Question

Find the area of the hexagon ABCDEF. here AD=8cm, AJ=6cm, AI=5cm, AH=3cm, AG=2.5cm and FG,BH,EI and CJ are perpendicular to diagonal AD.​

Answer 1

Answer:

32.5 is the answer of this question

Answer 2

Find the area of the given irregular hexagon for given values.

Explanation:

1. let there two more perpendicular lines FK(from F to EI) and CL (from C to BH).
2. Then the given hexagon can be broken into two rectangles and six triangles.
3. now we have, $\Delta_1AGF,\ \Delta_2AHB,\ \Delta_3DIE,\ \Delta_4DJC,\ \Delta_5EFK,\ \Delta_6CBL$               and rectangles, $FKIG\ and\ CJHB$  
4. given that , $AD=8,\ AJ=6,\ AI=5, AH=3,\ AG=2.5,\ FG=BH=3,\ EI=4,\ CJ=2$                                      we get, $GI=2.5, HJ=3,\ BL=1,\ EK=1$  
5. now area of individual sections are,                                                                    [tex]a(AGF)=\frac{1}{2}(AG)(GF)=\frac{1}{2}(2.5)(3)=3.75 \\ a(AHB)=\frac{1}{2}(AH)(HI)=\frac{1}{2}(3)(3)=4.5 \\ a(DIE)=\frac{1}{2}(DI)(IE)=\frac{1}{2}(3)(4)=6\\ a(DJC)=\frac{1}{2}(DJ)(JC)=\frac{1}{2}(2)(2)=2\\ a(EFK)=\frac{1}{2}(FK)(KE)=\frac{1}{2}(2.5)(1)1.25\\ a(CBL)=\frac{1}{2}(BL)(LC)=\frac{1}{2}(3)(1)=1.5\\ a(FKIG)=(FG)(GI)=(3)(2.5)=7.5\\ a(CJHB)=(HJ)(JC)=(3)(2)=6\\[/tex]          
6. now the sum of the hexagon is sum off all of the above individual sections ,                                                                                                                   [tex]A_{hex}=3.75+4.5+6+2+1.25+1.5+7.5+6\\ A_{hex}=32.5\ cm^2[/tex]------->ANSWER