From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 4 cm, 6 cm, and 2 cm. Find the area of the triangle.
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Answer: 48√3 cm2
Step by Step Explanation: - The following figure shows the required triangle,
![](https://www.edugain.com/egdraw/draw.php?num=5&sx=250&sy=240&x0=20&y0=40&A1=shape:polygon;points:0,0,200,0,100,170,0,0,120,50,100,170,200,0,120,50;textc:B,C,[0.15]A;texts:[0.-10]x,[0.20]x,[-10.20]x&A2=shape:polygon;points:0,0,200,0,100,170;textc:B,C,[0.15]A&A3=shape:line;x1:120;y1:50;x2:45;y2:75;tc1:O;tc2:P;ts1:6cm&A4=shape:line;x1:120;y1:50;x2:155;y2:75;tc2:[5.10]Q;ts1:2cm&A5=shape:line;x1:120;y1:50;x2:120;y2:0;tc2:R;ts1:4cm )
Let's assume the sides of the equilateral triangle ΔABC be x.
The area of the triangle ΔABC can be calculated using Heron's formula since all sides of the triangles are known.
S =
=
= cm.
The area of the ΔABC = √S(S - AB)(S - BC)(S - CA)
= √3x2(3x2−x)(3x2−x)(3x2−x)√3x2(3x2−x)(3x2−x)(3x2−x)
= √3x2(x2)(x2)(x2)√3x2(x2)(x2)(x2)
= √3x2(x2)3√3x2(x2)3
= √3(x2)4√3(x2)4
= √3(x2)2√3(x2)2
= (x)2 ------(1) - The area of the triangle AOB =
=
= - Similarly, the area of the triangle ΔBOC =
and the area of the triangle ΔAOC = . - The the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)
= + +
= -----(2) - By comparing equation (1) and (2), we get,
(x)2 =
⇒ x = - Now, Area(ΔABC) = (x)2
= ( )
2
=
= 48 √3 cm2 - Hence, the area of the triangle is 48√3 cm2.