Geometry Questions and Answers

Side of the square

$=sqrt{ 6.4 times 2.5}$$=\sqrt{6.4\times 2.5}$
$=sqrt{frac { 64 times 2.5}{100}} = sqrt{16} = 4 text m$

The volume of the metallic shell

$= frac{4}{3} pi (2)^3 - frac{4}{3} pi(1)^3 = frac{28pi}{3}$

But by question,

Volume of sphere $= frac{28pi}{3}$

$Rightarrow frac{4}{3}pi r^3 = frac{28pi}{3} Rightarrow r^3 = 7 Rightarrow r = 7^frac{1}{3}$

$text{ Volume of the cone } = frac{1}{3} times frac{22}{7} times 10^2 times 5$

$= frac{11000}{21}$

$text{ Volume of sphere } = frac{4}{3}pi times 125$

$= frac{11000}{21}$

For the bigger cylinder,
$text{ H } = 10 text{ cm }, 2 pi r = 8$

$therefore text{ V } = pi r^2 quadtext{ H } = pi (frac{16}{pi^2}) times 10 = frac{160}{pi} text{cu cm}$

Area of rhombus

$= frac{1}{2} times text{ product of diagonal }$

$= frac{1}{2} times 24 times 10$

$Big[ because text{ small diagonal } = 2sqrt{13^2 - 12^2 } = 2 times 5 = 10 Big]$

$= 120 text{ sq. m }$

$text{Volume of cylinder } = pi r^2h$

$= 1100 = pi times 5 times 5 times h$

$Rightarrow h = frac{1100}{pi times 25} text{m} = 14text{m.}$

$therefore text {Are of curved surface} = 2pi rh$

$= 2pi times 5 times 14 = 440 text{cm}^2$

Let side of square be $x$ and that of triangle is $y$

$begin{aligned} therefore 4x &= 3y x &= frac{3}{4}y text{Area of triangle } &= frac{sqrt3}{4} times y^2 16sqrt3 &= frac{sqrt3}{4} times y^2 y &= 8 text{ and } x &= frac{3}{4} times 8 = 6 text{ cm} end{aligned}$

Total surface are of cone $= pi rl + pi r^2$
$= pi 3(3 + 5) = 24pi text{ cm}^2$

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

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

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

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

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

$Rightarrow$ 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

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

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

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

Let the radius of the disc of area $text{A}_1$ be R. Then

$text{A}_1 = pi text{R}^2, text{A}_2 = frac{pitext{R}^2}{4}$

$text{A}_3 = pitext{R}^2 - frac{pitext{R}^2}{4} = frac{3pitext{R}^2}{4}$

$therefore text{A}_1text{A}_3 = frac{3}{4}pi^2text{R}^4 = 3pitext{R}^2 times text{A}_2$

$= 12 text{A}_2^2 lt 16 text{A}_2^2$

$c = pi quad r sqrt{r^2 + h^2}, v = frac{1}{3}pi r^2 h$

Now $3pi vh^3 = pi^2 r^2 h^4$

$c^2h^2 = pi r^2 (r^2 + h^2) h^2 text{ and } 9v^2 = pi^2 r^4 h^2$

$Rightarrow 3pi vh^3 - c^2h^2 + 9v^2 = 0$

Capacity of the bucket = Volume of cone ADO - Volume of cone BCO

$= frac{1}{3} times pi times (28)^2 times (45 + h) - frac{1}{3} times pi times (7)^2 times h quadtext{...(I)}$

Now from figure

$frac{text{PD}}{text{PO}} = frac{text{QC}}{text{QO}} Rightarrow frac{28}{45 + h} = frac{7}{h} Rightarrow h = 15$

Now on putting the value of $h$ in equation (I)

We get required Volume $= 48510 text{ cm}^3$

Length of diagonal  $(l) = sqrt{dfrac{13254}{6}}$

= 47

$therefore quad d = sqrt{3} l$

$= 47sqrt3 text{ cm }$

$[because text{Surface are = } 6l^2 text{ of a cube}]$

Area of circle  $= 2pi r$

$text{Given, } dfrac{2pi(r_1 - r)}{2pi r} times 100 = 15$

$Rightarrow quad r_1 = frac{23}{20}r$

$therefore$ % increase in area = $= frac{pi Big(frac{23}{20}rBig)^2 - pi r^2}{pi r^2} times 100 = 32.25$

$text{ Given the perimeter of polygon } = 2text{P}$

$text{ then each side of a polygon } = frac{2text{P}}{n}$

Area of polygon $= n times frac{1}{2} times frac{2text{P}}{n } times r$

= Pr

No. of Spherical bullets $= frac{text{volume of a cube}}{text{volume of each bullet}}$

$= dfrac{44 times 44 times 44}{frac{4}{3} times frac{22}{7} times 2 times 2 times 2}$

= 2541

Area of a bigger square $= a^2$

$therefore text{Diagonal } = sqrt{2}a = d_1$

Area of a smaller square $=frac{a^2}{2} = x^2$

$x$ =  side of smaller square ...  $x = frac{a}{sqrt2}$

Diagonal of  a smaller square$= d_2 = sqrt{2}x$

$d_2 = sqrt{2} cdot frac{a}{sqrt2} = a$

$therefore$  Required Ratio $= d_2 : d_1 = a:sqrt{2}a = 1 :sqrt{2}$

$text{AC} = 2d, x + 2x = 90degree Rightarrow x = 30degree frac{text{AB}}{text{AC}} = text{cos }30degree$

$therefore text{AB } = frac{sqrt3}{2} times 2d = sqrt{3}d$

$frac{text{BC}}{text{AC}} = sin 30degree$

$therefore text{BC } = text{AC }times frac{1}{2} = 2d times frac{1}{2} = d$

Area of a rectangle $= text{AB } times text{BC} = sqrt{3}d cdot d = sqrt{3}d^2$

$text{S } = dfrac{15+16+17}{2} = 24$

Area of a $triangle = sqrt{s(s - a)(s-b)(s-c)}$

$= sqrt{24(24-15)(24-16)(24-17)} = 24sqrt{21} text{ sq. cm}$

5 revolution in 5 sec
$therefore$ One revolution in 1 sec
Distance covered in 1 revolution

$=2pi r = 2 times frac{22}{7} times 84 =2 times 22 times 12 text{ cm}$

$text{Speed } = dfrac{text{D}}{text{T}} = dfrac{2 times 22 times 12}{1text{ sec}}text{ cm}$

$= frac{2 times 22 times 12}{100} times frac{18}{5} text{km/hr} =19 text{ km/hr app}$

$frac{(n - 2) 180 }{n} = 160degree Rightarrow n = 18$

No of sides = 18