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).

Question

Maverick Smith · Accepted Answer

a rhombus

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. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 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.

Wyatt Anderson · Answer

LMNP is a rhombus.

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.

Savannah Aguilar · Answer

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

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.

Caroline Aucoin · Answer

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

#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.

Claire Akin · Answer

undefined To determine the shape of the given figure based on the coordinates provided, we need to analyze its properties.

Given the coordinates:
L(7,5)
M(5,0)
N(3,5)
P(5,10)

We can start by plotting these points on a coordinate plane and then connecting them to form the shape. Once we have the shape, we can analyze its properties.

From the given coordinates, we can see that LM and NP are parallel, and LN and MP are parallel. Additionally, LM = NP and LN = MP. This implies that opposite sides are parallel and equal in length.

Based on these properties, we can conclude that the given shape is a parallelogram.

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).