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QUESTION IMAGE

nd the area of the shaded region:

Question

nd the area of the shaded region:

Explanation:

Step1: Identify the overall and non - shaded shapes

For each figure, identify the overall rectangle or other simple shapes and the non - shaded simple shapes within them. The area of the shaded region is the area of the overall shape minus the area of the non - shaded shape(s). The formula for the area of a rectangle is $A = l\times w$, where $l$ is the length and $w$ is the width.

Step2: Calculate areas for part (a)

The overall rectangle has length $l = 4$ cm and width $w = 4$ cm, so its area $A_{total1}=4\times4 = 16$ $cm^{2}$. The non - shaded square has side length $s = 2$ cm, so its area $A_{non - shaded1}=2\times2=4$ $cm^{2}$. The shaded area $A_{shaded1}=A_{total1}-A_{non - shaded1}=16 - 4=12$ $cm^{2}$.

Step3: Calculate areas for part (b)

The overall rectangle has length $l = 8$ cm and width $w = 6$ cm, so $A_{total2}=8\times6 = 48$ $cm^{2}$. The non - shaded rectangle has length $l_{1}=8 - 2=6$ cm and width $w_{1}=6 - 2 = 4$ cm, so $A_{non - shaded2}=6\times4 = 24$ $cm^{2}$. The shaded area $A_{shaded2}=A_{total2}-A_{non - shaded2}=48-24 = 24$ $cm^{2}$.

Step4: Calculate areas for part (c)

The overall rectangle has length $l = 5.4$ cm and width $w = 3.8$ cm (assuming the horizontal length is $3.8$ cm from the sum of the given lengths). $A_{total3}=5.4\times3.8=20.52$ $cm^{2}$. The non - shaded rectangle has length $l_{2}=3.8-(1.8 + 1.1)=0.9$ cm and width $w_{2}=0.7$ cm, so $A_{non - shaded3}=0.9\times0.7 = 0.63$ $cm^{2}$. The shaded area $A_{shaded3}=A_{total3}-A_{non - shaded3}=20.52-0.63 = 19.89$ $cm^{2}$.

Step5: Calculate areas for part (d)

The overall rectangle has length $l = 8$ cm and width $w = 9$ cm, so $A_{total4}=8\times9=72$ $cm^{2}$. The non - shaded L - shaped figure can be divided into two rectangles. One rectangle has length $l_{3}=4$ cm and width $w_{3}=2$ cm, and the other has length $l_{4}=4$ cm and width $w_{4}=2$ cm. The total non - shaded area $A_{non - shaded4}=4\times2+4\times2=16$ $cm^{2}$. The shaded area $A_{shaded4}=A_{total4}-A_{non - shaded4}=72 - 16=56$ $cm^{2}$.

Answer:

(a) $12$ $cm^{2}$; (b) $24$ $cm^{2}$; (c) $19.89$ $cm^{2}$; (d) $56$ $cm^{2}$