Answer:
Step-by-step explanation:
Formulae
Area of a triangle = 1/2 x base x height
Area of a rectangle = width x length
Question 1
Surface area of square based pyramid = area of base + 4 × area of triangle
⇒ SA = (6 × 6) + 4(1/2 × 6 × 14)
= 36 + 168
= 204 yd²
Question 2
Surface area of a cuboid = 2 × base area + 2 × end area + 2 × side area
⇒ SA = 2(10 × 6) + 2(6 × 3) + 2(10 × 3)
= 120 + 36 + 60
= 216 mm²
Extra credit question
Surface area of a triangular based prism = base area + 2 × triangle area + 2 × side areas
base area = 3 × 10 = 30 mm²
triangle area = 1/2 × 3 × 2 = 3 mm²
side area = 10 × 2.5 = 25 mm²
⇒ total SA = 30 + (2 × 3) + (2 × 25)
= 30 + 6 + 50
= 86 mm²
❍ Concept :-
We can solve these questions by using the formulas of area of triangle and rectangle, which will make our work easier. So, we know that,
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[tex] \blue{ \underline{ \boxed{ \begin{array}{cc} \sf1. ~Area_{Triangle} = \dfrac{1} {2} \times b \times h \qquad \: \: \: \: \: \: \: \\ \\ \\ \sf2. \: Area_{Rectangle} = Length \times Breadth \end{array}}}} \\ \\ [/tex]
Using these formulas and figuring out the number of triangles and rectangles in each shape, we will solve this question.
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✰ Solution :-
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1. In this square pyramid, if we closely observe, we will get to know that there are 1 rectangle and 4 triangular faces. Thus,
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[tex]\sf \hookrightarrow Area=6 \times 6 + 4( \dfrac{1}{2} \times 6 \times 14) \\ \\ \\ \sf \hookrightarrow Area=36 + 168 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \hookrightarrow Area= {204 \: yd}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ [/tex]
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2. Now, here, we can use the formula of total surface area for cuboid, i.e.
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TSA = 2( lb + bh + hl )
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[tex]\sf \hookrightarrow Area=2(10 \times 6 + 6 \times 3 + 10 \times 3) \\ \\ \\ \sf \hookrightarrow Area=2(60 + 18 + 30) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \hookrightarrow Area=2 \times 108 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \hookrightarrow Area= {216 \: mm}^{2} \: \: \qquad \qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ [/tex]
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Extra Credit Question.
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Here, we can find out the total surface area by finding,
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Base area i.e. length x breadth ( B.A. )
Side area i.e. length x breadth ( S.A. )
triangular area = 1/2 x b x h ( T.A. )
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Now,
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TSA = BA + 2 x S.A. + 2 x T.A.
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[tex]\sf \hookrightarrow Area=(10 \times 3) + 2(10 \times 2.5) + 2(\dfrac{1}{2} \times 3 \times 2) \\ \\ \\ \sf \hookrightarrow Area= 30 + 2 \times 25 + 2 \times 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \hookrightarrow Area=30 + 50 + 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \hookrightarrow Area= {86 \: mm}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ [/tex]