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frez [133]
3 years ago
7

TV−→− bisects ∠RTS. If the m∠RTV=(16x−6)° and m∠VTS=(13x+9)° , what is the value of x and the m∠RTV ?

Mathematics
1 answer:
natta225 [31]3 years ago
8 0

Answer:

x = 5

RTV = 74

Step-by-step explanation:

Given

RTV=16x - 6

VTS=13x+9

Required

Determine the values of x and RTV

Since, TV is a bisector, then

RTS = RTV + VTS --- (1)

and

RTV = VTS -- (2)

Substitute values of RTV and VTS in (2)

16x - 6 = 13x + 9

Collect Like Terms

16x - 13x = 9 + 6

3x = 15

Solve for x

x = 5

Substitute 5 for x in RTV=16x - 6

RTV = 16 * 5 - 6

RTV = 80 - 6

RTV = 74

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Answer:

V = 240π cm^3 , S= 168π cm^2

Step-by-step explanation:

The given figure is a combination of hemi-sphere and a cone

<u>Volume:</u>

For volume

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h = 8 cm

Volume\ of\ cone = \frac{1}{3}\pi r^2h\\= \frac{1}{3}\pi (6)^2*8\\=\frac{1}{3}\pi *36*8\\=\frac{288}{3}\pi\\=96\pi cm^3 \\\\Volume\ of\ hemisphere = \frac{2}{3}\pi r^3\\=\frac{2}{3}*\pi * (6)^3\\=\frac{2}{3}*\pi *216\\=\frac{432}{3}\\=144\pi cm^3 \\\\Total\ Volume= Volume\ of\ cone + Volume\ of\ hemisphere\\= 96\pi +144\pi \\=240\pi cm^3

<u>Surface Area:</u>

For this particular figure we have to consider the lateral area of the cone shape and surface area of the hemisphere

We have to find the lateral height

l = \sqrt{r^2+h^2}\\ l = \sqrt{(6)^2+(8)^2} \\l= \sqrt{36+64}\\ l = \sqrt{100}\\l = 10cm\\\\Surface\ area\ of\ cone = \pi rl\\= \pi (6)(10)\\=\pi *60\\=60 \pi\ cm^2\\\\Surface\ area\ of\ hemisphere = 2\pi r^2\\= 2 \pi * (6)\\= 2 \pi *36\\= 72 \pi\ cm^2\\\\Total\ surface\ Area = Surface\ area\ of\ cone + Surface\ area\ of \ hemisphere\\= 60 \pi + 72 \pi\\=132 \pi\ cm^2

Hence the first option is correct ..

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