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

In Δ AOB, Right angled at O

`tan 60=(P)/(B)\\\\ √(3)=(AO)/(15)\\\\ AO=15 * 1.732\\\\ AO=25.98`

`sin 60=(P)/(H)\\\\ (√(3))/(2)=(25.98)/(H)\\\\H=(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

`(1)/(2)* {\text{product of diagonals}\\\\=(1)/(2)*30*30√(3)=450√(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