Tag: bearing and drawings

Questions Related to bearing and drawings

Milli starts from A and travels $2$ km and then come back for $280$ m. Then, the distance between point A and Milli is ?

maths mapping your way mapping mapping space around us bearing and drawings

  1. $2$ km

  2. $1$ km $720$ m

  3. $1$ km $200$ m

  4. $1$ km $100$ m

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Correct Option: B
Explanation:

Let us take the point of start A and point at a distant $2$km be B.

$AB=2$km=$2000$m
As she come back 280m , we assume that point be C.
We have to find out the distance AC
$AC=AB-BC \therefore AC= 2000-280 = 1720= 1 $ km   $720$ m 

In a ground, the distance between two consecutive trees is $4$ m and distance between next $2$ trees is $5$ m. Then calculate the distance between first and third tree.

maths mapping your way mapping mapping space around us bearing and drawings

  1. $9$ m

  2. $5$ m

  3. $4$ m

  4. $3$ m

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Correct Option: A
Explanation:

Let the first Tree be at point A, second at point B and  third one a C.

$AB=4$ m and $BC=5$ m.
$AC= AB+BC$
$\therefore AC = 9$ m.

Distance between two houses is $40$ m. If a new house with area $4\times 4$ is constructed in the middle of $2$ houses, then find the distance between middle house and one of the corner house.

maths mapping your way mapping mapping space around us bearing and drawings

  1. $20$ m

  2. $19$ m

  3. $18$ m

  4. $17$ m

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Correct Option: C
Explanation:
The distance of middle point of middle house from the corner house is $\dfrac { 40 }{ 2 } =20$ and house to house distance will be $\therefore 20-2=18$
The distance between he Middle house and Corner house will be $18$ m.

In an Atlas a map occupies $\displaystyle \frac{2}{5}$th of a page with dimensions 25 cm and 30 cm respectively If the real area of the map is 10800 sq. m the scale to which the map is drawn is

maths mapping your way mapping mapping space around us bearing and drawings

  1. 1 cm = 36 m

  2. 1 cm = 26 m

  3. 1 cm = 33 m

  4. 1 cm = 6 m

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Correct Option: D
Explanation:

Area occupies by the map=$(2/5)*25*30cm^{2}=300cm^{2}$
Area of $300cm^{2}$ on the map represent $10800 m^{2}$ real area of the map.
So,Area of $1cm^{2}$ on the map represent =$10800m^{2}/300cm^{2} 
                                                                       =  36m^{2}:1 cm^{2}$
Ratio of length=
$R=\sqrt { \frac { Real\quad area\quad  }{ Map\quad area } \quad  } =\sqrt { \frac { 36m^{2} }{ 1cm^{2} }  } =\frac { 6m }{ 1cm } $
Scale =1cm=6m
Answer (D) 1 cm = 6 m

$ \Delta ABC ~ \Delta PQR $ for the correspondence $ABC \leftrightarrow PQR $ . If the perimeter of $ \Delta ABC $ is $12$ and the perimeter of $ \Delta PQR $ is $20 $ , then $AB : PQ = $ ______

maths mapping your way mapping mapping space around us bearing and drawings

  1. $\dfrac {6}{5}$

  2. $\dfrac {2}{5}$

  3. $\dfrac {3}{5}$

  4. $\dfrac {1}{5}$

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Correct Option: C
Explanation:
$ABC\leftrightarrow PQR$
$\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}=k(say)$

$AB=k(PQ);AC=k(PR);BC=k(QR)$
$[AB+BC+AC]=k[PQ+RQ+PR]$
Perimeter $(\Delta ABC)$ = perimeter $(\Delta PQR)\times k$
$12=20\times k$

$k=\dfrac{12}{20}=\dfrac{3}{5}$

$\therefore \dfrac{AB}{PQ}=k=\dfrac{3}{5}$

Perpendicular AL, BM are drawn from the vertices A,B of a triangle ABC to meet BC, AC at L, M. by proving the triangles ALC, BMC similar, or otherwise, then CM.CA=CL.CB

maths mapping your way mapping mapping space around us bearing and drawings

  1. True

  2. False

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Correct Option: A

A model of an aeroplane is made to a scale of $1:400$. Calculate the length, in cm, of the model; if the length of the aeroplane is $40$ m.

maths mapping your way mapping mapping space around us bearing and drawings

  1. $10$ cm

  2. $20$ cm

  3. $140$ cm

  4. none of the above

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Correct Option: A
Explanation:

Scale $= k =$ $\dfrac{1}{400}$
Now, $\dfrac{\text{Length of model}}{\text{Length of aeroplane}} = k$
$\text{Length of model }= \dfrac{40 \times 100}{400}$
$\text{Length of model} =10$ cm

The line segments joining the midpoints of the sides of a triangle form four triangles each of which is:

maths bearing and drawings mapping mapping space around us changing scale

  1. similar to the original triangle

  2. congruent to the original triangle

  3. an equilateral triangle

  4. an isosceles triangle

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Correct Option: A
Explanation:

Given: $\triangle ABC$, D, E and F are mid points of AB, BC, CA respectively.

In $\triangle ABC$
F is mid point of AC and D is mid point of AB. 
Thus, By Mid point theorem
$FD = \dfrac{1}{2} CB$ and $FD = CB$ or $FD = CE$ and $FD \parallel CE$ (1)

Similarly,
$DE = FC$ and $DE \parallel FC$ (2)
$FE = DB$ and $FE \parallel DB$ (3) 
From (1), (2) and (3)
$\Box ADEF$, $\Box DBEF$ and $\Box DECF$ are parallelograms
The diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Hence, $\triangle DEF \cong \triangle ADF$
$\triangle DEF \cong \triangle DBE$
$\triangle DEF \cong \triangle FEC$
or, $\triangle DEF \cong \triangle ADF \cong \triangle ECF \cong \triangle ADF$
thus, mid points divide the triangle into 4 equal parts.

Thus, the smaller triangles are congruent to each other and similar to the original triangle.

$D, E, F$ are the mid points of the sides $AB, BC,CA$ respectively of $\triangle ABC$. Then $\triangle DEF$ is congruent to 

maths bearing and drawings mapping mapping space around us changing scale

  1. $\triangle ABC$

  2. $\triangle AEF$

  3. $\triangle BDF , \triangle CDE $

  4. $\triangle ADF $

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Correct Option: D
Explanation:

Given: $\triangle ABC$, $D, E \ and \ F$ are mid points of $AB, BC, CA$ respectively.

In $\triangle ABC$

$F$ is mid point of $AC \ and \ D$ is mid point of $AB.$ 

Thus, By Mid point theorem

$FD = \dfrac{1}{2} CB$ and $FD = CB$ or $FD = CE$ and $FD \parallel CE$         ....(1)

Similarly,
$DE = FC$ and $DE \parallel FC$            ...(2)

$FE = DB$ and $FE \parallel DB$             ...3) 

From (1), (2) and (3)

$\Box ADEF$, $\Box DBEF$, $\Box DECF$ are parallelograms

The diagonal of a parallelogram divides the parallelogram into two congruent triangles.

Hence, $\triangle DEF \cong \triangle ADF$

$\triangle DEF \cong \triangle DBE$

$\triangle DEF \cong \triangle FEC$

$\triangle DEF \cong \triangle ADF \cong \triangle ECF \cong \triangle ADF$

In $\triangle ABC$, $AB=3cm, AC=4cm$ and $AD$ is the bisector of $\angle A$. Then $BD:DC$ is:

maths bearing and drawings mapping mapping space around us changing scale

  1. $9:16$

  2. $16:9$

  3. $3:4$

  4. $4:3$

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Correct Option: C
Explanation:

In $\triangle ABC$,
AD bisects $\angle A$
By angle bisector theorem,
$\dfrac{AB}{AC} = \dfrac{BD}{DC}$
$\dfrac{BD}{DC} = \dfrac{3}{4}$