Chapter 12.3, Problem 34PPS

To determine

Tofind:The surface area of given composite solid, round to the nearest tenth.

Answer to Problem 34PPS

The surface area of the composite solid is $427.3{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given information:

The diagram of composite solid is shown in Figure-1.

Formulaused:

The lateral surface area of a regular pyramid of perimeter P and slant height h is given by $L=\frac{1}{2}Pl$ .

Calculations:

The composite figure is made of joining two pyramids Zabcd of height $h=5\text{m}$ and BD and $Z^{\prime }{a}^{\prime }{b}^{\prime }{c}^{\prime }{d}^{\prime }$ of height $h=9\text{m}$ between them sandwiched a square (side $a=\text{10}\text{m}$ ) based parallelepiped as shown in Figure-2.

The surface area of composite solid is equal to the sum of lateral areas of two pyramids plus the lateral area of square based parallelepiped.

Now, the slant height of upper pyramidis $l=\sqrt{a^2+h^2}=\sqrt{{\text{5}}^{\text{2}}\text{+}{\text{5}}^{\text{2}}}\text{=5}\sqrt{\text{2}}\text{m}$ , and perimeter of its base is $P=4a\text{=4×10=40}\text{m}$ .

Therefore, lateral area of the upper pyramid is $S_1=\frac{1}{2}Pl\text{=}\frac{\text{1}}{\text{2}}\text{×40×5}\sqrt{\text{2}}{\text{m}}^{\text{2}}=\text{100}\sqrt{\text{2}}{\text{m}}^{\text{2}}$.... (1)

The slant height of lower pyramid is $l=\sqrt{a^2+h^2}=\sqrt{{\text{5}}^{\text{2}}\text{+}{\text{9}}^{\text{2}}}\text{=}\sqrt{\text{106}}\text{m}$ , and perimeter of its base is $P=4a\text{=4×10=40}\text{m}$ .

Therefore, lateral area of the upper pyramid is $S_2=\frac{1}{2}Pl\text{=}\frac{\text{1}}{\text{2}}\text{×40×}\sqrt{\text{106}}{\text{m}}^{\text{2}}=\text{20}\sqrt{\text{106}}{\text{m}}^{\text{2}}$ .

The lateral area of the parallelepiped $S_3=P\times h\text{=40×2}=\text{80}{\text{m}}^{\text{2}}$ .

Therefore, the lateral area of composite solid $S=\left(100\sqrt{2}\text{+20}\sqrt{106}\text{+80}\right){\text{m}}^{\text{2}}=427.3{\text{m}}^{\text{2}}$

Conclusion:

The surface area of the composite solid is $427.3{\text{m}}^{\text{2}}$ .