Question

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Answer 1

Answer:

Step-by-step explanation:

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Semi-perimeter of region I (triangle)=s=5+5+12=112=5.5 cm

Using Heron's formula, 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−−−−−−−−−−−−−−−−√=2.5 cm2 (approx.)          (1)

Area of region II (rectangle) = base × height =1×6.5=6.5 cm2         (2)Applying Pythagoras theorem on △ABC, we get

AB2=AC2+BC2

⇒AB2=62+(1.5)2

⇒AB2=36+2.25=38.25

⇒AB=6.2 cm2 (approx.)

Semi-perimeter of region IV (△ABC)=s=6+1.5+6.22=13.72=6.85 cm2

Using Heron's formula, Area of region IV (△ABC)=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−−−−−−−−−−−−−−−−−−−−√=4.5 cm2          (3)

Area of region V (triangle)=Area of region IV (△ABC)=4.5 cm2                  (4)

Region III is trapezium but we do not know height of this trapezium. Let's draw this trapezium separately so we can find its area.

Draw ST⊥PQ and RU⊥PQ. Doing so, we will have PT=0.5 cm and UQ=0.5 cm.

We can use Pythagoras theorem on △SPT to find height of trpazezium PQRS.

SP2=ST2+PT2

⇒12=ST2+(0.5)2

⇒ST2=1−0.25=0.75

⇒ST=0.87 cm   (approx.)

Area of region III (trapezium)=12(SR+PQ)×ST

=12(1+2)×0.87=1.305 cm2                (5)

Adding (1), (2), (3), (4) and (5), we get

Area of aeroplane picture =2.5+6.5+4.5+4.5+1.305=19.305 cm2

Answer 2

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