Find area of rhombus each side is measure 20cm and one of whose diagonal is 24 cm
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Answered by
5
diagonal= √4a²-24²
=√4×20²-24²
=√1600-576
=√1024
=32
area =pq/2 (p and q are diagonals)
=32×24÷2
=384 Sq.cm
=√4×20²-24²
=√1600-576
=√1024
=32
area =pq/2 (p and q are diagonals)
=32×24÷2
=384 Sq.cm
Anonymous:
i hope this explaination is sufficient for u
Answered by
3
Hey there!
Consider ABCD to be the required rhombus ,
BD = 24 cm.
AB= BC = CD = DA = 20 cm.
Now, we need
CO= 1/2CA
Now, Consider triangle COD,
In this triangle, we know that
CD = 20 cm.
DO = 12cm
Now , This is a right angled triangle.
Accordance to Pythagoras theorem,
CD² = DO² + CO²
20² = 12² + CO²
400-144 = CO²
CO = √256 = 16cm .
2CO = AC
AC = 2(16) = 32cm .
Now,
We have another diagonal = 32cm .
Finally, We have two diagonals 32 cm, 24cm
Area of Rhombus = 1/2d*D = 1/2 * (32)(24)
= 16 * 24
= 384 cm²
Consider ABCD to be the required rhombus ,
BD = 24 cm.
AB= BC = CD = DA = 20 cm.
Now, we need
CO= 1/2CA
Now, Consider triangle COD,
In this triangle, we know that
CD = 20 cm.
DO = 12cm
Now , This is a right angled triangle.
Accordance to Pythagoras theorem,
CD² = DO² + CO²
20² = 12² + CO²
400-144 = CO²
CO = √256 = 16cm .
2CO = AC
AC = 2(16) = 32cm .
Now,
We have another diagonal = 32cm .
Finally, We have two diagonals 32 cm, 24cm
Area of Rhombus = 1/2d*D = 1/2 * (32)(24)
= 16 * 24
= 384 cm²
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