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Question
Side of equilateral triangle PQR is 8 cm then find the area of triangle whose side is half of the side of triangle PQR.
Solution
Given: ΔPQR is an equilateral triangle with PQ = QR = PR = 8 cm and ΔABC is an equilateral triangle with AB = BC = AC = 4 cm
To find: A(ΔABC)
Construction: Draw seg AD ⊥ BC, B−D−C
In ΔABD,
∠ADB = 90° ...[Construction]
∠ABD = 60° ...[Angle of an equilateral triangle]
∠BAD = 30° ...[Remaining angle of a triangle]
∴ ΔABD is a 30°–60°–90° triangle.
∴ AD = `sqrt(3)/2` AB ...[Side opposite to 60°]
∴ AD = `sqrt(3)/2 xx 4`
∴ AD = `2sqrt(3)` ...(i)
Area of triangle = `1/2 xx "height" xx "base" `
∴ Area of ΔABC = `1/2 xx "AD" xx "BC"`
= `1/2 xx 2sqrt(3) xx 4`
= `2 xx 2sqrt(3)`
= `4sqrt(3)` ......[From (i)]
∴ Area of the triangle whose side is half of the side of ΔPQR is `4sqrt(3)` sq.cm
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