Question

Is this shape a kite, parallelogram, or a rhombus? The shape has coordinates: L(7,5) M(5,0) N(3,5) P(5,10).

Answer 1

a rhombus

Explanation:

The given coordinates:

L(7,5)
M(5,0)
N(3,5)
P(5,10).

The coordinates of the mid point of diagonal LN is
`(7+3)/2,(5+5)/2=(5,5)`

The coordinates of the mid point of diagonal MP is
`(5+5)/2,(0+10)/2=(5,5)`

So the coordinates of mid points of two diagonal being same they bisect each other, It is possible if the quadrilateral is a parallelogram.
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Now Checking the length of 4 sides

Length of LM =`sqrt((7-5)^2+(5-0)^2)=sqrt29`

Length of MN =`sqrt((5-3)^2+(0-5)^2)=sqrt29`

Length of NP =`sqrt((3-5)^2+(5-10)^2)=sqrt29`

Length of PL=`sqrt((5-7)^2+(10-5)^2)=sqrt29`

So the given quadrilateral is equilateral one and it would be a
rhombus
The second part is sufficient to prove everything required here.
Because equality in length of all sides also proves it a parallelogram as well as a special kite having all sides equal.

Answer 2

LMNP is a rhombus.

Explanation:

The points are `L(7,5)`, `M(5,0)`, `N(3,5)` and `P(5,10)`

Distance between

LM is `sqrt((5-7)^2+(0-5)^2)=sqrt(4+25)=sqrt29`

MN is `sqrt((3-5)^2+(5-0)^2)=sqrt(4+25)=sqrt29`

NP is `sqrt((5-3)^2+(10-5)^2)=sqrt(4+25)=sqrt29`

LP is `sqrt((5-7)^2+(10-5)^2)=sqrt(4+25)=sqrt29`

As all the sides are equal, it is a rhombus.

Note If opposite (or alternate) sides are equal it is a parallelogram and if adjacent sides are equal, it is a kite.

Answer 3

The diagonals bisect at 90° so the shape is a rhombus.

Explanation:

As proved by the contributor, dk_ch, the shape is not a kite, but is at least a parallelogram, because the diagonals have the same midpoint and therefore bisect each other.

Finding the lengths of all the sides is a rather tedious process.

Another property of a rhombus is that the diagonals bisect at 90°.

Finding the gradient of each diagonal is a quick method of proving whether or not they are perpendicular to each other.

From the coordinates of the four vertices, it can be seen that
PM is a vertical line `( x = 5)` (same `x` coordinates)
NL is a horizontal line `(y = 5)` (same `y` coordinates)

The diagonals are therefore perpendicular and bisect each other.

Answer 4

It's not a kite or a square or a parallelogram. It's a rhombus .

Explanation:

`L (7,5), M (5,0), N (3,5), P (5,10)`

To verify whether it's a kite.

For a kite, diagonals intersect each other at right angles but only one diagonal is bisected as against both in the case of rhombus and square.

`"Slope " = m_(ln) = (5-5) / (3 -7) = -0 "or" theta = 180^0`

`"Slope " = m_(mp) = (10-0) / (5-5) = oo " or ' theta_1 = 90^@`

`m_(ln) * m_(mp) = 0 * oo = -1`

Hence both diagonals are intersecting at right angles.

`"Mid point of " bar(LN) = (7+3)/2, (5 + 5)/2 = (5,5)`

`"Mid point of " bar(MP) = (5+5)/2, (0+10)/2 = (5,5)`

Since mid points of both the diagonals are the same, diagonals bisect each other at right angles and hence it's a rhombus or a square and not a kite.

`bar (LM) = sqrt((5-7)^2 + (0-5)^2) = sqrt29`

`bar (MN) = sqrt((3-5)^2 + (0-5)^2) = sqrt29`

`bar (LN) = sqrt((3-7)^2 + (5-5)^2) = sqrt16`

Since `(LM)^2 + (MN)^2 != (LN)^2`, it's not a right triangle and the given measurement does not form a square.

hence it's only a Rhombus.