A side of an equilatral triangle ABC is 4cm.Find the side of an equilatral triangle PQR whose area is double the area of ∆ABC
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Solution:
Area of triangle PQR = 2 x (Area of triangle ABC)
Since area of triangle ABC = √3/4 × side^2
= √3/4 × 4^2
= 6.928 cm^2.
Therefore, area of triangle PQR = √3/4 × side^2
2 × 6.928 = √3/4 × side^2
13.856 = 0.433 × side^2
Therefore, side = 5.656 cm
Hence, side of equilateral triangle PQR is 5.656 cm
Area of triangle PQR = 2 x (Area of triangle ABC)
Since area of triangle ABC = √3/4 × side^2
= √3/4 × 4^2
= 6.928 cm^2.
Therefore, area of triangle PQR = √3/4 × side^2
2 × 6.928 = √3/4 × side^2
13.856 = 0.433 × side^2
Therefore, side = 5.656 cm
Hence, side of equilateral triangle PQR is 5.656 cm
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