First of all we work out the length of a sides ab, bc, CD, & ad.
We know that ab = bc = CD = ad
also ae = ac/2
If a to e = 2 then ac = 4
so
ab2 + bc2 = ac2
2ab2 = 16
ab2 = 8
ab = 2.8284271247461900976033774484194
so the perimeter = ab * 4 = 11.31
All apart from square, rhombus, kite and arrowhead.
The formula for the perimeter of a square is P equals 4 times a. 'P' represents the perimeter, and 'a' represents a side of the square.
The diagonals of a square are perpendicular (they intersect and form right angles). But they are angles bisectors since they bisect each pair of opposite angles. A perpendicular bisector actually bisects a side of a figure.
This cannot be answered. For instance if your area is 169ft x 1ft that equals 169sq ftIf your area is 13ft x 13ft that also equals 169sq ft.The perimeter for scenario A is 340ftThe perimeter for scenario B is 52ftSolutions:-If the area of perimeter is 169 square feet, then the edge of the square is 13. The perimeter is then 4*13 feet = 52 feet.
A square, a rhombus and a kite are three examples of quadrilaterals that have perpendicular diagonals that intersect each other at right angles.
A square, a rhombus and a kite have diagonals that intersect each other at 90 degrees.
No but the diagonals of a square intersect at right angles
It's not possible
Square and Rhombus
It is a kite that has diagonals that intersect each other at 90 degrees.
No but the diagonals of a square, rhombus and a kite do intersect each other at 90 degrees
square and rectangle * * * * * No. Square and Kite but NOT rectangle.
kite,square and rhombus
Square, rhombus and a kite.
A square, a rhombus and a kite have diagonals that intersect each other at right angles.
A square, a rhombus and a kite all have perpendicular diagonals that intersect at right angles
Not always but they are perpendicular in a square, a rhombus and a kite in that the diagonals intersect each other at 90 degrees