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2009-II-17

Posted on 16-06-202121-06-2023 By app.cch No Comments on 2009-II-17
Ans: B
The length of a side of the base square

$\begin{array}{cl}
= & 48 \div 4 \\
= & 12 \text{ cm}
\end{array}$

Therefore, the base area

$\begin{array}{cl}
= & 12^2 \\
= & 144\text{ cm}^2
\end{array}$

Note that the base of the solid right pyramid is a square, then all the slant faces are identical. By applying the Heron’s formula to one of the slant faces, we have

$\begin{array}{rcl}
s & = & \dfrac{10 + 10 + 12}{2} \\
& = & 16 \text{ cm}
\end{array}$

Therefore, the area of a slant face

$\begin{array}{cl}
= & \sqrt{16(16-10)(16-10)(16-12) } \\
= & 48\text{ cm}^2
\end{array}$

Therefore, the total surface area of the pyramid

$\begin{array}{cl}
= & 48 \times 4 + 144 \\
= & 336\text{ cm}^2
\end{array}$

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2009, HKCEE, Paper 2 Tags:Mensuration

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3D Problems (41) Basic Functions (13) Basic Geometry (68) Binomial Theorem (7) Change of Subject (32) Complex Numbers (16) Coordinates (46) Differentiation (16) Equations of Circle (54) Equations of Straight Line (43) Estimations and Errors (35) Factorization (39) Graph of Functions (3) Inequality (39) Integration (15) Laws of Indices (43) Linear Programming (21) Locus (13) Logarithm (34) Mathematical Induction (7) Matrices (4) Mensuration (98) Numeral System (19) Percentage (42) Polynomials (49) Probability (85) Properties of Circles (56) Quadratic Equations and Functions (57) Rate and Ratio (30) Rational Functions (20) Sequences (66) Simultaneous Linear Equations (27) Statistics (122) System of Linear Equations (3) Transformations (44) Trigonometry (M2) (7) Trigonometry and Its Applications (67) Variations (38) Vectors (3)

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