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Question

A point O is taken inside an equilateral ΔABC. If OMAC, OLBC and ONAB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of ΔABC. (3 marks)

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Solution

Let each side of ΔABC be 'a' cm.

Then Area (ΔABC)=Area (ΔOBC)+Area (ΔAOC)+Area (ΔAOB) (1 mark)

We know area of a triangle with base 'b' and height 'h' id given by '12×base×height'
So, Area (ΔABC)=12a(OL+OM+ON)

=12a(14+10+6)

=15a (1 mark)

Given, ΔABC is equailateral.
Area of an equilateral triangle with side 'a' is given by 34a2.

34a2=15a

a=15×43

a=60×33=203

Area (ΔABC)=15×203=3003 cm2 (1 mark)


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