`
3 years ago

```							Let abcd be a rhombus.Side of rhombus=6 cmAltitude 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 cm2Half into d1 ×d2 =24d1×d2=24×28× d2=48d2 =48 upon 8d2=6cm
```
3 years ago
```							Divide the rhombus into two triangles:-Area of triangle= ½ * b * h ½ * b * h + ½ * b * h½ * 6 * 4 + ½ *6 * 4= 12 +1224 sq. cmarea of rhombus = ½ * product of its diagonals 24 = ½ * 8 * d224= 4 * d2d2 = 6 cm
```
2 years ago
```							first we will divide the rhombus into two triangleas rhombus abcd in triangle abc and triangle cdaarea of rhombus = area of triangle abc and area of triangle cdaarea of triangle abc= ½ * 6 * 4                              = 12 cm area of triangle cda = ½ * 6 * 4                              = 12 cmon adding                   12cm + 12cm = 24cmhence the area of rhombus is 24cm so now by appliying the formula we can find the diagonald1 = 8 cmarea = 24cmarea of rhombus = ½ * d1 * d2                          24  = ½ * 8 * d2                        24*2/8   = d2                        6 cm = d2hence the other diagonal is of 6 cm
```
2 years ago
```							Area of rhombus=altitude*side                              =6*4sq.m                              =24sq.mArea of rhombus=half into (product of diagonal)                     24     =1\2*8*x                       X      =3m
```
one year ago
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