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4 years ago

A base of a triangle measures 32 centimeters . what is the measure of the corresponding midsegment of a triangle.

Mathematics

2 answers:

raketka [301]4 years ago

4 0

B.16CM

I hope it helped

soldier1979 [14.2K]4 years ago

3 0

The Midsegment Theorem states that the midsegment of a triangle is half the length of its base. Based on this, the midsegment of a triangle with a base of 32 cm would be B. 16 cm.

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Shkiper50 [21]

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Parameterize the paths

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from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

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