Question

A garden is designed in the shape of a rhombus formed from 4 identical 30°-60°-90° triangles. The shorter distance across the middle of the garden measures 30 feet
60 ft
60 sq rt 3 ft
120 ft
120 sq rt 3 ft

Answer 1

Answer:

In Δ AOB, Right angled at O

$tan 60=\frac{P}{B}\\\\ \sqrt{3}=\frac{AO}{15}\\\\ AO=15 \times 1.732\\\\ AO=25.98$

$sin 60=\frac{P}{H}\\\\ \frac{\sqrt{3}}{2}=\frac{25.98}{H}\\\\H=\frac{51.96}{1.732}\\\\ H=30$

So, side of rhombus = 30 cm

AO=15√3 cm

Length of another diagonal= 25.98 + 25.98= 51.96 cm, Because diagonals of rhombus bisect each other.

Area of Rhombus

   $\frac{1}{2}\times {\text{product of diagonals}\\\\=\frac{1}{2}*30*30\sqrt{3}=450\sqrt{3}$

=450√3 cm²

Answer 2

legs of the triangles

Each triangle 30-60-90 is:

         one leg: 15 ft => short diagonal = 2 * 15ft = 30ft

         other leg, x:

         tan(30) = 15 / x => x = 15 ft / tan(30) = 25.98 ft

=> long diagonal = 2 * 25.98ft = 51.96 ft

side of the rhoumbus =  hypotenuse of one triangle

side of the rhombus = √ [ (15)^2 + (25.98)^2 ] = √(900) = 30 ft

Area of the rhombus:

4 * area of one triangle = 4 [base*height/2] = 4*15*25.98/2 = 779.43 ft^2

The shortest distance accross the garden is equal to the side of the rhombus = 30 ft