Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

1. the perimeter of a rhombus is 52cm. one of its diagonals is 10cm. it…

Question

  1. the perimeter of a rhombus is 52cm. one of its diagonals is 10cm. its area is ____ cm².
  2. as shown in the figure, the diagonals of rectangle abcd, $overline{ac}$ and $overline{bd}$, intersect at point o. point e is the mid - point of $overline{bc}$. $bc = 4sqrt{3}$. if $oe = 2$, then the length of $overline{bd}$ is ____. a. 4 b. 6 c. 8 d. 10

Explanation:

Step1: Find side - length of rhombus

Since the perimeter of a rhombus is \(52\mathrm{cm}\) and all sides of a rhombus are equal, if the side - length is \(a\), then \(4a = 52\), so \(a=\frac{52}{4}=13\mathrm{cm}\).

Step2: Use the property of rhombus diagonals

The diagonals of a rhombus bisect each other at right - angles. Let the given diagonal \(d_1 = 10\mathrm{cm}\), then half of it \(x=\frac{d_1}{2}=5\mathrm{cm}\). Using the Pythagorean theorem in one of the four right - triangles formed by the diagonals of the rhombus, if the other half of the second diagonal is \(y\), then \(y=\sqrt{a^{2}-x^{2}}\). Substituting \(a = 13\mathrm{cm}\) and \(x = 5\mathrm{cm}\), we get \(y=\sqrt{13^{2}-5^{2}}=\sqrt{169 - 25}=\sqrt{144}=12\mathrm{cm}\). So the second diagonal \(d_2=2y = 24\mathrm{cm}\).

Step3: Calculate the area of the rhombus

The area formula of a rhombus is \(A=\frac{1}{2}d_1d_2\). Substituting \(d_1 = 10\mathrm{cm}\) and \(d_2 = 24\mathrm{cm}\), we get \(A=\frac{1}{2}\times10\times24 = 120\mathrm{cm}^{2}\).

Step4: Solve for the length of the diagonal of the rectangle

In rectangle \(ABCD\), \(O\) is the intersection of the diagonals \(AC\) and \(BD\), so \(O\) is the mid - point of \(AC\) and \(BD\). Since \(E\) is the mid - point of \(BC\) and \(OE\) is parallel to \(AB\) (by the mid - point theorem in a triangle), and \(OE = 2\). In right - triangle \(ABC\), \(OE\) is a mid - segment parallel to \(AB\). Let \(AB = 2OE=4\). Using the Pythagorean theorem in right - triangle \(ABC\) with \(BC = 4\sqrt{3}\) and \(AB = 4\), then \(AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{4^{2}+(4\sqrt{3})^{2}}=\sqrt{16 + 48}=\sqrt{64}=8\). Since the diagonals of a rectangle are equal, \(BD = AC=8\).

Answer:

  1. \(120\)
  2. C. \(8\)