Area and volume help?

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Area and volume help?

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Answer 1 · 2018-02-16T13:43:26

Area $A = 42$ $A=42$

Explanation:

In the figure, we can split the given closed polygon into rectangles and triangles

Triangles are 4 in number

$b_{1} = 3, h_{1} = 2, T_{1} = \frac{1}{2} (b_{1} h_{1}) = \frac{1}{2} (3) (2) = 3$ $b_1=3, h_1=2, T_1=1/2(b_1h_1)=1/2(3)(2)=3$
$b_{2} = 4, h_{2} = 2, T_{2} = \frac{1}{2} (b_{2} h_{2}) = \frac{1}{2} (4) (2) = 4$ $b_2=4, h_2=2, T_2=1/2(b_2h_2)=1/2(4)(2)=4$
$b_{3} = 3, h_{3} = 2, T_{1} = \frac{1}{2} (b_{1} h_{1}) = \frac{1}{2} (3) (2) = 3$ $b_3=3, h_3=2, T_1=1/2(b_1h_1)=1/2(3)(2)=3$
$b_{4} = 4, h_{4} = 2, T_{1} = \frac{1}{2} (b_{1} h_{1}) = \frac{1}{2} (4) (2) = 4$ $b_4=4, h_4=2, T_1=1/2(b_1h_1)=1/2(4)(2)=4$

Total area of triangles $T = T_{1} + T_{2} + T_{3} + T_{4}$ $T=T_1+T_2+T_3+T_4$
$T = 3 + 4 + 3 + 4 = 14$ $T=3+4+3+4=14$

Rectangle is 1 in number

$b = 7, h = 4, R = b_{1} h_{1} = (7) (4) = 28$ $b=7, h=4, R=b_1h_1=(7)(4)=28$

Total area $A = T + R = 14 + 28 = 42$ $A=T+R=14+28=42$

Area $A = 42$ $A=42$

Answer 2 · 2018-02-16T13:58:35

$area = 42 square units$ $"area "=42" square units"$

Explanation:

$we can split the figure into 'regular' figures$ $"we can split the figure into 'regular' figures"$

$the central region is a rectangle$ $"the central region is a "color(blue)"rectangle"$

$the regions above and below the rectangle are triangles$ $"the regions above and below the rectangle are "color(blue)"triangles"$

$∙ area of rectangle = length \times breadth$ $• " area of rectangle "="length "xx" breadth"$

$∙ area of triangle = \frac{1}{2} \times base \times height$ $• " area of triangle "=1/2xx" base "xx" height"$

$area of rectangle = 7 \times 4 = 28$ $"area of rectangle "=7xx4=28$

$area of 2 congruent triangles = 2 \times \frac{1}{2} \times 7 \times 2 = 14$ $"area of 2 congruent triangles "=2xx1/2xx7xx2=14$

$area of figure = 28 + 14 = 42 square units$ $"area of figure "=28+14=42" square units"$

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Area and volume help?

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2 Answers

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