Geometry Questions and Answers

$begin{aligned} text{Volume of cone } &= frac{1}{3}pi r^2h text{New volume of cone } &=frac{1}{3}pi Bigg(frac{3}{2}Bigg)^2 r^2 frac{3}{2}h &= frac{1}{3}pifrac{9}{8}r^2h = frac{27}{8}v text{ increase in volume } &= frac{27}{8} v - v frac{19}{8}v$

$begin{aligned} text{Volume of the sphere } &= frac{4}{3}pifrac{3}{4} times frac{3}{4} times frac{3}{4} text{ m}^3 text{By question } pi r^2h &= frac{4}{3}pifrac{3}{4} times frac{3}{4} times frac{3}{4} r^2 &= frac{3}{4} times frac{3}{4} (h = 1, text{ given}) therefore r &= frac{3}{4} = 0.75 text{ m } end{aligned}$

$dfrac{V_0}{V_n} = dfrac{pi r^3h}{pibiggl(dfrac{r}{2}biggr)^2 . dfrac{3}{2}h} = dfrac{8}{3}$

$text{Here AB } = sqrt{(3 - 0)^2 + (0 - 0)^2 } = 3$

$text{Similary BC } = 5 text{ and CA } = 4$

$Rightarrow$ The given triangle is  a right angle and hence all the three altitudes intersect at the point A  (0, 0)

$therefore text{The orthocentre } = (0,0)$

ABCD is a rhombus, then
by pythagoras theorem

$text{AC}^2 + text{BD}^2 = (text{OA} + text{OC})^2 (text{OB} + text{OD})^2$

$= 4(text{OA}^2 + text{OB}^2) = 4 text{AB}^2$

$text{Since } triangle text{ABC and } triangle text{DEF } text{ are similar ,}$

$therefore frac{h}{text{EF}} = frac{2.1}{text{BC}} quadquad text{ ...(I)}$

$text{ Also } frac{1}{2} times text{BC} times 2.1 = 12$

$Rightarrow text{ BC } = frac{24}{2.1} text{ and } frac{1}{2} times text{ EF } times h =48$

$Rightarrow text{ EF } times h = 96$

$text{Now by equation (I)}$

$dfrac{h}{dfrac{96}{h}} = dfrac{2.1}{frac{24}{2.1}} Rightarrow frac{h^2}{96} = frac{(2.1)^2}{24} Rightarrow h = 4.2$

The centre of the circle (4, 3) and its radius is $r = sqrt{16 + 9 - 16} = 3$
If the circle

If the circle touches the $x$-axis, then at that point $y$ must be zero.

$therefore quad$ Equation of the circle becomes $x^2 - 8x + 16 = 0$

$Rightarrow (x - 4)^2 = 0 Rightarrow x = 4$

If the circle touches the y axis, then at that point $x = 0$.

$therefore quad$ Equation of the circle becomes

$y^2 - 6y + 16 = 0$

$Rightarrow y$ is imaginary

$therefore quad$ The given circle touches the $x$-axis only at the point (4, 0)

Diagonal $= sqrt{3} text{ cm}$

Area of the circle

$= pi r^2 = Big(frac{sqrt{3}}{2}Big)^2$

$=frac{3pi}{4}$

Let the side of an equilateral triangle be$x$, then perimeter $=3x$ and area of triangle $=frac{sqrt{3}}{4}x^2$ and side of a square = y then perimeter$= 4y text{ and area} = y^2$

$=4y\text{ and area}=y^2$

By question

$3x = 4y Rightarrow frac{x}{y} = frac{4}{3}$$frac{text{Area of triangle}}{text{Area of square}$

$frac{text{Area of triangle}}{text{Area of square}} =dfrac{frac{sqrt{3}}{4}x^2}{y^2} = dfrac{sqrt{3}}{4}timesBig(frac{4}{3}Big)^2$

$= frac{9sqrt{3}}{64}lt 1$

Required area $= frac{1}{2} times 2r times r = r^2$