Question

area: composite figures
consider this figure.
enter the total area of figure abcd in square centimeters.
a =

Explanation:

Step1: Find length of diagonal BD

First, use the Pythagorean theorem in $\triangle AFB$:
$BD = 2 \times \sqrt{8^2 - 6^2} = 2 \times \sqrt{64 - 36} = 2 \times \sqrt{28} = 2 \times 2\sqrt{7} = 4\sqrt{7}$ cm
Wait, correction: Actually, split the figure into 4 triangles: $\triangle AFB$, $\triangle AFD$, $\triangle CFB$, $\triangle CFD$.

Step1: Calculate area of $\triangle AFB$

Area formula: $\frac{1}{2} \times base \times height$
$\text{Area}_{AFB} = \frac{1}{2} \times 6 \times \sqrt{8^2 - 6^2} = \frac{1}{2} \times 6 \times \sqrt{28} = 3 \times 2\sqrt{7} = 6\sqrt{7}$ cm²

Step2: Calculate area of $\triangle AFD$

$\text{Area}_{AFD} = \frac{1}{2} \times 6 \times \sqrt{8^2 - 6^2} = 6\sqrt{7}$ cm²

Step3: Calculate area of $\triangle CFB$

$\text{Area}_{CFB} = \frac{1}{2} \times 15 \times \sqrt{8^2 - 6^2} = \frac{1}{2} \times 15 \times 2\sqrt{7} = 15\sqrt{7}$ cm²

Step4: Calculate area of $\triangle CFD$

$\text{Area}_{CFD} = \frac{1}{2} \times 15 \times \sqrt{8^2 - 6^2} = 15\sqrt{7}$ cm²

Step5: Sum all 4 triangle areas

$\text{Total Area} = 6\sqrt{7} + 6\sqrt{7} + 15\sqrt{7} + 15\sqrt{7} = 42\sqrt{7}$ cm²
Alternative simpler method: The figure is a quadrilateral split by diagonal AC (length $6+15=21$ cm), and the perpendicular height from B and D to AC is the same, $h = \sqrt{8^2 - 6^2} = 2\sqrt{7}$ cm.

Step1: Find total base AC

$AC = 6 + 15 = 21$ cm

Step2: Find height from B/D to AC

$h = \sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt{7}$ cm

Step3: Calculate total area (2 triangles ABC + ADC)

$\text{Total Area} = \frac{1}{2} \times 21 \times 2\sqrt{7} + \frac{1}{2} \times 21 \times 2\sqrt{7} = 21\sqrt{7} + 21\sqrt{7} = 42\sqrt{7}$ cm²
Decimal approximation: $42\sqrt{7} \approx 42 \times 2.6458 \approx 111.12$ cm²

Answer:

$42\sqrt{7}$ or approximately 111.12 square centimeters