FIND THE AREA OF A RHOMBUS EACH SIDE OF WHICH MEASURES 20cm AND ONE OF WHOSE DIAGONALS IS 24cm
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Area of Rhombus =2(area of isosceles triangle)
2×[b/4underroot 4a^2-b^2]
where a=20 cm and b =24cm
Now put the value and get the result
2×[b/4underroot 4a^2-b^2]
where a=20 cm and b =24cm
Now put the value and get the result
Answered by
3
Answer:
A = 384cm²
a = Side = 20cm
p = Diagonal = 24 cm
a = pq / 2
a = p² + q² / 2
a = 1 / 2 p √4² - p²
= 1 / 2 • 24 • √ 4.20² - 24² = 384 cm³
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