Question

1. find the area, perimeter, and length of bd

(diagram: rectangle abcd with right angles at a, b, c, d; dc = (2sqrt{3}), bc = (sqrt{27}))

Explanation:

Step1: Simplify the side lengths

First, simplify \(\sqrt{27}\) and \(2\sqrt{3}\). We know that \(\sqrt{27}=\sqrt{9\times3} = 3\sqrt{3}\), and the length of \(DC = 2\sqrt{3}\), \(BC=3\sqrt{3}\). Wait, no, actually, in a rectangle (since all angles are right angles), the area of a rectangle is given by \(A = length\times width\). Here, \(DC = 2\sqrt{3}\) (length) and \(BC=\sqrt{27}=3\sqrt{3}\) (width)? Wait, no, wait, let's check again. Wait, the figure is a rectangle with \(DC = 2\sqrt{3}\) and \(BC=\sqrt{27}\). Let's simplify \(\sqrt{27}\): \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). So the length and width of the rectangle are \(2\sqrt{3}\) and \(3\sqrt{3}\) respectively.

Step2: Calculate the area

The area \(A\) of a rectangle is \(A = l\times w\), where \(l\) is length and \(w\) is width. So \(A=(2\sqrt{3})\times(3\sqrt{3})\). Using the property \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) and \(a\sqrt{b}\times c\sqrt{d}=ac\sqrt{bd}\), here \(a = 2\), \(b = 3\), \(c = 3\), \(d = 3\). So \(A=2\times3\times\sqrt{3\times3}=6\times3 = 18\).

Step3: Calculate the perimeter

The perimeter \(P\) of a rectangle is \(P = 2(l + w)\). So \(l = 2\sqrt{3}\), \(w = 3\sqrt{3}\). Then \(l + w=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\). Then \(P = 2\times5\sqrt{3}=10\sqrt{3}\). Wait, no, wait, maybe I made a mistake. Wait, actually, in the figure, \(DC = 2\sqrt{3}\) and \(BC=\sqrt{27}\). Wait, \(\sqrt{27}=3\sqrt{3}\), so the sides are \(2\sqrt{3}\) and \(3\sqrt{3}\). So perimeter is \(2\times(2\sqrt{3}+3\sqrt{3})=2\times5\sqrt{3}=10\sqrt{3}\).

Step4: Calculate the length of \(BD\)

\(BD\) is the diagonal of the rectangle. By the Pythagorean theorem, in a rectangle, the diagonal \(d\) is given by \(d=\sqrt{l^{2}+w^{2}}\). So \(l = 2\sqrt{3}\), \(w = 3\sqrt{3}\). Then \(l^{2}=(2\sqrt{3})^{2}=4\times3 = 12\), \(w^{2}=(3\sqrt{3})^{2}=9\times3 = 27\). Then \(d=\sqrt{12 + 27}=\sqrt{39}\)? Wait, no, wait, that can't be. Wait, maybe I mixed up the sides. Wait, maybe \(DC = 2\sqrt{3}\) and \(BC=\sqrt{27}\), but maybe it's a square? Wait, no, \(\sqrt{27}=3\sqrt{3}\) and \(2\sqrt{3}\) are not equal. Wait, maybe the figure is a rectangle, so \(AB = DC = 2\sqrt{3}\) and \(AD = BC=\sqrt{27}=3\sqrt{3}\). Then diagonal \(BD\): in triangle \(BCD\), which is a right triangle with legs \(BC = 3\sqrt{3}\) and \(DC = 2\sqrt{3}\)? Wait, no, \(BCD\) is a right triangle with legs \(BC\) and \(DC\)? Wait, no, \(B\), \(C\), \(D\): \(BC\) is vertical, \(DC\) is horizontal, so \(BD\) is the diagonal from \(B\) to \(D\), so the legs are \(BC\) and \(DC\)? Wait, no, \(BC\) is length \(\sqrt{27}\), \(DC\) is length \(2\sqrt{3}\), so \(BD=\sqrt{(2\sqrt{3})^{2}+(\sqrt{27})^{2}}=\sqrt{12 + 27}=\sqrt{39}\)? Wait, that seems odd. Wait, maybe I made a mistake in simplifying \(\sqrt{27}\). Wait, \(\sqrt{27}=3\sqrt{3}\), correct. \(2\sqrt{3}\) is correct. Then \( (2\sqrt{3})^2 = 4*3=12\), \( (3\sqrt{3})^2=9*3=27\), sum is 39, square root of 39. But let's check again.

Wait, maybe the figure is a rectangle, so area is length times width: \(2\sqrt{3}\times\sqrt{27}\). Let's compute that: \(\sqrt{27}=3\sqrt{3}\), so \(2\sqrt{3}\times3\sqrt{3}=6\times3=18\), that's correct. Perimeter: \(2\times(2\sqrt{3}+\sqrt{27})=2\times(2\sqrt{3}+3\sqrt{3})=2\times5\sqrt{3}=10\sqrt{3}\). Diagonal \(BD\): using Pythagoras, \(BD=\sqrt{(2\sqrt{3})^2+(\sqrt{27})^2}=\sqrt{12 + 27}=\sqrt{39}\). Wait, but maybe the figure is a rectangle, so that's the calculation.

Wait, maybe I made a mistake in the sides. Let's re-express:

- \(DC = 2\sqrt{3}\)
- \(BC=\sqrt{27}=3\sqrt{3}\)
- So, rectangle with length \(2\sqrt{3}\) and width \(3\sqrt{3}\)

Area: \(A = length \times width = 2\sqrt{3} \times 3\sqrt{3} = 6 \times (\sqrt{3})^2 = 6 \times 3 = 18\)

Perimeter: \(P = 2 \times (length + width) = 2 \times (2\sqrt{3} + 3\sqrt{3}) = 2 \times 5\sqrt{3} = 10\sqrt{3}\)

Length of \(BD\): In right triangle \(BCD\), \(BD\) is the hypotenuse, so \(BD = \sqrt{(DC)^2 + (BC)^2} = \sqrt{(2\sqrt{3})^2 + (3\sqrt{3})^2} = \sqrt{12 + 27} = \sqrt{39}\)

Wait, but let's check if the figure is a square. If it were a square, then \(2\sqrt{3}\) should equal \(\sqrt{27}\), but \(2\sqrt{3}\approx3.464\), \(\sqrt{27}\approx5.196\), so not a square, it's a rectangle.

Answer:

• Area: \(18\)
• Perimeter: \(10\sqrt{3}\)
• Length of \(BD\): \(\sqrt{39}\)