The Rhombus Area is one of quadrilateral which has all the four sides equal. The opposite sides of rhombus are parallel to each other and the opposite angles are also equal to one another. The rhombus is also called as a diamond. Square is nothing but the rhombus with right angles. The rhombus midpoint is the point at equidistant from both the ends.
Rhombus Midpoint : Problem 1
Let PQRS is a rhombus and A, B, C and D are the rhombus midpoint of PQ, RS, QR, and SR. Prove that ABCD is a rectangle.
Solution:
Given: PQRS is a rhombus.
A, B, C and D are the rhombus midpoint of PQ, RS, SP, and QR.
To prove: ABCD is a rectangle.
Proof:
Construct the diagonals PR and QS.
Let them intersect at O. Join DA, AB, BC and CD.
Let DA intersect PR at L and let AB intersect QS at N.
DC || AB and DC || AB
Therefore ABCD is a parallelogram (One pair of opposite sides are parallel and equal)
Now consider quadrilateral ONAL
In which LON = 90 degrees [Since opposite angles of a parallelogram are equal]
Hence, DAB = 90 degrees
Therefore, ABCD is a rectangle.
Algebra is widely used in day to day activities watch out for my forthcoming posts on Perimeter of a Square Formula and simple math problems for kids. I am sure they will be helpful.
Rhombus Midpoint : Problem 2
The diagonals of a quadrilateral PQRS are perpendicular. Show that the quadrilateral, formed by the rhombus midpoint of its sides is a rectangle.
Solution:
PQRS is a quadrilateral and SQ PR
To prove: ABCD is a rectangle.
Proof:
DC || AB and DC = AB (since both DC and AB are parallel to PR and equal to (1/2)PR)
ABCD is a parallelogram. [According to rhombus midpoint theorem]
Considering the quadrilateral ALOK,
AK || LO [Since SP || BD]
AL || KO [Since PQ || AC]
And KOL = 90 degrees
Therefore ALOK is a parallelogram.
KAL = 90 degrees
Therefore, DAB = 90 degrees
In parallelogram ABCD, CAB = 90 degrees
Therefore, ABCD is a rectangle.
Is this topic algebra questions hard for you? Watch out for my coming posts.
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