Tag: bearing and drawings

Questions Related to bearing and drawings

A model of an aeroplane is made to a scale of $1:400$. Calculate the length, in m, of the aeroplane, if the length of its model is $16$ cm.

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

  1. $68$ m

  2. $64$ m

  3. $54$ m

  4. none of the above

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Correct Option: B
Explanation:

Scale $= k =$ $\dfrac{1}{400}$
Now, $\dfrac{\text{Length of model}}{\text{Length of aeroplane}} = k$
$\text{Length of aeroplace} = 400 \times 16$
$\text{Length of model }= 6400$ cm
$\text{Length of model }= 64$ m

Triangle $ABC$ is such that $AB=3cm, BC=2cm$ and $CA=2.5cm$. Triangle $DEF$ is similar to $\triangle ABC$. If $EF=4cm$, then the perimeter of $\triangle DEF$ is:

maths bearing and drawings mapping mapping space around us changing scale

  1. $7.5cm$

  2. $15cm$

  3. $22.5cm$

  4. $30cm$

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Correct Option: B
Explanation:

Given: $AB = 3 cm$, $BC = 2 cm$ and $CA = 2.5 cm$ and $EF = 4 cm$ 

Also, $\triangle ABC \sim \triangle DEF$

Thus, $\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

$\dfrac{3}{DE} = \dfrac{2}{4} = \dfrac{2.5}{DF}$

Hence, $DE = 6 cm$ and $DF = 5 cm$

Perimeter of $\Delta$ DEF = $DE + EF + EF$

Perimeter of $\Delta$ DEF = $4 + 5 + 6$

Perimeter of $\Delta$ DEF = $15$ cm

$\triangle ABC\sim \triangle DEF$. IF $BC=4cm$, $EF=5cm$ and area $(\triangle ABC)=32{cm}^{2}$, determine the area of $\triangle DEF$.

maths bearing and drawings mapping mapping space around us changing scale

  1. $50{m}^{2}$

  2. $40{m}^{2}$

  3. $40{cm}^{2}$

  4. None of these

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Correct Option: D
Explanation:

Ar. ($\triangle ABC$) = $32 cm^2$
$BC = 4 cm$
$EF = 5 cm$
For similar triangles the ratio of areas is equal to the ratio of square of its sides.
Thus, $\frac{A(\triangle ABC)}{A(\triangle DEF)} = \frac{BC^2}{EF^2}$
$\frac{32}{A(\triangle DEF)} = \frac{4^2}{5^2}$
$A(\triangle DEF) = \frac{32 \times 25}{16}$
$A(\triangle DEF) = 50 cm^2$

The areas of two similar triangles are $48{cm}^{2}$ and $75{cm}^{2}$ respectively. If the altitude of the first triangle be $3.6cm$, find the corresponding altitude of the other.

maths bearing and drawings mapping mapping space around us changing scale

  1. $4cm$

  2. $4.5cm$

  3. $5cm$

  4. $5.5cm$

Show answer
Correct Option: B
Explanation:

Ar. ($\triangle _{1}$) = $48 cm^2$
Ar. ($\triangle _{2}$) = $75 cm^2$
$a _{1} = 3.6 cm$
For similar triangles the ratio of areas is equal to the ratio of square of their altitudes.
Thus, $\frac{A(\triangle _{1})}{A(\triangle _{2})} = \frac{(a _1)^2}{(a _2)^2}$
$\frac{48}{75} = \frac{(3.6)^2}{(a _2)^2}$
$(a _2)^2 = \frac{12.96 \times 75}{48}$
$(a _2)^2 = 20.25$
$a _2 = 4.5$cm

$\triangle ABC$ and $\triangle PQR$ are similar triangle such that area $(\triangle ABC)=49{cm}^{2}$ and Area $(\triangle PQR)=25{cm}^{2}$. If $AB=5.6cm$, find the length of $PQ$.

maths bearing and drawings mapping mapping space around us changing scale

  1. $4cm$

  2. $5cm$

  3. $5.6cm$

  4. $7cm$

Show answer
Correct Option: A
Explanation:

Ar. ($\triangle ABC$) = $49 cm^2$
Ar. ($\triangle PQR$) = $25 cm^2$
$AB = 5.6 cm$
For similar triangles the ratio of areas is equal to the ratio of square of its sides.
Thus, $\dfrac{A(\triangle ABC)}{A(\triangle PQR)} = \dfrac{AB^2}{PQ^2}$
$\dfrac{49}{25} = \dfrac{(5.6)^2}{PQ^2}$
$PQ^2 = \dfrac{31.36 \times 25}{49}$
$PQ^2 = 16$
$PQ = 4$cm

State true or false:
On a map drawn to a scale of $1: 25,0000$; a triangular plot of land has the following measurements AB=$3 cm$, $BC=4$ cm, and angle ABC=$\displaystyle 90^{\circ}.$
The actual lengths of AB and BC in km are $7.5$ km and $10$ km.
maths bearing and drawings mapping mapping space around us changing scale

  1. True

  2. False

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Correct Option: B
Explanation:
Scaling factor = $\dfrac{Length \quad of \quad model}{Length \quad of \quad building}$

$\dfrac{1}{250000} = \dfrac{AB}{Length _{AB}}$

$Length _{AB} = 250000 \times 3$

$Length _{AB} = 750000$

$Length _{AB} = 0.75 $ km

$\dfrac{1}{250000} = \dfrac{BC}{Length _{BC}}$

$Length _{BC} = 250000 \times 4$

$Length _{BC} = 1000000$

$Length _{BC} = 1 $ km

Given statement is false.

The dimensions of the model of a multistorey building are 1.2 m$\displaystyle \times 75cm\times 2m.$
If the scale facor is 1 : 30; find the actual dimesions of the building.

maths bearing and drawings mapping mapping space around us changing scale

  1. $\displaystyle 23m\times 22.5m\times 60m$

  2. $\displaystyle 12m\times 22.5m\times 60m$

  3. $\displaystyle 17m\times 22.5m\times 60m$

  4. $\displaystyle 36m\times 22.5m\times 60m$

Show answer
Correct Option: D
Explanation:

Scaling factor = $\frac{Length \quad of \quad model}{Length \quad of \quad building}$
$\frac{1}{30} = \frac{1.2}{Length \quad of \quad building}$
Length of building = $1.2 \times 30$
Length of building = $36$ m
Similarly, Breadth of building = $75 \times 30$ = 22.5 m
Height of building = $30 \times 2$ = 60 m
hence, dimensions of building are : $36 \times 22.5 \times 60$

State true or false:
On a map drawn to a scale of 1: 25,0000; a triangular plot of land has the following measurements AB=3 cm, BC=4 cm, and angle ABC=$\displaystyle 90^{\circ}.$
The area of the plot in sq. km is 37.5 sq. km.

maths bearing and drawings mapping mapping space around us changing scale

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Scaling factor = $\dfrac{Length \quad of \quad model}{Length \quad of \quad building}$
$\dfrac{1}{250000} = \dfrac{AB}{Length _{AB}}$
$Length _{AB} = 250000 \times 3$
$Length _{AB} = 750000$
$Length _{AB} = 0.75 $ km

$\dfrac{1}{250000} = \dfrac{BC}{Length _{BC}}$
$Length _{BC} = 250000 \times 4$
$Length _{BC} = 1000000$
$Length _{BC} = 1 $ km

Area of plot = $\dfrac{1}{2} \times Length _{AB} \times Length _{BC}$ (Area of triangle =$ \dfrac{1}{2} base \times height$)
Area of plot = $\dfrac{1}{2} \times (0.75) \times 1$
Area of plot = $0.375$ $km^2$

On a scale map $0.7\ cm$ represents $8.4\ km$. If the distance between two points on the map is $46.5\ cm$, what is the actual distance between the points?

maths bearing and drawings mapping mapping space around us changing scale

  1. $56\ km$

  2. $55.80\ km$

  3. $62.80\ km$

  4. $72\ km$

Show answer
Correct Option: B
Explanation:

Given $0.7$cm represents $8.4$ km 

$46.5$ cm represents $x$
By proportionality 

$\dfrac{0.7}{46.5} = \dfrac{8.4}{x}$

$x = \dfrac{8.4\times46.5}{0.7}$
 
$x = 55.80km$

The perimeter of two similar triangles $ABC$ and $PQR$ are $36\ cm$ and $24\ cm$ respectively. If $PQ = 10\ cm$ then the length of $AB$ is

maths bearing and drawings mapping mapping space around us changing scale

  1. $18\ cm$

  2. $12\ cm$

  3. $15\ cm$

  4. $30\ cm$

Show answer
Correct Option: C
Explanation:

$\triangle ABC = \triangle PQR$, then
$\dfrac {\text {Perimeter of}\triangle ABC}{\text {Perimeter of}\triangle PQR} = \dfrac {AB}{PQ} = \dfrac {BC}{QR} = \dfrac {AC}{PR}$
$\dfrac {36}{24} = \dfrac {AB}{10}; AB = \dfrac {36}{24}\times 10 = 15\ cm$.