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