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Answers
Answer:
how much is papers length it is not shown only e then how we can say the answer paper
Answer:
Consider the problem
Semi perimeter of region I (triangle)
s=
2
5+5+1
=
2
11
=5.5cm
By using Hero's formula, then the area of region I (triangle)
=
s(s−5)(s−5)(s−1)
=
5.5(5.5−5)(5.5−5)(5.5−1)
=
5.5×0.5×0.5×4.5
=2.5cm
2
(approx)..........(1)
And,
Area of region II (rectangle)
Area of rectangle =length×Breadth
so,
1×6.5
=6.5cm
2
..........(2)
Now applying Pythagoras theorem on ΔABC
AB
2
=AC
2
+BC
2
⇒AB
2
=6
2
+(1.5)
2
⇒AB
2
=36+2.25=38.25
⇒AB=6.2cm
2
(approx)
Semi perimeter of region IV ΔABC
s=
2
6+1.5+6.2
=
2
13.7
=6.85cm
2
Using Hero's formula, Area of region IV ΔABC
=
s(s−6)(s−1.5)(s−6.184)
=
6.85(6.85−6)(6.85−1.5)(6.85−6.2)
=
6.85×0.85×5.35×0.65
=4.5cm
2
..............(3)
Area of region V triangle =AreaofregionIV(ΔABC)=4.5cm
2
.......(4)
Here it seems Region III is a trapezium and we do not know the height of this trapezium, so we have to draw this trapezium separately so we can find its area.
So, Draw ST⊥PQandRU⊥PQ. doing so, we have PT=0.5cmandUQ=0.5cm
Now, we can use Pythagoras theorem on ΔSPT to find the height of trapezium PQRS.
SP
2
=ST
2
+PT
2
⇒1
2
=ST
2
+(0.5)
2
⇒ST
2
=1−0.25=0.75
⇒ST=0.87cm(approx)
So, Area of region III (Trapezium)
=
2
1
(SR+PQ)×ST
=
2
1
(1+2)×0.87
=1.305cm
2
.........(5)
Now, Adding (1),(2),(3),(4) and (5) to find the total area of aeroplane picture.
=2.5+6.5+4.5+4.5+1.305
=19.305cm
2
(approx)
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