Answer:
29.39 cm²
Step-by-step explanation:
To find the area of triangle \(\bold{\sf \Delta ABC}\) with the given conditions, we need to calculate the base \(\bold{\sf AB}\) and the height \(\bold{\sf CD}\). Here’s the detailed solution:
Identify the Triangle Structure
Given:
- \(\bold{\sf AC = BC = 10 \textsf{ cm}}\)
- \(\bold{\sf \angle A = 72^\circ}\)
- \(\bold{\sf CD \perp AB}\), where \(\bold{\sf D}\) is the foot of the perpendicular from \(\bold{\sf C}\) to \(\bold{\sf AB}\)
Since \(\bold{\sf \Delta ABC}\) is isosceles with \(\bold{\sf AC = BC}\), the perpendicular \(\bold{\sf CD}\) also bisects \(\bold{\sf \angle ACB}\) and \(\bold{\sf AB}\) into two equal segments. Let’s denote:
- \(\bold{\sf AD = DB = x }\)
- \(\bold{\sf CD = h}\), the height of the triangle
Calculate \(\bold{\sf AB}\)
To find \(\bold{\sf x}\), we use the fact that \(\bold{\sf CD}\) bisects \(\bold{\sf \angle ACB}\), making \(\bold{\sf \Delta ADC}\) a right triangle with:
\(\sf \angle CAD = \dfrac{\angle ACB}{2}\)
Since \(\bold{\sf AC = BC}\) and \(\bold{\sf \angle A + \angle C + \angle B = 180^\circ}\), we have:
\(\sf \angle C = 180^\circ - 2\angle A = 180^\circ - 2 \times 72^\circ = 36^\circ\)
Thus:
\(\sf \angle CAD = \dfrac{36^\circ}{2} = 18^\circ\)
In \(\bold{\sf \Delta ADC}\):
\(\sf \cos(18^\circ) = \dfrac{AD}{AC}\)
So:
\(\sf \cos(18^\circ) = \dfrac{x}{10}\)
\(\sf x = 10 \times \cos(18^\circ)\)
\(\sf x = 10 \cos(18^\circ) = 9.510565163\)
The total base \(\bold{\sf AB}\) is:
\(\sf AB = 2x \\\\ = 2 \times 10 \cos(18^\circ)\\\\= 20 \cos(18^\circ) \\\\ = 19.02113033 \textsf{ cm}\)
Calculate the Height \(\bold{\sf CD}\)
Using the Pythagorean theorem in \(\bold{\sf \Delta ADC}\):
\(\sf AC^2 = AD^2 + CD^2\)
Thus:
\(\sf 10^2 = (9.510565163)^2 + h^2\)
\(\sf 100 = 90.45084972 + h^2\)
\(\sf h^2 = 100 - 90.45084972\)
\(\sf h^2 = 9.549150281\)
\(\sf h = \sqrt{9.549150281} \approx 3.090169944 \textsf{ cm}\)
Calculate the Area of \(\bold{\sf \Delta ABC}\)
The area \(\bold{\sf A}\) of \(\bold{\sf \Delta ABC}\) is:
\(\sf A = \dfrac{1}{2} \times AB \times CD\)
So:
\(\sf A = \dfrac{1}{2} \times 19.02113033 \times 3.090169944\)
\(\sf A \approx 29.38926261 \textsf{ cm}^2\)
\(\sf A \approx 29.39 \textsf{ cm}^2 \textsf{(in 2 d.p.)}\)
Final Answer
The area of \(\bold{\sf \Delta ABC}\) is approximately \(\bold{\sf 29.39}\) cm².
