Find the area of hexagon ABCDEF in which $B L ⊥ A D$ , $C M ⊥ A D$ , $E N ⊥ A D$ and $F P ⊥ A D$ such that 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.

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Solution

The given details are,
$B L ⊥ A D$ , $C M ⊥ A D$ , $E N ⊥ A D$ and $F P ⊥ A D$
$AP=6 cm, PL= 2 cm, LN=8 cm, NM=2 cm, MD=3 cm, FP=8 cm, EN=12 cm. BL= 8 cm ,
$C M = 6 c m .$
$A L = A P + P L = 6 + 2 = 8 c m$
$P N = P L + L N = 2 + 8 = 10 c m$
$L M = L N + N M = 8 + 2 = 10 c m$
$N D = N M + M D = 2 + 3 = 5 c m$
By Using the formula,
Area (hex. ABCDEF) $= a r e a (△ A P F) + a r e a (△ D E N) + a r e a (△ A B L) + a r e a (△ C M D) + a r e a (T r a p . P N E F) + a r e a (T r a p . L M C B)$
Area of triangle $= 1 / 2 \times b a s e \times h e i g h t$
Area of trapezium $= 1 / 2 \times (s u m o f p a r a l l e l s i d e s) \times h e i g h t$
$∴$ lets calculate,
$A r e a (△ A P F) = 1 / 2 (A P) \times (F P) = 1 / 2 \times 6 \times 8 = 24 c m^{2}$
$A r e a (△ D E N) = 1 / 2 (N D) \times (E N) = 1 / 2 \times 5 \times 12 = 30 c m^{2}$
$A r e a (△ A B L) = 1 / 2 (A L) \times (B L) = 1 / 2 \times 8 \times 8 = 32 c m^{2}$
$A r e a (△ C M D) = 1 / 2 (M D) \times (C M) = 1 / 2 \times 3 \times 6 = 9 c m^{2}$
$A r e a (T r a p . D M N E) = 1 / 2 \times (F P + E N) \times O N = 1 / 2 \times (18 + 12) \times 1 = 100 c m^{2}$
$A r e a (T r a p . D M N E) = 1 / 2 \times (B L + C M) \times L M = 1 / 2 \times (B L + C M) \times L M = 1 / 2 \times (8 + 6) \times 10 = 70 c m^{2}$
$∴ A r e a (h e r . A B C D E F) = a r e a (△ A P F) + a r e a (△ D E N) + a r e a (△ A B L) + a r e a (△ C M D) + a r e a (T r a p . P N E F) + a r e a (T r a p . L M C B)$
$= 24 + 30 + 32 + 9 + 100 + 70$
$= 265 c m^{2}$

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Find the area of hexagon ABCDEF in which BL⊥AD, CM⊥AD, EN⊥AD and FP⊥AD such that 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.

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