Yes. A rectangle is a parallelogram with four right angles. A rhombus is a parallelogram with four equal sides. Therefore, if a rectangle has four equal sides, then it is a rhombus. If a rhombus has four right angles, it is a rectangle. In this case, the rectangle and the rhombus are a special type of parallelogram called a square.
No. That is because there is no such word as "rhobus". A rhombus, on the other hand, IS a polygon. It is a quadrilateral.
No, other way round...
They are both quadrilaterals.
Yes, a square is a special kind of rhombus.
It has two lines of symmetry.
A square IS always a special kind of rhombus.
two circles and arectangle
Arectangle is a quadrilateral that is equiangular. In other words all of its angles are the same measure.
Yes, if it the special kind of rectangle called a square.
12*10 = 120 square feet
There are many possible shapes with two acute angles.
It is the right angle triangle which has 3 sides whereas all quadrilaterals have 4 sides.
A kite or a rhobus. It only looks like a rhombus if you turn it on its side.
A diamond, like the one seen in playing cards is often a rhombus although in some cases the sides are slightly curved (inwards). A diamond, as in a lump of carbon, is not a rhombus.
The area (A) or a rectangle can be found by multiplying the length (l) by the width (w). It looks like this:Arectangle = l x wBefore performing the simple multiplication, be sure that both dimensions are in the same units. Then the area will appear in those units squared.
To find the area (A) of any rectangle, we multiply the length (l) by the width (w). In the case of a rectangle that is 11 cm by 9 cm, we do this:Arectangle = l x w = 11 cm x 9 cm = 99 cm2or 99 square centimeters.
Let the other diagonal be x:- If: 0.5*x*12 = 54 Then: x = 54/6 => 9 The rhombus will consist of 4 right angles: base 4.5 cm and height 6 cm Using Pythagoras: hypotenuses = 7.5 cm Therefore perimeter: 4*7.5 = 30 cm
Rectangle: A quadrilateral with 4 right angles, diagonals congruent/bisecting, and opposite sides congruent, BUT ADJACENT SIDES ARE NOT CONGRUENT. Rhobus: A quadrilateral with opposite congruent angles, but adjacent angles are Not congruent, perpendicular bisecting diagonals and 4 congruent sides. Square: A quadrilateral that is a rectangle and a square with 4 right angles, diagonals congruet/bisecting that ar perpendicular, and opposites sides congruent.
The length of the rectangle in question is 12 meters, and the width of the rectangle is 6 meters.We solve this by knowing that the area of a rectangle is equal to the length (l) times the width (w). That formula looks like this:Arectangle = l x wWe are told that the length is twice the width. Here's that idea in the form of an equation:l = 2wSince the length is twice the width, we can substitute the 2w for the l in the first equation. It looks like this:Arectangle = 2w x wNow we add the fact that the area was given as 72 meters2. and put it all together. It looks like this:72 meters2 = 2w x w72 meters2 = 2w236 meters2 = w2sqrt 36 meters2 = w6 meters = wWe have our width as 6 meters. The length is twice that, or 6 meters x 2 = 12 meters. There are our length and width. Let's check our work.Area = length (12 meters) times width (6 meters) = 12m x 6m = 72 m2Our work checks.
The rectangle is 15 feet long by 7 feet wide. But let's do the math so you come away with something other than just some numbers for an answer. Length (l) times width (w) will get us the area of a rectangle, as you know. These are the variables in the problem, and they can have different values (hence their being called variables). But we also have one of them expressed in terms of the other one. And we have the area. Let's take that to the "machine" we set up, which is the expression we will create (the "formula" if you prefer) that will lead us to the answer. Roll up your sleeves and let's do this. Arectangle = l x wl = w + 8 [we have l in terms of w, so we can put that back into the original expression] Arectangle = (w + 8) x w [put in the known area (105) and do the multiplication] 105 = w2 + 8w [we used the distributive property of multiplication, as you see, and now we subtract 105 from each side] w2 + 8w - 105 = 0 [yes, we have a second degree equation, but no panic as we'll factor] (w + 15) x (w - 7) = 0 [the two factors, when multiplied, give the original expression] Here's the deal. If we have two numbers that when multiplied together give us a product of zero, then either one of the numbers must be zero, or the othernumber must be zero, or both numbers must be zero. There are two answers for w here, so let's find both of them by setting each number equal to zero and then solving for it. w + 15 = 0 , w = -15 [we subtracted 15 from each side] w - 7 = 0 , w = 7 [we added 7 to each side] We were solving for the width of a rectangle. Can a rectangle have a width with negative length (-15) as solved? No, it can't. But it can have a width of 7 as solved. Let's take the 7 and plug it back into the original expression and solve for l(the length) there. Arectangle = l x w 105 = lx 7 l = 105 / 7 = 15 [105 divided by 7, which is the width, gives us the length] The length of the rectangle is 15 feet as discovered. The 7 x 15 = 105, and 7 + 8 = 15, so we've checked it and found it to be good. Piece of cake.