Tag: constructing line segment

Questions Related to constructing line segment

Steps of constructing a line segment equal to the length of given segment is written in jumbled form below:
1. Draw a line $l$. Mark a point $A$ on line $l$. Without changing compass's setting, place the compass at $A$.
2. Make an arc on the line $l$ which cuts $l$ at $B$. Now, $AB$ is a copy of $CD$.
3. Draw a line segment $CD$ of any length.
4. Fix the compass's end on $C$ and pencil on $D$. This gives the length of $CD$.

Which of the above comes first.

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: C
Explanation:

Correct sequence is :

Step 1. Draw a line segment $CD$ of any length.
Step 2 .Fix the compass's end on $C$ and a pencil on $D$. This gives length $CD.$
Step 3. Draw a line $l$. Mark a point $A$ on line $l$.Without changing compass's setting place the compass at $A$
Step 4. Make an arc on the line $l$ whcih cuts $l$ at $B$  Now $AB$ is a copy of $CD$
So the first step is $3$
Option $C$ is correct.

Choose the correct answer from the alternatives given.
Water is flowing at the rate of $5$ km/hr through a pipe of diameter $14$ cm into a rectangular tank which is $50$ m long, $44$ m wide. The time taken (in hours) for the rise in the level of water in the tank to be $7$ cm is

  1. $2$

  2. $1\dfrac{1}{2}$

  3. $3$

  4. $2\dfrac{1}{2}$


Correct Option: A
Explanation:

Water
flowed by the pipe in lh = $\pi r^2h$
= $\dfrac{22}{7} \times$ $\dfrac{7\times 7}{100\times100}$ $\times 5000 m^3 =77m^3$
Volume
of expected water in the tank = $\frac{50 \times 44 \times 7}{100} = 154
m^3$ 
Required
time= $154/77 = 2 hrs$.

In $\triangle ABC,$ if $AB=7 \ cm$ and $BC=8 \ cm$, then which of the following cannot be the length of $AC$?

  1. $17\ cm$

  2. $16\ cm$

  3. $14\ cm$

  4. $None\ of\ these$


Correct Option: D

Four distinct points $(2K , 3K), (1 , 0), (0 , 1)$ and $(0 , 0)$ lie on a circle when

  1. all values of $K$ are integral

  2. $0 < K < 1$

  3. $K < 0$

  4. For one values of $K$


Correct Option: D
Explanation:

Let A = $(0,0)$ , B$(0,1)$ , C$(2k,3k)$ , D$(1,0)$

As we can see

$AD \perp AB$

$\angle A = 90$

$\angle C = 90$

$\implies m _{BC} \times m _{DC} = -1$

$\dfrac{3k - 1}{2k} \times {3k}{2k - 1} = -1$

$\implies k(9k - 3) = -(4k - 2)k$

$k = 0 , 9k + 4k = 2+ 3 \implies 13k = 5 \implies k = \dfrac{5}{13}$

But if $k = 0$, C will be $(0,0)$ which is A

$k = \dfrac{5}{13} $ only one point.

A ray of light passing through the point A$(1, 2, 3)$, strikes the plane $x + y + z = 12$ at B on reflection passes through point C$(3, 5, 9)$. The coordinates of point B are 

  1. $(2, 5, 5)$

  2. $(-4, 6, 10)$

  3. $(-7, 0, 19)$

  4. $(0, -5, 17)$


Correct Option: C
Explanation:
Let image of point $A(1,2,3)$ about $x+y+z=12$ be $D(p,q,r)$ then,
$\cfrac{p-1}{1}=\cfrac{q-2}{1}=\cfrac{r-3}{1}=-2\cfrac{1+2+3-12}{1^2+1^2+1^2} \\ D(p,q,r)=D(5,6,7) $

Line joining $CD$
$\cfrac{x-5}{2}=\cfrac{y-6}{1}=\cfrac{z-7}{-2}=\lambda$
Coordinate of $B(2\lambda+5,\lambda+6,-2\lambda+7)$
This lies on plane $2\lambda+5+\lambda+6-2\lambda+7=12 \Rightarrow \lambda=-6$
$B(2\lambda+5,\lambda+6,-2\lambda+7)=B(-7,0,19)$