If the measutes of one side and one diagonal of a rhombus are 10cm and 16cm resp then its area in
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area is diagonal 1 into diagonal 2
If side is 10 cm , it remains same so the other diagonal is also 16 cm.
So, the AREA = 16 * 16 = 256 cm^2
If side is 10 cm , it remains same so the other diagonal is also 16 cm.
So, the AREA = 16 * 16 = 256 cm^2
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Solution:-
let, Rhombus ABCD.
given:-
•The sides of a rhombus are 10cm and one diagonal is 16cm .
let, DO = OB = 8 cm, BD = 16 cm
and AO = OC = ?
1) we know diagonal of rhombus are equally bisect each other and they are perpendicular to each other.
2) All sides of rhombus are equal.
so,
by Pythagoras theorem.
=> (AB)² = ( AO )² + ( OB )²
=> (10)² = (AO)² + (8)²
=> 100 = (AO)² + 64
=> 100 - 64 = (AO)²
=> 36 = (AO)²
i.e.
=> (AO)² = 36
=> AO = √36
=> AO = 6 cm
so, we know AO = OC = 6 cm
hence, AC = 12 cm
Area of rhombus = [(AC)×(BD)]/2
Area of rhombus = [ 12 × 16]/2
Area of rhombus = [192]/2
Area of rhombus= 99.84 cm²
Hence area of rhombus is
96 cm².
i hope it helps you.
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