Find the area of Rhombus whose one side is 13 cm and one diagonal is 26 cm.
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hi
there is some mistake in your question.
The length if diagonal need to be 24cm.
here is your answer
Area of rhombus = 1/2× product of diagonals.
now, As diagonal of rhombus bisect each other.
Hence,by Pythagoras theorem,
12^2+ oc^2= 13^2
= 144+ OC^2 = 169
=OC= 5 cm
and AC = 5×2=10cm
Now area = 1/2× 10 × 24
= 120cm^2
hope it helps
plz mark brainliest answer
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A=pq/2
a=√p^2+q^2/2
Solving forA
A=1/2p√4a^2﹣p^2
= 1/2*24*√4*13^2﹣24^2
=120cm²
a=√p^2+q^2/2
Solving forA
A=1/2p√4a^2﹣p^2
= 1/2*24*√4*13^2﹣24^2
=120cm²
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