Geometry Questions and Answers

We know that Area of trapezium = $frac{1}{2} (text{sum of 11 side}) times text{ distance between them }$

$Rightarrow quad 336 = frac{1}{2} (5x +7x ) times 14$

$Rightarrow quad x = 4$

Smaller side = 20

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

Let the sides of rectangular field be $2x$ and $x$ metre. Then

Area $= 2x times x$

$162 = 2x times x$

$2x^2 = 162$

$x^2 = frac{162}{2} = 81$

$x = 9 text{ m}$

Thus the sides of rectangular field are of 18 m and 9 m. By making a pavement of 1 m wide, the remaining size of length and breadth will now be 16 m and 7 m.

$therefore$ Interior area of field $= 16 times 7 = 112 text{ m}^2$

Reduction in area $= 162-112=50 text{ m}^2$

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

4 hrs. 40 min $= 4 + frac{40}{60} = frac{14}{3} text{ hrs }$

Reqd. angle $= frac{360}{24} times frac{14}{3} = 70degree$