OUR LADY OF MERCY SCHOOL

Name:    Eric M., Francesco G.,                             Grade: 9th
              João Marcelo S. & Natássia N.

Teacher: Ms. Freire & Ms. Oliveira                   Date: Monday, June 11th, 2001

Assignment: Brazilian Social Studies and Math Project

 

 

Considered one of the Wonders of the Modern World, it is a famous statue of Christ with open arms, making a shape of a cross. It is the highest monument in Rio de Janeiro and is located on top of the Corcovado Mt. (710m high). The statue is 30m high, plus a pedestal of 8m. It weighs 1.145 tons.

                The statue was first conceived in 1921, when the Catholic Church and the magazine “O Cruzeiro” gathered funds to construct a statue of Christ to celebrate Brazil’s centenary of independence, which would be in 1922. The first sketches were done by Carlos Oswald, who pictured Christ carrying a cross and holding a globe, while standing over a pedestal symbolizing the world. Rio’s population preferred the statue as it is today, with open arms, embracing all people. It was planned by Heitor de Silva Costa and the Polish sculptor Paul Landowski constructed the head and hands. It was inaugurated in October 12th, 1931. Christ the Redeemer is a religious and tourist symbol of Rio de Janeiro. There, many ecumenical encounters happen and many tourists come to visit.

                Access is by car or on a trolley, that departs every 30 minutes. After using the car or trolley, people have to go up a stairway of 222 steps to see the statue. The government of Rio is planning to put a panoramic elevator.

 

Problem:

Christ the Redeemer will be restored. For this, it will have to be covered with cloth. Considering the information in the figure above, find out:

 a)       Measurements of each of the four triangles in the Christ.

b)       Total area of the Christ the Redeemer.

If each m2 of cloth costs $2.30, how much will be spent to cover it?

 

Solution:
a)       Observation: diagonals are perpendicular. Therefore, Pythagorean theorem is applied.

top triangles:

 

x2=42+152

x2=225+676

x2=241                                                                             

x@15.5m                                                                      

perimeter: 34.5m

area: ½∙b∙h

½∙15∙4=30m2

bottom triangles

 

 

x2=152+262      

x2=16+225

x2=901

x@30m

perimeter: 71m

area: ½∙b∙h

½∙15∙26=195m2

b) Base Area = 2(top ▲ area) + 2(bottom ▲ area)    Lateral Area=2(3∙15.5) + 2(3∙30)    Surface Area = 2 Base Area + Lateral Area (Prism)

 

Base Area                                            Lateral Area                             Surface Area

 = 2(30) + 2(195)                                 = 2(46.5) + 2(90)                    = 2(450) + 273

 = 60 + 390                                          = 93 + 180                              = 900 + 273

 = 450m2                                              = 273m2                                  = 1173m2

c)  $2.30 x 1173m2 = $2697.90


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Bibliography:

    http://www.cidadedorio.com.br/subcorcovado.htm

    http://www.corcovado.com.br/prin_eng.htm