Tag: maths

Questions Related to maths

The ratio of the areas of two  similar triangles is equal to the

  1. ratio of corresponding medians

  2. ratio of corresponding sides

  3. ratio of the squares of corresponding sides

  4. none of these


Correct Option: C
Explanation:

The area of triangle is proportional to the square of the side of the triangle.
ratio of areas of two similar triangles= ratio of the squares of corresponding sides 

In two similar triangles ABC and PQR, if their corresponding altitudes AD and Ps are in the ratio 4:9, find the ratio of the areas of $\triangle ABC$ and $\triangle PQR$.

  1. $16:81$

  2. $9:16$

  3. $81:16$

  4. $16:9$


Correct Option: A
Explanation:
Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.

$\therefore $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) } =\dfrac { { AD }^{ 2 } }{ { PS }^{ 2 } } $

$\Rightarrow $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) } ={ \left( \dfrac { 4 }{ 9 }  \right)  }^{ 2 }=\dfrac { 16 }{ 81 } $              [$\because AD:PS=4:9$]

$\Rightarrow $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle PQR) }$ = $\dfrac{16}{81}$

If $\triangle ABC$ is similar to $\triangle DEF$ such that BC=3 cm, EF=4 cm and area of $\triangle ABC=54 {cm}^{2}$. Determine the area of $\triangle DEF$.

  1. $40\ cm^2$

  2. $59\ cm^2$

  3. $69\ cm^2$

  4. $96\ cm^2$


Correct Option: D
Explanation:
Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\therefore $ $\dfrac { Area(\triangle ABC) }{ Area(\triangle DEF) } =\dfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } $

$\Rightarrow $ $\dfrac { 54 }{ Area(\triangle DEF) } =\dfrac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } $

$\Rightarrow $ $Area(\triangle DEF)=\dfrac { 54\times 16 }{ 9 } =96{ cm }^{ 2 }$

Two $\triangle sABC $ and DEF are similar. If $ar(DEF)= 243\ cm^2, ar(ABC)=108\ cm^2$ and $BC= 6\ cm$. Find $EF$.

  1. $9$

  2. $81$

  3. $3$

  4. $72$


Correct Option: A
Explanation:
Given:-
$\triangle{ABC} \simeq \triangle{DEF}$
$ar \left( DEF \right) = 243 {cm}^{2}$
$ar \left( ABC \right) = 108 {cm}^{2}$
$BC = 6 cm$

To Find:- $EF = ?$

As we know that,
$\because \; \triangle{ABC} \simeq \triangle{DEF}$

$\cfrac{ar \left( \triangle{ABC} \right)}{ar \left( \triangle{DEF} \right)} = {\left( \cfrac{AB}{DE} \right)}^{2} = {\left( \cfrac{BC}{EF} \right)}^{2} = {\left( \cfrac{AC}{DF} \right)}^{2}$

$\therefore \; \cfrac{ar \left( \triangle{ABC} \right)}{ar \left( \triangle{DEF} \right)} = {\left( \cfrac{BC}{EF} \right)}^{2}$

$\Rightarrow \; \cfrac{108}{243} = \cfrac{{6}^{2}}{{EF}^{2}}$

$\Rightarrow \; {EF}^{2} = \cfrac{243}{108} \times 36$

$\Rightarrow \; EF = \sqrt{81}$

$\Rightarrow \; EF = 9$

Hence, the correct answer is $9$.

$\Delta ABC$ and $\Delta DEF$ are similar and $\angle A=40^\mathring \ ,\angle E+\angle F=$

  1. $140$

  2. $40$

  3. $80$

  4. $180$


Correct Option: A
Explanation:

Since the triangles are similar.

$\angle A=\angle D$
$\angle D=40^{\circ}$
In triangle $\Delta DEF$
$\angle D+\angle E+\angle F=180^{\circ}$
$\angle E+\angle F=180^{\circ}-40^{\circ}=140^{\circ}$

STATEMENT - 1 : If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.
STATEMENT - 2 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

  1. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

  2. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

  3. Statement - 1 is True, Statement - 2 is False

  4. Statement - 1 is False, Statement - 2 is True


Correct Option: B
Explanation:

 If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
Both the statements are correct but $2$ is not the reason for $1$
If two corresponding angles are equal then the third corresponding become also equal , so the triangles are similar.
Option $B$ is correct

If $\triangle ABC $ and $BDE$ are similar triangles such that $2AB = DE$ and $BC= 8$ cm, then $EF$ is

  1. $16$ cm

  2. $17$ cm

  3. $4$ cm

  4. $8$ cm


Correct Option: A

Is the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians?

  1. True

  2. False


Correct Option: A

The areas of two similar triangles $\triangle{ABC}$ and $\triangle{DEF}$ are $144\ cm^{2}$ and $81\ cm^{2}$ respectively. If the longest side of larger $\triangle{ABC}$ be $36\ cm$, then, the largest side of the similar triangle $\triangle{DEF}$ is

  1. $20\ cm$

  2. $26\ cm$

  3. $27\ cm$

  4. $30\ cm$


Correct Option: A

The correspondence $ABC\rightarrow PQR$ is a similarity in $\Delta ABC$ and $\Delta PQR$. If the perimeter of $\Delta ABC$ is $24$ and the perimeter of $\Delta PQR$ is $40$, then $AB=PQ=$

  1. $4:3$

  2. $3:4$

  3. $5:3$

  4. $3:5$


Correct Option: A