Question

The figure the Floor Plan of a balcony ,it consists of △ABC and semi-circle ,

Answer 1

Step-by-step explanation:

(a) In right triangle ABC,

$AC^2 =AB^2 +BC^2 \\ = {12}^{2} + {5}^{2} \\ = 144 + 25 \\ = 169 \\ AC = \sqrt{169} \\ \therefore \: AC = 13\: m \\ \\ AC \: is \: the \: diameter \: of \: semicircle \\ \therefore \: r = \frac{1}{2} \times AC = \frac{1}{2} \times 13 = 6.5 \: m \\ Circumference \:of \:semi-circle \\ C = \pi \: r = 3.14 \times 6.5 = 20.41 \:m\:$

Perimeter of Balcony = 12 + 5 + 20.41 = 37.41 m

(b) Area of balcony

$=\frac{1}{2}\times 12\times 5+\frac{1}{2} \pi r^2 \\\\=\times 6\times 5+\frac{1}{2} \times 3.14\times (6.5)^2 \\\\=30+\frac{1}{2} \times 3.14\times42.25\\\\= 30+\frac{1}{2} \times 132.665\\\\= 30 + 66.3325\\\\= 96.3325 \:m^2 \\\\\approx 96. 33\:m^2$