Question
The diagram, given below, shows two paths drawn inside a rectangular field $80\ m$ long and $45\ m$ wide. The widths of the two paths are $8\ m$ and $15\ m$ as shown. Find the area of the shaded portion.
✓
Answer
Consider the following figure.
Thus, the area of the shaded portion
$=$ Area$(\text {ABCD} ) +$ Area$(\text{ EFGH} ) - $Area$(\text{IJKL} ) \dots...(1)$
Dimensions of $\text{ABCD}: 45 \mathrm{~m} \times 15 \mathrm{~m}$
Thus, the area of $\text{ABCD}=45 \times 15=675 \mathrm{~m}^2$
Dimensions of $\text{EFGH}: 80 \mathrm{~m} \times 8 \mathrm{~m}$
Thus, the area of $\text{EFGH} =80 \times 8=640 \mathrm{~m}^2$
Dimensions of $\text{IJKL}: 15 \mathrm{~m} \times 8 \mathrm{~m}$
Thus, the area of $\text{IJKL} =15 \times 8=120 \mathrm{~m}^2$
Therefore, from equation $(1),$
the area of the shaded portion $=675+640-120$
$=1195 \mathrm{~m}^2$
Thus, the area of the shaded portion
$=$ Area$(\text {ABCD} ) +$ Area$(\text{ EFGH} ) - $Area$(\text{IJKL} ) \dots...(1)$
Dimensions of $\text{ABCD}: 45 \mathrm{~m} \times 15 \mathrm{~m}$
Thus, the area of $\text{ABCD}=45 \times 15=675 \mathrm{~m}^2$
Dimensions of $\text{EFGH}: 80 \mathrm{~m} \times 8 \mathrm{~m}$
Thus, the area of $\text{EFGH} =80 \times 8=640 \mathrm{~m}^2$
Dimensions of $\text{IJKL}: 15 \mathrm{~m} \times 8 \mathrm{~m}$
Thus, the area of $\text{IJKL} =15 \times 8=120 \mathrm{~m}^2$
Therefore, from equation $(1),$
the area of the shaded portion $=675+640-120$
$=1195 \mathrm{~m}^2$
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