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3 years ago

Which of the following sets of possible side lengths forms a triangle?

1 answer:

7 0

We know that
The triangle inequality states that for any triangle, the sum of the lengths of any two sides of a triangle is greater than the length of the third side

Part 1) Which of the following sets of possible side lengths forms a triangle?
case A) 6, 9, and 15
6+9 is not > 15

case B) 4, 10, and 15
4+10 is not > 15

case C) 10, 15, and 25
10+15 is not > 25

case D) 12, 12, and 23
12+12 is > 23-----> ok
12+23 is > 12----> ok

the answer part 1) is
12, 12, and 23

Part 2) Which of the following sets of possible side lengths forms a right triangle?

case A) 12, 35, and 38
if the side lengths forms a right triangle
then
12²+35²=38²
12²+35²------> 1369
38²------> 1444

1369 is not equal to 1444

case B) 11,60, and 61
if the side lengths forms a right triangle
then
11²+60²=61²
11²+60²-------> 3721
61²------> 3721
3721 is equal to 3721------> the sides forms a right triangle

case C) 9,40 and 45
if the side lengths forms a right triangle
then
9²+40²=45²
9²+40²------> 1681
45²------> 2025
1681 is not 2025

case D) 6, 12, and 13
if the side lengths forms a right triangle
then
6²+12²=13²
6²+12²------> 180
13²--------> 169
180 is not 169

the answer part 2) is
case B) 11,60, and 61 forms a right triangle

Part 3) Which of the following sets of sides does not form a right triangle?

case A) 6, 8, 10
if the side lengths forms a right triangle
then
6²+8²=10²
6²+8²------> 100
10²-------> 100
100 is equal to 100 ------> the side lengths forms a right triangle

case B) 8,15,17
if the side lengths forms a right triangle
then
8²+15²=17²
8²+15²------> 289
17²-------> 289
289 is equal to 289-----> the side lengths forms a right triangle

case C) 7,24,26
if the side lengths forms a right triangle
then
7²+24²=26²
7²+24²------> 1176
26²--------> 676
1176 is not 676-----> the sides lengths does not form a right triangle

case D) 5,12,13
if the side lengths forms a right triangle
then
5²+12²=13²
5²+12²-------> 169
13²-------> 169
169 is equal to 169------> the side lengths forms a right triangle

the answer Part 3) is
the option 7,24,26 does not form a right triangle

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3 years ago

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2 years ago

Please help w this! Its a calculus question! look at the picture for the problem,

neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18$

Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is

$\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18$

so the total area is, once again, $36\pi-72$ .

An even simpler way is to subtract the area of the square from the area of the circle.

$\pi6^2-(6\sqrt2)^2=36\pi-72$

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3 years ago