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Veseljchak [2.6K]
3 years ago
15

what is the area of a rectangle which is 57 cm long and its breadth is 34 cm? One weird question for you to answer, what's my na

me?
Mathematics
2 answers:
BaLLatris [955]3 years ago
6 0
1938 cm for the first question
For the second...Ruth?
GarryVolchara [31]3 years ago
3 0
Area of a rectangle is length*width.
If length is is 57 cm and width is 34 cm, then area is 
(57 cm)(34 cm) = 1938 cm^2
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mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are  equal, thus

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GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5

Now, mid point of JH=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

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Mid point of GI=(\frac{5}{2},1)

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

(GI)^{2}=(IJ)^{2}+(JG)^{2}                             (1)

Now, JI=\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20} and GJ=\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}

Substituting these values in (1), we get

5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }

25=20+5

25=25

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI=\sqrt{(4-1)^{2}+(3+1)^{2}}=5  and HJ=\sqrt{(0-5)^2+(1-1)^2}=5 are equal, thus, GHIJ is a rectangle.

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