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Tems11 [23]
3 years ago
15

A tile installer plots an irregular shape on grid paper. Each square on the grid represents 1 square centimeter. What is the are

a of the irregular shape

Mathematics
1 answer:
Gelneren [198K]3 years ago
6 0

Answer:

34cm²

Step-by-step explanation:

Find diagram in the attached file.

Since each square on the grid represent 1cm², the length of the each square will be 1cm

Area of the irregular shape can be gotten by breaking the irregular shape into three shapes which are the triangle, rectangle and a parallelogram.

Area of the irregular shape = area of triangle + area of rectangle + area of parallelogram

For TRIANGLE

Area of triangle = 1/2×base×height

Since the base of triangle is on 4square grid,

Base = 4cm

Height = 2cm

Area = 1/2×4×2

Area of ∆ = 4cm²

FOR RECTANGLE

Area of rectangle = Length ×Breadth

Length = 3cm

Breadth = 5cm

Area of the rectangular shape = 3×5

= 15cm²

FOR PARALLELOGRAM

Area of a parallelogram = Base × Height

Base = 5cm

Height = 3cm

Area = 5cm×3cm

Area = 15cm²

Area of the irregular shape = 4cm²+15cm²+15cm²

=34cm²

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3 regions are defined in the figure find the volume generated by rotating the given region about the specific line
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The volume generated by rotating the given region R_{3} about OC is \frac{4}{g}  \pi

<h3>Washer method</h3>

Because the given region (R_{3}) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

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\int\limits^a_b {\pi } \, (R_{o^{2} }  - R_{i^{2} } )       dy

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R_{i} = x= (\frac{y}{2}) ^{4}

a=0, b=2

v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} )  dy

v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }}  ] dy

v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]

v=\pi [\frac{1}{4} \frac{y^{3} }{3}  \int\limits^2_0 - \frac{1}{2^{8} }  \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

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