Answer:
Diagonals of a rhombus bisect each other at right angles.
Let ABCD be the rhombus, Diagonal AC=16 cm and side AB=10 cm.
In right △AOB, AB=10 cm, AO=8 cm
By Pythagoras theorem, AB
2
=AO
+BO
⇒10
=8
⇒BO
=100−64=36
∴BO=6cms
Diagonal BD=2×6=12cm
Area of Rhombus=
1
×(product of the diagonals)
=
×16×12=8×12=96sq.cm
Hence length of other diagonal =12 cm;
And Area of rhombus =96sq.cm
please brainliest mark
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Answers & Comments
Answer:
Diagonals of a rhombus bisect each other at right angles.
Let ABCD be the rhombus, Diagonal AC=16 cm and side AB=10 cm.
In right △AOB, AB=10 cm, AO=8 cm
By Pythagoras theorem, AB
2
=AO
2
+BO
2
⇒10
2
=8
2
+BO
2
⇒BO
2
=100−64=36
∴BO=6cms
Diagonal BD=2×6=12cm
Area of Rhombus=
2
1
×(product of the diagonals)
=
2
1
×16×12=8×12=96sq.cm
Hence length of other diagonal =12 cm;
And Area of rhombus =96sq.cm
please brainliest mark