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3 years ago

11

The length of a rectangle is more than three times its width. If the perimeter of the rectangle is 92 meters, find the dimension

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1 answer:

Juliette [100K]3 years ago

4 0

The dimensions are 35 m x 11 m

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You work at a pioneer historical site. On this site you have handcarts. One cart has a handle that connects to the center of the

Gelneren [198K]

Answer:

a) see below

b) radius = 16.4 in (1 d.p.)

c) 18°. Yes contents will remain. No, handle will not rest on the ground.

d) Yes contents would spill. Max height of handle = 32.8 in (1 d.p.)

Step-by-step explanation:

<u>Part a</u>

A chord is a <u>line segment</u> with endpoints on the <u>circumference</u> of the circle.

The diameter is a <u>chord</u> that passes through the center of a circle.

Therefore, the spokes passing through the center of the wheel are congruent chords.

The spokes on the wheel represent the radii of the circle. Spokes on a wheel are usually evenly spaced, therefore the congruent central angles are the angles formed when two spokes meet at the center of the wheel.

<u>Part b</u>

The <u>tangent</u> of a circle is always <u>perpendicular</u> to the <u>radius</u>.

The tangent to the wheel touches the wheel at point B on the diagram. The radius is at a right angle to this tangent. Therefore, we can model this as a right triangle and use the <u>tan trigonometric ratio</u> to calculate the radius of the wheel (see attached diagram 1).

$\sf \tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle

Given:

$\theta$ = 20°
O = radius (r)
A = 45 in

Substituting the given values into the tan trig ratio:

$\implies \sf \tan(20^{\circ})=\dfrac{r}{45}$

$\implies \sf r=45\tan(20^{\circ})$

$\implies \sf r=16.37866054...$

Therefore, the radius is 16.4 in (1 d.p.).

<u>Part c</u>

The measure of an angle formed by a secant and a tangent from a point outside the circle is <u>half the difference</u> of the measures of the <u>intercepted arcs</u>.

If the measure of the arc AB was changed to 72°, then the other intercepted arc would be 180° - 72° = 108° (since AC is the diameter).

$\implies \sf new\: angle=\dfrac{108^{\circ}-72^{\circ}}{2}=18^{\circ}$

As the handle of the cart needs to be no more than 20° with the ground for the contents not to spill out, the contents will remain in the handcart at an angle of 18°.

The handle will not rest of the ground (see attached diagram 2).

<u>Part d</u>

This can be modeled as a right triangle (see diagram 3), with:

height = (48 - r) in
hypotenuse ≈ 48 in

Use the sin trig ratio to find the angle the handle makes with the horizontal:

$\implies \sf \sin (\theta)=\dfrac{O}{H}$

$\implies \sf \sin (\theta)=\dfrac{48-r}{48}$

$\implies \sf \sin (\theta)=\dfrac{48-45\tan(20^{\circ})}{48}$

$\implies \theta = 41.2^{\circ}\:\sf(1\:d.p.)$

As 41.2° > 20° the contents will spill out the back.

To find the <u>maximum height</u> of the handle from the ground before the contents start spilling out, find the <u>height from center of the wheel</u> (setting the angle to its maximum of 20°):

$\implies \sin(20^{\circ})=\dfrac{h}{48}$

$\implies h=48\sin(20^{\circ})$

Then add it to the radius:

$\implies \sf max\:height=48\sin(20^{\circ})+45\tan(20^{\circ})=32.8\:in\:(1\:d.p.)$

(see diagram 4)

------------------------------------------------------------------------------------------

<u>Circle Theorem vocabulary</u>

<u>Secant</u>: a straight line that intersects a circle at two points.

<u>Arc</u>: the curve between two points on the circumference of a circle

<u>Intercepted arc</u>: the curve between the two points where two chords or line segments (that meet at one point on the other side of the circle) intercept the circumference of a circle.

<u>Tangent</u>: a straight line that touches a circle at only one point.

7 0

2 years ago

Anybody yooo plzzzzzz

Elena-2011 [213]

Answer:

what do you have to do this?

Step-by-step explanation:

what grade and type of math is this?

6 0

3 years ago

A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra

Ksivusya [100]

The Quadrilateral is JKLM,

let $M_{JK}, M_{KL}, M_{LM}, M_{JM},$ be the midpoints of JK, KL, LM and JM respectively.

---------------------------------------------------------------------------------------------------------

Given any 2 point P(m,n) and Q(k,l),<span>

the coordinates of the midpoint of the line segment PQ are given by the formula:

$M_{PQ}=( \frac{m+k}{2} ,
\frac{n+l}{2})$ , </span>

-------------------------------------------------------------------------------------------------

thus the coordinates of points $M_{JK}, M_{KL}, M_{LM}, M_{JM},$

are as follows:

$M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)$

------------------------------------------------------------------------------------------------

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

$|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}
}$ </span>

------------------------------------------------------------------------------------------------

thus the distances connecting the opposite entrances can be calculated as follows:

$|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5$

$|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39$

Thus the total distance of the paths joining the opposite entrances is

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)

Answer: 104 m

8 0

3 years ago

im thinking of a 3 digit number when i divide it by 5 i am left with a remainder of 3 when i divide it by 10 i am left with a re

UNO [17]

9514 1404 393

Answer:

198

Step-by-step explanation:

The ones digit must be 8. The tens digit must be 5 or greater. Since it only has one hundred, it could be any of ...

158, 168, 178, 188, 198

6 0

3 years ago

The parent function of the function g(x) = (x - 5)2 + k is f(x) = x2. The vertex of the function g(x) is located at (9,-8).

likoan [24]

Answer:

h = 5 and k = -8.

Step-by-step explanation:

The parent equation is f(x) = x², which is the equation of a parabola having the vertex at point (0,0).

The original equation is given by g(x) = (x - 5)² + k, ⇒ y = (x - 5)² + k

⇒ y - k = (x - 5)²

So, the vertex of the original equation is at (5, k) which is given to be (5, -8)

Therefore, h = 5 and k = -8. (Answer)

6 0

3 years ago