Quadrilaterals - Quadrilaterals with Equal Sidelengths

The diagonals of a rhombus are $24\text{ cm}$ and $10\text{ cm}$. Find the length of each side and its perimeter.

Let the diagonals be $d_1 = 24\text{ cm}$ and $d_2 = 10\text{ cm}$. Since the diagonals of a rhombus bisect each other at right angles, the half-lengths are: $\frac{d_1}{2} = 12\text{ cm}, \quad \frac{d_2}{2} = 5\text{ cm}$ Using Pythagoras theorem for one of the triangles formed by the diagonals: $s = \sqrt{12^2 + 5^2}$ $s = \sqrt{144 + 25}$ $s = \sqrt{169} = 13\text{ cm}$ Perimeter: $P = 4 \times 13 = 52\text{ cm}$

We used the property that diagonals of a rhombus are perpendicular bisectors of each other to form a right-angled triangle, then applied the Pythagorean theorem to find the side.

Calculate the area of a square whose diagonal is $10\sqrt{2}\text{ cm}$.

Method 1 (Finding side first): Let the side be $s$. In a square, $d = s\sqrt{2}$. $10\sqrt{2} = s\sqrt{2} \implies s = 10\text{ cm}$ $Area = s^2 = 10^2 = 100\text{ cm}^2$ Method 2 (Directly using diagonal): $Area = \frac{1}{2} \times d^2$ $Area = \frac{1}{2} \times (10\sqrt{2})^2$ $Area = \frac{1}{2} \times (100 \times 2) = 100\text{ cm}^2$

The area of a square can be found either by squaring the side length or by taking half the square of its diagonal.

In a rhombus $ABCD$, the measure of $\angle BAC = 35^\circ$. Find $\angle ADC$.

In rhombus $ABCD$, $AB = BC$, so $\triangle ABC$ is isosceles. Therefore, $\angle BAC = \angle BCA = 35^\circ$. In $\triangle ABC$: $\angle ABC = 180^\circ - (35^\circ + 35^\circ) = 180^\circ - 70^\circ = 110^\circ$ Since opposite angles of a rhombus are equal: $\angle ADC = \angle ABC = 110^\circ$

We utilized the property that all sides are equal to identify an isosceles triangle, then used the angle sum property of a triangle and the property that opposite angles of a rhombus (a type of parallelogram) are equal.

1. In a square, diagonals are equal and bisect each other.
2. Given $PO = 6\text{ cm}$, then $PR = 2 \times PO = 2 \times 6 = 12\text{ cm}$.
3. Since diagonals are equal, $QS = PR = 12\text{ cm}$.
4. In $\triangle POQ$, $\angle POQ = 90^\circ$ (diagonals bisect at $90^\circ$).
5. By Pythagoras theorem in $\triangle POQ$: $PQ^2 = PO^2 + QO^2$
6. Since $QO = PO = 6\text{ cm}$: $PQ^2 = 6^2 + 6^2 = 36 + 36 = 72$ $PQ = \sqrt{72} = 6\sqrt{2}\text{ cm}$

We use the properties that square diagonals are equal, bisect each other, and meet at right angles to apply the Pythagorean theorem.

1. Perimeter $= 4 \times s = 52\text{ cm}$, so side $s = \frac{52}{4} = 13\text{ cm}$.
2. Let diagonals be $d_1 = 10\text{ cm}$ and $d_2$.
3. The diagonals bisect at $90^\circ$, forming right-angled triangles with legs $\frac{d_1}{2}$ and $\frac{d_2}{2}$ and hypotenuse $s$.
4. Using the relation: $s^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2$
5. $13^2 = (\frac{10}{2})^2 + (\frac{d_2}{2})^2$
6. $169 = 5^2 + (\frac{d_2}{2})^2$
7. $169 = 25 + (\frac{d_2}{2})^2$
8. $(\frac{d_2}{2})^2 = 144$
9. $\frac{d_2}{2} = \sqrt{144} = 12\text{ cm}$
10. $d_2 = 12 \times 2 = 24\text{ cm}$.

Calculate the side from the perimeter, then use the half-diagonals and side in a right-angled triangle (Pythagorean theorem) to find the missing diagonal.