Hexagon inside a triangle

Geometry Level pending

In $\triangle PQR, PR=30, QR=40$ and $PQ=50$ . Points $A$ and $B$ lies on $\overline {PQ}$ , points $C$ and $D$ lies on $\overline {QR}$ , and points $E$ and $F$ lie on $\overline {PR}$ , with $PA=QB=QC=RD=RE=PF=10$ . Find the area of hexagon $ABCDEF$ .

120 square units 480 square units 240 square units 360 square units

1 solution

Saya Suka
Mar 28, 2021

Area of hexagon ABCDEF
= { Area of right triangle PQR } – { Total area of triangles ∆PAF, ∆BQC & ∆EDR }
= (1/2) × 30 × 40 – [ (10² / 2) × (sin P + sin Q + sin R) ]
= 30 × 40 / 2 – [ 50 × (4/5 + 3/5 + 1) ]
= 600 – 50 × (12 / 5)
= 600 – 120
= 480 unit²

Hexagon inside a triangle

1 solution

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