Question

Find
a. the length of the missing dimension and b. theperimeter of each figure.
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Answer 1

`\large\boxed{a)\ (31)/(32)\ in}\\\boxed{b)\ 16(5)/(16)\ in}`

Explanation:

Look at the picture.

`a)\\3(9)/(32)\ in-2(5)/(16)\ in=3(9)/(32)\ in-2(5\cdot2)/(16\cdot2)\ in=3(9)/(32)\ in-2(10)/(32)\ in\\\\=\bigg((3-2)+(9-10)/(32)\bigg)\ in=\bigg(1+(-1)/(32)\bigg)\ in=\bigg((32)/(32)-(1)/(32)\bigg)\ in\\\\=(32-1)/(32)\ in=(31)/(32)\ in`

b)

The perimeter of given figure is equal to the perimeter of the rectangle (look at the second picture).

`P=2\left(3(9)/(32)\ in+2(1)/(2)\ in+2(3)/(8)\ in\right)=2\left(3(9)/(32)+2(1\cdot16)/(2\cdot16)+2(3\cdot4)/(8\cdot4)\right)\ in\\\\=2\bigg(3(9)/(32)+2(16)/(32)+2(12)/(32)\bigg)\ in=2\left((3+2+2)+(9+16+12)/(32)\right)\ in\\\\=2\left(7+(37)/(32)\right)\ in=\bigg((2)(7)+(2)\left((37)/(32)\right)\bigg)\ in=\left(14+(37)/(16)\right)\ in\\\\=\bigg(14+2(5)/(16)\bigg)\ in=16(5)/(16)\ in`