Question

From apoint in interior of an equilateral triangle, perpendiculars drawn to 3 sides are 8cm, 10 cm and 11cm respectively.
Find area of triangle to nearest cm.

Answer 1

Here is the answer.
Choose as brainliest if possible.

Answer 2

see in figure O is a interior point of equilateral ∆PQR . where , three perpendiculars are x = 8 cm
y = 10 cm and z = 11 cm
because ∆PQR is an equilateral triangle
so, PQ = QR = RS = L ( let )

now,
area of ∆POR = 1/2 × OU × PR
ar∆PQR = 1/2 × y × L
ar∆PQR = 1/2× 10 × L = 5L
similarly,

area of ∆POQ = 1/2×OS × PQ
ar∆POQ = 1/2 × x × L
ar ∆PQQ = 1/2×8× L = 4L

area of ∆QOR =1/2×OT × QR
ar∆QOR = 1/2× z × L
ar∆QOR = 1/2×11× L = 11L/2

now,
area of ∆PQR = ar∆POQ + ar∆POR + ar∆QOR

√3/4 × side length² = 5L + 4L + 5.5L

√3/4 L² = (14.5)L

√3/4 L = 14.5

L = 14.5×4/√3 = 58/√3 cm

hence,
area of ∆PQR = √3/4 ×(58/√3)²
= √3/4 × 58²/3 cm²
= 29²/√3 cm²
= 485.56 cm²