Tag: maths

Questions Related to maths

If the sides of two similar triangles are in the ratio $1:7$, find the ratio of their areas.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $7:1$

  2. $1:7$

  3. $1:49$

  4. $1:14$

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Correct Option: C
Explanation:

We know that the relation between area of two similar triangle:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. 

Given, sides of two similar triangles are in the ratio $1:7$.
So, the ratio of their areas $= 1:49$.

The corresponding sides of two similar triangles are in the ratio $a : b$. What is the ratio of their areas?

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $a : b$

  2. $2a : 2b$

  3. $a^{2} : b^{2}$

  4. $\dfrac {1}{a} : \dfrac {1}{b}$

Show answer
Correct Option: C
Explanation:

Given two triangles are similar, then the ratio of the areas $=a^2:b^2$

Eg: The ratio of the sides of a similar triangle is $4:9$
Scale factor for the sides of these triangles $k=\cfrac 49$
$\therefore$ Ratio of area will be:
$k^2=\cfrac {area of \triangle A}{area of \triangle B}=(\cfrac 49)^2=\cfrac {16}{81}$

The ratio of areas of two similar triangles is $81 : 49$. If the median of the smaller triangle is $4.9\ cm$, what is the median of the other?

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $4.9\ cm$

  2. $6.3\ cm$

  3. $7\ cm$

  4. $9\ cm$

Show answer
Correct Option: B
Explanation:

Area of $\triangle ABC= \cfrac 12 \times base \times height$

In similar triangles, $\cfrac {base 1}{base 2}=\cfrac {height 1}{height 2}=\cfrac {side 1}{side}$
$\therefore \cfrac {Area 1}{Area 2}= (\cfrac {Median 1}{Median 2})^2$
Ratio of Medians $=\sqrt{\cfrac {81}{49}}=\cfrac 97 >1$ 
$\therefore$ Altitude of smaller triangle $=4.9 \times \cfrac 97=6.3$

$\triangle ABD \sim \triangle DEF$ and the perimeters of $\triangle ABC$ and $\triangle DEF$ are $30 cm$ and $18 cm$ respectively. If $BC = 9 cm$, calculate measure of $EF$.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $6.3\ cm$

  2. $5.4\ cm$

  3. $7.2\ cm$

  4. $4.5\ cm$

Show answer
Correct Option: B
Explanation:

Similar triangles are triangle with similar shape but can have different sizes. 

Since there is something common about them, then to establish the relationship we have something called linear scale factor which is used to get the length of the other when the length of the other similar triangles is known. 
$LSF=\cfrac {30}{18}=\cfrac 53$
$\cfrac {BC}{EF}=\cfrac 53 \Rightarrow \cfrac 9{EF}=\cfrac 53$
$\therefore EF=9 \times \cfrac 35 = 5.4cm$

Two isosceles triangles have their corresponding angles equal and their areas are in the ratio $25 : 36$. Find the ratio of their corresponding heights

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $25 : 35$

  2. $36 : 25$

  3. $5 : 6$

  4. $6 : 5$

Show answer
Correct Option: C
Explanation:
We know, Ratios of areas of similar triangles is equal to ratio of squares of their corresponding sides.
Hence,

$\dfrac{Area \ of \ \triangle _1}{Area \ of \ \triangle _2}=\dfrac{(height \ of \ \triangle _1)^2}{(height \ of \ \triangle _2)^2}$

Taking square root on both sides,

$\dfrac{(height \ of \ \triangle _1)}{(height \ of \ \triangle _2)}=\sqrt{\dfrac{Area \ of \ \triangle _1}{Area \ of \ \triangle _2}}$

$\dfrac{(height \ of \ \triangle _1)}{(height \ of \ \triangle _2)}=\sqrt{\dfrac{25}{36}}=\dfrac{5}{6}$

Option C

In similar triangles $\triangle ABC$ and $\triangle FDE, DE = 4 cm, BC = 8 cm$ and area of $\triangle FDE = 25 cm^2$. What is the area of $\Delta ABC$?

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. 144 cm$^2$

  2. 121 cm$^2$

  3. 100 cm$^2$

  4. 81 cm$^2$

Show answer
Correct Option: C
Explanation:

  $DE=4\,cm,\,BC=8\,cm$ and $ar(\triangle FDE)=25\,cm^2$                  [ Given ]


$\Rightarrow$  $\triangle ABC\sim\triangle FDE$             [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle FDE)}=\dfrac{(BC)^2}{(DE)^2}$                       [ By area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{25}=\dfrac{(8)^2}{(4)^2}$

$\Rightarrow$  $ar(\triangle ABC)=\dfrac{64}{16}\times 25$

$\Rightarrow$  $ar(\triangle ABC)=4\times 25$
$\therefore$  $ar(\triangle ABC)=100\,cm^2$

The areas of two similar triangles are $81\ cm^{2}$ and $49\ cm^{2}$. If the altitude of the bigger triangle is $4.5\ cm$, find the corresponding altitude of the smaller triangle.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $3 cm$

  2. $2.5 cm$

  3. $4 cm$

  4. $3.5 cm$

Show answer
Correct Option: D
Explanation:

Given: Area of $two$ similar triangle $81{cm}^{2}$ and $49{cm}^{2}$

Altitude of bigger triangle $=4.5cm$
For similar triangle,
${\text{Ratio on sides}}^{2}=\text {Ratio of their Area}$
$\therefore$ $\cfrac { { 4.5 }^{ 2 } }{ { x }^{ 2 } } =\cfrac { 81 }{ 49 } $
$\cfrac { 45\times 45 }{ { x }^{ 2 }\times 100 } =\cfrac { 81 }{ 49 } $
$100{x}^{2}=25\times 49$
${x}^{2}=\cfrac{25\times 49}{100}$
${x}^{2}=\cfrac{49}{4}$
$x=\cfrac{7}{2}$
$x=3.5cm$

If $\triangle ABC$ and $\triangle PQR$ are similar and $\dfrac {BC}{QR} = \dfrac {1}{3}$ find $\dfrac {area (PQR)}{area (BCA)}$

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $9$

  2. $3$

  3. $\dfrac {1}{3}$

  4. $\dfrac {1}{9}$

Show answer
Correct Option: A
Explanation:

$\triangle ABC$ & $\triangle PQR$ are similar.

$\therefore \cfrac{AB}{PQ}=\cfrac{BC}{QR}=\cfrac{1}{3}$
$\therefore \cfrac{\text{Area}(PQR)}{\text{Area}(BCA)}={(\cfrac{QR}{BC}})^{2}$
$\cfrac { { Area }(PQR) }{ { Area }(BCA) } ={ (\cfrac { 3 }{ 1 } ) }^{ 2 }=9$

What is the ratio of the areas of two similar triangles whose corresponding sides are in the ratio 15:19?

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $\sqrt{15} : \sqrt{19}$

  2. $15 : 19$

  3. $225 : 361$

  4. $125 : 144$

Show answer
Correct Option: C
Explanation:

We know ratio  of areas of two similar triangles is the ratio of the square of their corresponding sides.

Ratio of sides $\dfrac{15}{19}$
Ratio of areas $={ \left( \dfrac { 15 }{ 19 }  \right)  }^{ 2 }=\dfrac { 225 }{ 361 } $
So option $C$ is correct.

The areas of two similar triangles are 100 $cm^2$ and 64 $cm^2$. If the median of greater side of first triangle is 13 cm, find the corresponding median of the other triangle.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 16 cm

Show answer
Correct Option: C
Explanation:
Given area of two similar triangles are $100$ sq cm and $64$ sq cm
The areas of two Similar-Triangles are in the ratio of the squares of the corresponding medians
The ratio of area of triangle $=\dfrac{100}{64}=\dfrac{25}{16}$
Median of greater triangle is $13$ cm and let other median is $x$ cm
$\therefore \dfrac{(13)^{2}}{(x)^{2}}=\dfrac{25}{16}$

$\Rightarrow \dfrac{169}{x^{2}}=\dfrac{25}{16}$

$\Rightarrow 25x^{2}=169\times 16$

$\Rightarrow x^{2}=\dfrac{2704}{25}=108.16$

$\Rightarrow x=10cm$