A rhombus has perimeter 100m and one of its diagonals is 40m . Find the area of the rhombus
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Area = ½×product of diagonals. One digonal is given. Find other y using pythagores . Ans =600 m² .. Mark as brainliest.
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bt i have a confusion.. that see perm is 100 ok... so perm= 4×sides of a rhombus...i.e 100=4a 100/4=a..i.e a=25.. means the sides of a rhombus is 25 ok....so by pytha. we can find the another diagonal...ri8..
ohh sorry sorry....now i gotted....for finding the other diagonal i have to take the length of side...bt also the half of other diagonal for being pytha. ri8..i have understand now.
Yep. U r right
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See the figure of the rhombus ABCD above . Let the lengths of the diagonal be d1 and d2. Now let us find the area of rhombus ABCD.
Consider the triangles AOB and AOC. They are congruent(AB = AC, AO is common, angle ABO = angle ACO as triangle ABC is isosceles).
So angle AOB = angle AOC =180/2 = 90. So BO = OC = d2/2 and similarly AO = DO = d1/2.
The area of rhombus ABCD is the sum of the area of triangles AOB, AOC, BOD, DOC. Note that the area of each of these triangle is (1/2)*(d1/2)(d2/2).
So area of rhombus ABCD is 4*(1/2)*(d1/2)(d2/2) = (d1*d2)/2
We already know d1 = 40 cm. If we find d2 we are done.
For finding d2, consider the right angled triangle AOC. Apply Pythagoras theorem and we get:
(d1/2)^2 + (d2/2)^2 = 25^2, d1 = 40
We get d2 = sqrt((625 - 400)* 4) =30 cm
Hence area of rhombus ABCD = (d1*d2)/2 = 40*30/2 = 600 cm^2
I hope my answer will help you
Consider the triangles AOB and AOC. They are congruent(AB = AC, AO is common, angle ABO = angle ACO as triangle ABC is isosceles).
So angle AOB = angle AOC =180/2 = 90. So BO = OC = d2/2 and similarly AO = DO = d1/2.
The area of rhombus ABCD is the sum of the area of triangles AOB, AOC, BOD, DOC. Note that the area of each of these triangle is (1/2)*(d1/2)(d2/2).
So area of rhombus ABCD is 4*(1/2)*(d1/2)(d2/2) = (d1*d2)/2
We already know d1 = 40 cm. If we find d2 we are done.
For finding d2, consider the right angled triangle AOC. Apply Pythagoras theorem and we get:
(d1/2)^2 + (d2/2)^2 = 25^2, d1 = 40
We get d2 = sqrt((625 - 400)* 4) =30 cm
Hence area of rhombus ABCD = (d1*d2)/2 = 40*30/2 = 600 cm^2
I hope my answer will help you
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