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3 years ago

What is the area of the polygon

Mathematics

2 answers:

Paul [167]3 years ago

8 0

Split this figure into 3 shapes: 2 triangles and 1 trapezoid

Area of top triangle = 1/2(7)(2) = 7 square units
Area of bottom triangle = 1/2(3)(7) = 10.5 square units
Area of trapezoid = 1/2(3 + 6)(4) = 18 square units

Area of the polygon = 7 + 10.5 + 18 = 35.5 square units

vlada-n [284]3 years ago

6 0

Look at the picture.
$A_I=\dfrac{7\cdot2}{2}=7\\\\A_{II}=\dfrac{4\cdot3}{2}=6\\\\A_{III}=\dfrac{7\cdot3}{2}=10.5\\\\A_{IV}=3\cdot4=12\\\\The\ area\ of\ the\ polygon\\\\A=A_I+A_{II}+A_{III}+A_{IV}\\\\A=7+6+10.5+12=35.5\ un^2$

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Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

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Answer:

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Answer:

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