Tag: changing scale

Questions Related to changing scale

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,
$\dfrac{AB}{AC} = \dfrac{BD}{DC}$
$\dfrac{BD}{DC} = \dfrac{3}{4}$

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 = $0.375$ $km^2$