find the area of hexagon abcdef in which bl perpendicular to ad cm perpendicular to ad en perpendicular to ad fp perpendicular to ad such that ap is 6 cm pl is 2cm ln is 8cm nm is 2cm md is 3cm fp is 8 en is 12 cm bl is 8cm and cm is 6cm
Answers
The area of hexagon ABCDEF is 265 cm².
Step-by-step explanation:
Required Formulas:
- Area of triangle = ½ × (base) × (height)
- Area of trapezium = ½ × (sum of parallel sides) × height
It is given that,
In a hexagon ABCDE, we have
BL ⊥ AD
CM ⊥ AD
EN ⊥ AD
FP ⊥ AD
And
AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.
Referring to the figure attached below, we get
AL = AP + PL = 6 + 2 = 8 cm
PN = PL + LN = 2 + 8 = 10 cm
LM = LN + NM = 8 + 2 = 10 cm
ND = NM + MD =2 + 3 = 5 cm
Also we can see that,
Area (Hexagon ABCDEF) is given by,
= area ( ΔAPF) + area ( ΔDEN) + area ( ΔABL) + area ( ΔCMD) + area (Trap. PNEF) + area (Trap. LMCB) …… (i)
Now,
Area ( ΔAPF) = ½ × (AP) × (FP) = ½ × (6) × (8) = 24 cm² ….. (ii)
Area ( ΔDEN) = ½ × (ND) × (EN) = ½ × (5) × (12) = 30 cm² …….. (iii)
Area ( ΔABL) = ½ × (AL) × (BL) = ½ × (8) × (8) = 32 cm² …….. (iv)
Area ( ΔCMD) = ½ × (MD) × (CM) = ½ × (3) × (6) = 9 cm² …….. (v)
Area (Trap. PNEF) = ½ × (FP + EN) × PN = ½ × (8 + 12) × 10 = 100 cm² … (vi)
Area (Trap. LMCB) = ½ × (BL + CM) × LM = ½ × (8 + 6) × 10 = 70 cm²… (vii)
Thus, substituting values from (ii), (iii), (iv), (v), (vi), (vii) in (i), we get
The area of hexagon ABCDEF as ,
= 24 + 30 + 32 + 9 + 100 + 70
= 265 cm².
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