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jolli1 [7]
3 years ago
13

The figure shows a circle inscribed in a triangle. To construct the inscribed circle, angle bisectors were first constructed at

each angle of the triangle. Which happened next? A circle was constructed using the intersection of the angle bisectors as the center of the circle and the obtuse vertex as a point on the circumference of the circle. A circle was constructed using a vertex as the center of the circle and the intersection of the angle bisectors as a point on the circumference of the circle. Segments perpendicular to the sides of the triangle through the intersection of the angle bisectors were constructed. Segments bisecting each side of the triangle were constructed through the intersection of the angle bisectors.
Mathematics
2 answers:
klemol [59]3 years ago
7 0

Answer:

Segments perpendicular to the sides of the triangle through the intersection of the angle bisectors were constructed.

Step-by-step explanation:

The above choice represents a bit of excess work. Actually, only one such perpendicular line segment needs to be constructed in order to determine the radius of the inscribed circle.

Once you know the center and radius, you can construct the inscribed circle.

bearhunter [10]3 years ago
5 0

Answer:

C.Segments perpendicular to the sides of the triangle through the intersection of the angle bisectors were constructed.

Step-by-step explanation:

100%

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lilavasa [31]
Use the Pythagorean theorem:

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From a circular cylinder of diameter 10 cm and height 12 cm are conical cavity of the same base radius and of the same height is
Nataliya [291]
<h3>Volume of the remaining solid = 628 cm^2</h3>

<h3>Whole surface area = 659.4 cm^2</h3>

Step-by-step explanation:

Now, Given that:-

Diameter (d) = 10 cm

So, Radius (r) = 10/2 = 5cm

Height of the cylinder = 12cm.

volume \: of \: the \: cylinder \:  =  \pi {r}^{2} h

=  > \pi \times  {5}^{2} \times  12 {cm}^{3}   = 300\pi {cm}^{3}

Radius of the cone = 5 cm.

Height of the cone = 12 cm.

slant \: height \: of \: the \: cone \:  =  \sqrt{ {h}^{2}  + \:  {r}^{2} }

=  >  \sqrt{ {5}^{2}+{12}^{2} } cm \:  = 13cm

Volume of the cone = 1/3 *πr^2h

=  >  \frac{1}{3} \pi \times  {5}^{2}   \times 12 {cm}^{3}  = 100\pi {cm}^{3}

therefore, the volume of the remaining solid

= 300\pi {cm}^{3}  - 100\pi {cm}^{3}  \\  = 200 \times 3.14 {cm}^{3}  = 628 {cm}^{3}

Curved surface of the cylinder =

2\pi \: rh \:  = 2\pi \times 5 \times 12 {cm}^{2}  \\  = 120\pi {cm}^{2} .

curved \: surface \: of \: the \: cone \:  = \pi \: rl \\  = \pi \times 5 \times 13 {cm}^{2}  \\  = 65\pi {cm }^{2} \\ area \: of \: (upper)circular \: base \: \\  of \: cylinder \:  =  \\ =  \pi \:  {r}^{2}  = \pi \times  {5}^{2}

therefore, The whole surface area of the remaining solid

= curved surface area of cylinder + curved surface area of cone + area of (upper) circular base of cylinder

= 120\pi {cm}^{2}  + 65\pi {cm }^{2}  + 25 \pi {cm}^{2}  \\  = 210 \times 3.14 {cm}^{2}  = 659.4 {cm}^{2}

<h3>Hope it helps you!!</h3>

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Afina-wow [57]

The probability of choosing cards either Q or R when a card is drawn from a deck of 8 cards is 0.25.

Given that a card is randomly chosen from 8 cards shown in figure.

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