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Find the area of hexagon, ABCDEF in which BL ⊥ AD, CM ⊥ AD and EN ⊥ AD and FP ⊥ AD such that AP=6cm, PL=2cm, LN=8cm, NM=2cm, MD=3cm, FP=8cm, EN=12cm, BL=8cm, CM=6cm. |
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Answer» The given details are, BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD AP=6cm, PL=2cm, LN=8cm, NM=2cm, MD=3cm, FP=8cm, EN=12cm, BL=8cm, CM=6cm. AL=AP+PL =6+2 = 8cm PN= PL+LN= 2+8 =10cm LM=LN+NM=8+2=10cm ND=NM+MD=2+3=5cm By using the formula, Area (hex.ABCDEF) = area (△APF) + area (△DEN) + area (△ABL) + area (△CMD) + area (Trap.PNEF) + area (Trap.LMCB) Area of triangle= ½ × base × height Area of trapezium = ½ × (sum of parallel sides) × height ∴ lets calculate, Area (△APF) = ½ (AP) × (FP) = ½ × 6 × 8 = 24cm2 Area (△DEN) =½ (ND) × (EN) = ½ × 5 × 12 = 30cm2 Area (△ABL) =½ (AL) × (BL) = ½ × 8 × 8 = 32cm2 Area (△CMD) =½ (MD) × (CM) = ½ × 3 × 6 = 9cm2 Area (Trap.PNEF) =½ × (FP + EN) × PN = ½ × (18+12) × 10 = 100cm2 Area (Trap.LMCB) =½ × (BL + CM) × LM = ½ × (8+6) × 10 = 70cm2 ∴ Area (hex.ABCDEF) = area (△APF) + area (△DEN) + area (△ABL) + area (△CMD) + area (Trap.PNEF) + area (Trap.LMCB) = 24 + 30 + 32 + 9+100+70 = 265cm2 |
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