Question

please solve this problem​

Answer 1

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$PQ = 10 m \: \\ \: PS = 300m \\ EH = 10m \: \: \\ EF = 700m$

$And \: KL = 10m , \: KN = 10m$

Area of Roads = Area of PQRS + Area of EFGH - Area of KLMN

[: KLMN is Taken Twice Which is to be Subtracted]

$= PS×PQ+EF×EH-KL×KN$

$= (300 \times 10) + (700 \times 10) - (10 \times 10)$

$= 3000 + 7000 - 100$

$= 9 ,900 {m}^{2}$

Area of Road in Hectares,

$1 {m}^{2} = \frac{1}{10000} \: hectares$

$: \: 9 ,900 {m}^{2} = \frac{9900}{10000} = 0.99 \: hectares$

Now, Area of Park Excluding Cross Roads = Area of Park - Area of Road

$= ( AB \times AD) - 9 ,900$

$= (700 \times 300) - 9,900$

$= 2.00,10 ,100 \: {m}^{2}$

$= \frac{200100}{10000} \: hectares$

$= 20.01 \: hectares$

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Answer 2

Answer:

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\begin{gathered}PQ = 10 m \: \\ \: PS = 300m \\ EH = 10m \: \: \\ EF = 700m\end{gathered}

PQ=10m

PS=300m

EH=10m

EF=700m

And \: KL = 10m , \: KN = 10mAndKL=10m,KN=10m

Area of Roads = Area of PQRS + Area of EFGH - Area of KLMN

[: KLMN is Taken Twice Which is to be Subtracted]

= PS×PQ+EF×EH-KL×KN=PS×PQ+EF×EH−KL×KN

= (300 \times 10) + (700 \times 10) - (10 \times 10)=(300×10)+(700×10)−(10×10)

= 3000 + 7000 - 100=3000+7000−100

= 9 ,900 {m}^{2}=9,900m

2

Area of Road in Hectares,

1 {m}^{2} = \frac{1}{10000} \: hectares1m

2

=

10000

1

hectares

: \: 9 ,900 {m}^{2} = \frac{9900}{10000} = 0.99 \: hectares:9,900m

2

=

10000

9900

=0.99hectares

Now, Area of Park Excluding Cross Roads = Area of Park - Area of Road

= ( AB \times AD) - 9 ,900=(AB×AD)−9,900

= (700 \times 300) - 9,900=(700×300)−9,900

= 2.00,10 ,100 \: {m}^{2}=2.00,10,100m

2

= \frac{200100}{10000} \: hectares=

10000

200100

hectares

= 20.01 \: hectares=20.01hectare

Step-by-step explanation:

hope it helps you!

thank you!