Question

Find the coordinates of the point equidistant from three given points A(5,1), B(-3.-7) and
C(7,-1).​

Answer 1

Answer:

The given three points are A (5, 1) , B (–3, –7 ) and C (7, –1)

Let P (x, y) be the point equidistant from these three points.

So, PA = PB = PC

⇒ x2 + 25 – 10x + y2 + 1 – 2y = x2 + 9 + 6x + y2 + 49 + 14y = x2 + 49 – 14x + y2 + 1 + 2y

⇒ 25 – 10x + 1 – 2y = 9 + 6x + 49 + 14y = 49 – 14x + 1 + 2y

25 – 10x + 1 – 2y = 9 + 6x + 49 + 14y

⇒ 26 = 16x + 58 + 16y

⇒ 16x + 32 + 16y = 0

⇒ x + 2 + y = 0 … (1)

25 – 10x + 1 – 2y = 49 – 14x + 1 + 2y

⇒ 26 + 4x – 4y = 50

⇒ 4x – 4y = 24

⇒ x – y = 6 … (2)

Solving (1) and (2)

x = 2, y = –4

Thus, the required point is (2, –4)

Answer 2

GIVEN:-

•Radius of the cylinder=5cm

•Height of the cylinder =12 cm

TO FIND OUT:--

• The volume and the surface area of the remaining solid

SOLUTION :-

$\textsf{formula used}\begin{cases} \bf volume \: _{cylinder} = \pi r {}^{2} h\\ \bf CSA_{cylinder} = 2 \pi rh\\ \bf CSA\: _{cone}= \pi rl\\ \bf \: volume \: _{cone} = \frac{1}{3} \pi r {}^{2} h</p><p></p><p></p><p>\end{cases}$

Now ,

$\bf volume \: _{cylinder} \: = {\bigg (} \frac{22}{7} \times 5 \times 5 \times 12{ \bigg)cm {}^{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf \frac{6600}{7} cm {}^{3}$

$\bf \: volume \: _{cone} \: ={ \bigg(} \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 { \bigg)}cm {}^{3} \\ \\ \: \: \: \: \: \: \: \: = \bf \frac{2200}{7} cm {}^{3}$

•Volume of the remaining solid =(volume of the cylinder) -(volume of the cone)

$\bf \implies { \bigg(} \frac{6600}{7} - \frac{2200}{7}{ \bigg)} cm {}^{3} = \frac{4400}{7} cm {}^{3} = 628.57 \: cm {}^{3}$

•Slant height of the cone(l)

$\bf \: l = \sqrt{r {}^{2} + h {}^{2} }$

$\bf \implies l = \sqrt{5 {}^{2} + (12) {}^{2} } = \sqrt{169} = 13cm$

$\textsf{CSA} \bf_{cone}={\bigg (} \frac{22}{7} \times 5 \times 13 { \bigg)}cm² = \frac{1430}{7} cm²$

$\textsf{CSA} \bf_{cylinder}= {\bigg (} 2 \times \frac{22}{7} \times 5 \times 12{ \bigg)}= \frac{2640}{7} cm²$

•Now, area of upper circular base of base of cylinder =

$= \bf \pi r {}^{2} sq. \: unit \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf{ \bigg(} \frac{22}{7} \times 5 \times 5 { \bigg)}cm {}^{2} = \frac{550}{7} cm {}^{2}$

•Whole surface area of the remaining solid =CSA(cylinder)+CSA(cone)+area of upper base of cylinder

$\implies\bf { \bigg(} \frac{2640}{7} + \frac{1430}{7} + \frac{550}{7} { \bigg)}cm {}^{2} = 660cm {}^{2}$