Mensuration: Area and Perimeter - Use Brahmagupta formula for cyclic quadrilaterals and connect it to Heron formula

Find the area of a quadrilateral $ABCD$ in which $AB = 3\text{ cm}, BC = 4\text{ cm}, CD = 4\text{ cm}, DA = 5\text{ cm}$ and $AC = 5\text{ cm}$.

Step 1: The quadrilateral $ABCD$ is divided into two triangles, $\triangle ABC$ and $\triangle ADC$ by diagonal $AC$. Step 2: For $\triangle ABC$, the sides are $a=3, b=4, c=5$. Since $3^2 + 4^2 = 5^2$ ($9+16=25$), it is a right-angled triangle. $Area(\triangle ABC) = \frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2$. Step 3: For $\triangle ADC$, the sides are $a=5, b=4, c=5$. $s = \frac{5+4+5}{2} = \frac{14}{2} = 7\text{ cm}$. $Area(\triangle ADC) = \sqrt{7(7-5)(7-4)(7-5)} = \sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} \approx 9.17\text{ cm}^2$. Step 4: Total Area $= Area(\triangle ABC) + Area(\triangle ADC) = 6 + 9.17 = 15.17\text{ cm}^2$.

We divide the quadrilateral using the given diagonal $AC$. We use the simple right-angle area formula for one triangle and Heron's formula for the other, then add them together.

A park is in the shape of a quadrilateral $ABCD$ has $\angle C = 90^\circ, AB = 9\text{ m}, BC = 12\text{ m}, CD = 5\text{ m}$ and $AD = 8\text{ m}$. How much area does it occupy?

Step 1: Join $BD$ to form two triangles. In right $\triangle BCD$, using Pythagoras: $BD = \sqrt{BC^2 + CD^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ m}$. Step 2: $Area(\triangle BCD) = \frac{1}{2} \times 12 \times 5 = 30\text{ m}^2$. Step 3: For $\triangle ABD$, sides are $a=9, b=8, c=13$. $s = \frac{9+8+13}{2} = 15\text{ m}$. $Area(\triangle ABD) = \sqrt{15(15-9)(15-8)(15-13)} = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{1260} = 6\sqrt{35} \approx 35.5\text{ m}^2$. Step 4: Total Area $= 30 + 35.5 = 65.5\text{ m}^2$.

Since one angle is $90^\circ$, we use Pythagoras to find the diagonal length $BD$. This diagonal allows us to split the quadrilateral into a right triangle and a general triangle, calculating their areas separately.

1. Find the semi-perimeter $s$: $s = \frac{7 + 15 + 20 + 24}{2} = \frac{66}{2} = 33\text{ cm}$
2. Apply Brahmagupta's Formula: $Area = \sqrt{(33-7)(33-15)(33-20)(33-24)}$ $Area = \sqrt{26 \times 18 \times 13 \times 9}$
3. Simplify the expression: $Area = \sqrt{(13 \times 2) \times (9 \times 2) \times 13 \times 9}$ $Area = \sqrt{13^2 \times 9^2 \times 2^2}$ $Area = 13 \times 9 \times 2 = 234\text{ cm}^2$

To find the area of a cyclic quadrilateral, we first calculate the semi-perimeter $s$. Then we use the product of the differences between $s$ and each side length under a square root.

1. Semi-perimeter $s$: $s = \frac{10 + 12 + 14 + 8}{2} = \frac{44}{2} = 22\text{ cm}$
2. Area using Brahmagupta's Formula: $Area = \sqrt{(22-10)(22-12)(22-14)(22-8)}$ $Area = \sqrt{12 \times 10 \times 8 \times 14}$
3. Calculate the product: $Area = \sqrt{13440}$
4. Simplify: $Area = \sqrt{64 \times 210} = 8\sqrt{210}\text{ cm}^2 \approx 115.93\text{ cm}^2$

Since the quadrilateral is inscribed in a circle, it is cyclic. We compute the semi-perimeter and then substitute the values into Brahmagupta's formula to find the area.