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PilotLPTM [1.2K]
3 years ago
7

John says that because rhombuses do not have perpendicular sides, they cannot be rectangles. Explain his error in thinking

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
7 0
Even though most rhombuses are not rectangles(their angles are not all 90)
Some are. The only one that is, is a square. It is a rhombus, and it is a rectangle.
ankoles [38]3 years ago
3 0

For a rhombus to be a rectangle, a quadrilateral must have 4 equal sizes. However, some rhombuses, particularly squares have perpendicular sides and they are also rectangle.

<h2>Further Explanation</h2>

Therefore, a rhombus can be a rectangle only if all the angles are 90 degree and also a rectangle can be Rhombus only if the sides of the rectangle have equal length.  

A definition of a square also fits for a rectangle and Rhombus, because all the angles are 90 degree in rectangle and in rhombus, all the size has equal length.  

A rectangle has four sides with flat shape and all the angle is a right angle. Since all the sides are equal, a rectangle is called equiangular quadrilateral. It is also called parallelogram due to the fact that all the lines are parallel, a parallelogram is a quadrilateral where the opposite sides are equal and parallel.  

A rhombus has 4 equal straight sides with a flat shape. All the sides of a rhombus have equal length. The opposite sides are equal and parallel. The adjacent sides of a rhombus are supplementary.

Also by definition, all the sides in rhombuses are congruent and it has all the properties of a parallelogram particularly, the parallel sides, opposite angles and consecutive angles

LEARN MORE:

  • Marvin says that all rhombuses are squares brainly.com/question/1289560
  • What do rectangles and rhombuses have in common brainly.com/question/1929946

KEYWORDS:

  • perpendicular sides
  • rhombuses
  • rectangles
  • parallel sides
  • parallelogram
  • consecutive angles
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karen amad an hans have a total of 113 in their wallets? if hans has 4 times the amount of karen an karen has 7 dollars more tha
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Amount in Han wallet = $ 80

Amount in amad wallet = $ 13

Amount in karen wallet = $ 20

<u>Solution:</u>

Let "h" be the amount in Han's wallet

Let "k" be the amount in karen wallet

Let "a" be the amount in amad wallet

<em><u>They have a total of 113 in their wallet</u></em>

Therefore,

h + k +a = 113 ------- eqn 1

<em><u>If hans has 4 times the amount of karen</u></em>

Amount with Han = 4 times the amount with Karen

h = 4k ------- eqn 2

<em><u>And karen has 7 dollars more than amad</u></em>

k = 7 + a

a = k - 7 ---------- eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

4k + k + k - 7 = 113

6k - 7 = 113

6k = 113 + 7

6k = 120

k = 20

<em><u>Substitute k = 20 in eqn 3</u></em>

a = 20 - 7

a = 13

<em><u>Substitute k = 20 in eqn 2</u></em>

h = 4k = 4(20) = 80

h = 80

<em><u>Summarizing the results:</u></em>

Amount in Han wallet = $ 80

Amount in amad wallet = $ 13

Amount in karen wallet = $ 20

6 0
4 years ago
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