Question

103. In a pentagon ABCDE, DP is drawn
perpendicular to AB and is perpendicular to
CE also at point Q. If AP = BP = 12 cm,
EQ = CQ = 8 cm, DE = DC = 10 cm and
DP = 18 cm, find the area of the pentagon
ABCDE.​

Answer 1

Answer;

Area of the pentagon is 288 cm sq.

Step-by-step explanation:

[ First write the given info. i.e. the lengths of sides that r given ] Now...

In triangle DQE by using pythagoras theorem,

  a^2  + b^2 = c^2

⇒a^ 2  + 8^ 2  = 10^ 2  

⇒a^2  = 10^2  −8^ 2

 ⇒a ^2  =36

a=6cm

Now, DQ = 6 cm

So, PQ = DP - DQ  

            = 18 - 6   = 12 cm

Since, there are two triangles that measures 6cm and 8cm. Their area together would be 6×8=48 cm sq.

There are two rectangles that measure 12cm×8cm×2=192 cm sq.

 

There are two triangles at the base that measure = (4×12)=48 cm sq.

∴  Total area =48 + 192 + 48 = 288 cm sq.

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Answer 2

Answer:

288cm2

Step-by-step explanation:

a2+b2=c2

⇒a2+82=102

⇒a2=102−82

⇒a2=36

a=6cm

Since there are two triangles  that measure 6cm. and 8cm., their area together would be

6×8=48cm2

 There are two rectangles that measure 12cm×8cm.=192cm2

 There are two triangles at the base that measure 4×12=48cm2

 48+48+192=288cm2