Tag: similarity of triangles

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Questions Related to similarity of triangles

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, which is also equal to the square of the ratio of their corresponding medians.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

The ratio of the angles in $\triangle ABC$ is $2 : 3 : 4$. Which one of the following triangles is similar to $\triangle ABC ?$

  1. $ \triangle DEF $ has angles in the ratio $4 : 3 : 2.$
  2. $ \triangle PQR $ has angles in the ratio $1 : 2 : 3.$
  3. $ \triangle LMN $ has angles in the ratio $1 : 1 : 1.$
  4. $ \triangle STW $ has sides in the ratio $1 : 1 : 1.$
  5. $ \triangle XYZ $ has sides in the ratio $4 : 3 : 2.$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Similar triangles must have the same ratio of angles. The ratio 2:3:4 is equivalent to 4:6:8 or any scalar multiple, but the order of the ratio matters for similarity. Option A provides the same ratio 4:3:2, which represents the same set of interior angles as 2:3:4.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

If $A={30}^{\circ},\,a=100,\,c=100\sqrt{2}$, find the number of triangles that can be formed.

  1. $1$
  2. $2$
  3. $3 $
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Here $a, c$ and $A$ are given, $\therefore$ we will have to examine whether two triangle are possible or not. For two triangles
$(i)\,a>c\sin{A}$ and $(ii)a<c$
$\Rightarrow 100>100\sqrt{2}\sin{{30}^{\circ}}$
$\Rightarrow 100>100\sqrt{2}\times\dfrac{1}{2}$
$\Rightarrow 100>50\sqrt{2}$
and $a<c$
i.e., $100<100\sqrt{2}$
$\Rightarrow $ Two triangles can be formed.
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In triangle ABC, AB = AC = 8 cm, BC = 4 cm and P is a point in side AC such that AP = 6 cm. Prove that $\Delta\,BPC$ is similar to $\Delta\,ABC$. Also, find the length of BP.

  1. BP = 4 cm

  2. BP = 8 cm

  3. BP = 6 cm

  4. BP = 12 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $\triangle ABC$, $AB = AC = 8$, $BC = 4$ and $AP = 6$

In $\Delta\,ABC$,
$\displaystyle\,\frac{AB}{BC}\,=\,\frac{8}{4}\,=\,2$,
In $\Delta\,BPC$,
$\displaystyle\,\frac{BC}{CP}\,=\,\frac{4}{2}\,=\,2$

Now, in $\triangle ABC$ and $\triangle BPC$
$\displaystyle\,\dfrac{AB}{BC}\,= \displaystyle\,\dfrac{BC}{CP}$
$\angle\,ABC\,=\,\angle\,C.$
Therefore, by SAS, $\Delta\,ABC \sim \Delta\,BPC$

Thus, $\dfrac{AB}{BP} = \dfrac{AC}{BC}$


$\dfrac{8}{BP} = \dfrac{8}{4}$
$BP = 4$ cm

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In the given figure, $DE$ is parallel to $BC$ and the ratio of the areas of $\triangle ADE$ and trapezium $BDEC$ is $4:5.$ What is $DE : BC: ?$

  1. $1:2$
  2. $2:3$
  3. $4:5$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ratio of area(ADE) to area(ABC) is 4/(4+5) = 4/9. Since the ratio of areas of similar triangles is the square of the ratio of their corresponding sides, DE/BC = sqrt(4/9) = 2/3.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

If in $\triangle $s $ABC$ and $DEF,$ $\angle A=\angle E=37^{\circ}, AB:ED=AC:EF$ and $\angle F=69^{\circ},$ then what is the value of $\angle B: ?$

  1. $69^{\circ}$
  2. $74^{\circ}$
  3. $84^{\circ}$
  4. $94^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ABC$ and $\triangle DEF$
$\angle A = \angle E =  37^{o}$
$\dfrac{AB}{ED} = \dfrac{AC}{EF}$
Thus, $\triangle ABC \sim \triangle EDF$ ....... (By SAS rule)
Thus, $\angle B = \angle D$

Now, $\triangle DEF$
$\angle D + \angle E + \angle F = 180$
$\angle D + 37 + 69 = 180$
$\angle D = 74^{\circ}$
Hence, $\angle B = \angle D = 74^{\circ}$