Question

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Answer 1

Answer:

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Answer 2

Answer:

✍${\huge{\purple{\mathfrak{\underline{\underline{Question}}}}}}$✍

Find the Area of given figure ABCDEF

✍${\huge{\purple{\mathfrak{\underline{\underline{Formula\ Used}}}}}}$✍

• Area of triangle=$\frac{1}{2}×base×height$

• Area of trapezium=$\frac{1}{2}×(sum of parallel sides)×height$

✍${\huge{\purple{\mathfrak{\underline{\underline{Answer}}}}}}$✍

✈${\pink{\underline{\underline{Given}}}}$✈

• AD=8cm
• AJ=6cm
• AI=5cm
• AH=3cm
• AG=2.5 cm

FG,BH,EI And CJ are perpendicular to the diagonal AD

• FG=3cm
• BH=3cm
• EI=4cm
• CJ=2cm

✈${\pink{\underline{\underline{To\ Find}}}}$✈

• Area of whole figure ABCDEF=????

✈${\pink{\underline{\underline{Solution}}}}$✈

Area of whole figure ABCDEF;

=(Area of triangle AGF)+(Area of trapezium GFEI)+(Area of triangle EID)+(Area of triangle ABH)+(Area of trapezium HBCJ)+(Area of triangle JDC)

• Area of triangle AGF;

Here,base=AG=2.5cm

Height=FG=3cm

So, Area of triangle=$\frac{1}{2}×2.5×3$

$=\frac{7.5}{2}$

$=3.75{cm}^{2}$

•Area of trapezium GFEI;

Here,parallel sides=FG and EI=3cm and 4cm

height=GI=(AI-AG)=(6-2.5)=3.5cm

So,Area of trapezium=$\frac{1}{2}×(3+4)×3.5$

$=\frac{24.5}{2}$

=12.25${cm}^{2}$

• Area of triangle EID;

Here,base=ID=AD-AI=3cm

Height=EI=4cm

So, Area of triangle=$\frac{1}{2}×3×4$

$=\frac{12}{2}$

$=6{cm}^{2}$

• Area of triangle ABH;

Here,base=AH=3cm

Height=BH=3cm

So, Area of triangle=$\frac{1}{2}×3×3$

$=\frac{9}{2}$

$=4.5{cm}^{2}$

•Area of trapezium HBCJ;

Here,parallel sides=BF and JC=3cm and 2cm

height=HJ=(AJ-AH)=(6-3)=3cm

So,Area of trapezium=$\frac{1}{2}×(3+2)×3$

$=\frac{15}{2}$

=7.5${cm}^{2}$

• Area of triangle JDC;

Here,base=JD=(AD-AJ)=(6-8)=2cm

Height=JC=2cm

So, Area of triangle=$\frac{1}{\cancel2}×\cancel2×2$

$=2{cm}^{2}$

So,Total area=3.75+12.25+24.5+6+4.5+2

$=53{cm}^{2}\ is\ answer$