in the rectangular solid in Figure 1, find the distance from vertex $R$ to vertex $T$ .

2 \sqrt{19}

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2 \sqrt{17}

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8 \sqrt{2}

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10 \sqrt{2}

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Solution

The correct option is A $2 \sqrt{19}$
From the given figure:
The length of diagonal of rectangular base is $\sqrt{{(3 \sqrt{2)}}^{2} + {(5 \sqrt{2})}^{2}} = \sqrt{18 + 50} = \sqrt{68}$
$∴ R T = \sqrt{{\sqrt{68}}^{2} + {(2 \sqrt{2})}^{2}}$
$= \sqrt{68 + 8}$
$= \sqrt{76}$
$= 2 \sqrt{1} 9$

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in the rectangular solid in Figure 1, find the distance from vertex R to vertex T.

The correct option is A 2√19From the given figure:The length of diagonal of rectangular base is √(3√2)2+(5√2)2=√18+50=√68∴RT=√√682+(2√2)2=√68+8=√76=2√19

in the rectangular solid in Figure 1, find the distance from vertex $R$ to vertex $T$ .

The correct option is A $2 \sqrt{19}$
From the given figure:
The length of diagonal of rectangular base is $\sqrt{{(3 \sqrt{2)}}^{2} + {(5 \sqrt{2})}^{2}} = \sqrt{18 + 50} = \sqrt{68}$
$∴ R T = \sqrt{{\sqrt{68}}^{2} + {(2 \sqrt{2})}^{2}}$
$= \sqrt{68 + 8}$
$= \sqrt{76}$
$= 2 \sqrt{1} 9$