The diagonal of rectangle is 12cm. What is the perimeter of the figure formed by joining the mid points of the sides of the rectangle in order?
I think the figure formed is Rhombus.
Please explain with out the use of sin.
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You should come up with the notation for the length and width of rectange such that: 2L expresses the length and 2w expresses the width.
The segment that joins the midpoint of two consecutive sides expresses the hypotenuse of right triangle that has as lengths of legs `(2L)/2` and `(2w)/2` .
All these right triangles are alike, hence the hypotenuses are equal and the geometric shape that you form linking the midpoints of consecutive segments represents a rhombus.
You may find the length of side of rhombus using pythagorean theorem in a right triangle such that:
`s^2 = L^2 + w^2 =gt s = sqrt(L^2 + w^2)`
The perimeter of rhoumbus is 4 times the side s such that:
`P = 4s =gt P = 4sqrt(L^2 + w^2)`
The problem provides the length of diagonal of rectangle, hence, you may use pythagorean theorem to link this length to the dimension of rectangle.
`4L^2 + 4w^2 = 144`
Dividing by 4 both sides yields:
`L^2 + w^2 = 36 =gt sqrt(L^2 + w^2) = sqrt 36`
`sqrt(L^2 + w^2) = 6`
You should notice that the formula of perimeter of rhombus contains the factor `sqrt(L^2 + w^2), ` hence, substituting 6 for `sqrt(L^2 + w^2)` yields:
`P = 4*6 cm = 24 cm`
Hence, evaluating the perimeter of rhombus under given conditions yields `P = 24 cm` .
Let the sides of the rectangle be x and y.
==> x^2 + y^2 = 12^2
==> x^2 + y^2 = 144
Now, we need to find the perimeter of the shape formed by joining midpoints of the rectangle.
Each side of the new shape ( Rhombus) is the hypotenuse of a right triangle whose sides are (x/2) and (y/2)
Let the sides be h.
==> h^2 = (x/2)^2 + (y/2)^2 = (x^2+y^2)/4
But x^2 + y^2 = 144
==> h^2 = 144/4 = 36
==> h= 6
Then, each side of the rhombus is 6 cm.
==> Then the perimeter is P= 4*6 = 24 cm
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