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n200080 [17]
3 years ago
15

Point Z is equidistant from the sides of ΔRST. Point Z is equidistant from the sides of triangle R S T. Lines are drawn from the

point of the triangle to point Z. Lines are drawn from point Z to the sides of the triangle to form right angles and line segments Z A, Z B, and Z C. Which must be true? Line segment S Z is-congruent-to line segment T Z Line segment R Z is-congruent-to line segment B Z AngleCTZ Is-congruent-to AngleASZ AngleASZ Is-congruent-to AngleZSB
Mathematics
2 answers:
sukhopar [10]3 years ago
6 0

Answer:

AngleASZ Is-congruent-to AngleZSB

Step-by-step explanation:

The incenter of a triangle is a point inside a triangle that is equidistant from all the sides of a triangle. The incenter is the point formed by the intersection of all the three angles of the triangle bisected. The lines drawn from the incenter to the sides of the triangle forming right angles to the sides are congruent.

If Point Z is equidistant from the sides of ΔRST, point Z is the incenter of triangle RST. Lines are drawn from point Z to the sides of the triangle to form right angles and line segments Z A, Z B, and Z C. This lines are therefore congruent to each other, i.e. ZA = ZB = ZC.. Since the angles of the sides of the triangles are bisected to form the incenter, therefore:

AngleASZ Is-congruent-to AngleZSB

navik [9.2K]3 years ago
6 0

Answer:

AngleASZ Is-congruent-to AngleZSB

Step-by-step explanation:

D is the correct answer

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The volume of a rectangular prism is 48 cubic meters. It is 6 meters long and 2 meters wide.
horsena [70]

Answer:

12m^2

Step-by-step explanation:

volume = 48 m^3

l*b*h = 48

12*h = 48

h = 4 m

baSe of prism will be made of l and b

so area is 12 m^2

8 0
2 years ago
CUALES SON LOS GRADOS PARA HACER UNA GRAFICA DE PASTEL, CON LOS SIGUIENTES PORCENTAJES( 20%,50%,10%)
svlad2 [7]

Answer:

Lo primero que debemos saber, es que la gráfica de pastel nos permite dividir a una población, la cual sería el 100%, en porcentajes menores que representan algo.

En este caso tenemos los porcentajes 20%, 50%, 10%  (notar que no suman 100%, entonces nos falta algún porcentaje)

Bueno, como sabemos el gráfico (el cual es un círculo) representa el 100%.

Y en un círculo, tenemos un ángulo de 360°

Entonces comenzamos con la equivalencia:

360° = 100%

Entonces, ¿a que grado corresponde el 50%? Planteamos la ecuación:

x = 50%

Si tomamos el cociente entre las ecuaciones:

x = 50% y  360° = 100%

obtenemos:

x/360° = 50%/100%

resolviendo para x, obtenemos:

x = (50%/100%)*360° = 0.5*360° = 180°

Para los otros porcentajes hacemos lo mismo.

Para el de 20% tenemos:

y = 20%

planteamos la ecuación:

y/360° = 20%/100%

y = 0.2*360° = 72°

Finalmente, para el de 10% tendremos:

z = 10%

planteando la ecuación como antes:

z/360° = 10%/100%

z = 0.1*360 = 36°

Asi, hemos encontrado los grados correspondientes para cada porcentaje.

8 0
3 years ago
What are 2 binomials that are factors of this trinomial? x^2-x-20
S_A_V [24]

Answer:

(x-5) (x+4)

Step-by-step explanation:

x^2-x-20

What two terms multiply to -20 and add to -1

-5*4 = -20

-5+4 = -1

(x-5) (x+4)

8 0
3 years ago
Read 2 more answers
Division of fraction 3/5 divided by 10/11
Tema [17]
3/5 ÷ 10/11
multiply by the inverse

3/5 x 11/10 = 33/50
3 0
2 years ago
Read 2 more answers
Given:
diamong [38]

Answer:

10-5\sqrt{2}

Step-by-step explanation:

As per the attached figure, right angled \triangle MDL has an inscribed circle whose center is I.

We have joined the incenter I to the vertices of the \triangle MDL.

Sides MD and DL are equal because we are given that \angle M = \angle L = 45 ^\circ.

Formula for <em>area</em> of a \triangle = \dfrac{1}{2} \times base \times height

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10\sqrt{2}

Points P,Q and R are at 90 ^\circ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of  \triangleMIL, \triangleMID and \triangleDIL.

Also,

\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL

\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2

So, radius of circle = 10-5\sqrt2

8 0
3 years ago
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