M.C.Q (1 Marks)

MCQ 511 Mark

The height of an equilateral triangle having each side $12\ cm,$ is:

A
$6\sqrt{2}\text{ cm}$
✓
$6\sqrt{3}\text{ cm}$
C
$3\sqrt{6}\text{ cm}$
D
$6\sqrt{6}\text{ cm}$

Answer

Correct option: B.

$6\sqrt{3}\text{ cm}$

Let $\triangle\text{ABC}$ be the equilateral triangle and $AD$ be the height.
The height of an equilateral triangle is the same as its median.
So, $AD = 6m$
$\triangle\text{ABC}$ is a right$-$angled triangle.
By Pythagoras theorem,
$A C^2=A C^2+A D^2$
$\Rightarrow D C^2=A C^2-A D^2$
$\Rightarrow D C^2=12^2-6^2$
$\Rightarrow D C^2=144-36$
$\Rightarrow D C^2=108$
$\Rightarrow\text{DC}=\sqrt{3\times4\times9}$
$\Rightarrow\text{DC}=6\sqrt{3}\text{ cm}$
So, the height is $6\sqrt3\text{cm}.$

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MCQ 521 Mark

The areas of two similar triangles are $25 \ cm^2$ and $36 \ cm^2$ respectively. If the altitude of the first triangle is $3.5 \ cm$ then the corresponding altitude of the other triangle is:

A
$5.6\ cm$
B
$6.3\ cm$
✓
$4.2\ cm$
D
$7\ cm$

Answer

Correct option: C.

$4.2\ cm$

We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,
$\frac{25}{36}=\frac{(3.5)^2}{\text{h}^2}$
$\Rightarrow\text{h}^2=\frac{(3.5)^2\times36}{25}$
$\Rightarrow\text{h}^2=17.64$
$\Rightarrow\text{h}=4.2\text{cm}$

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MCQ 531 Mark

In $\triangle\text{ABC}$ it is given that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}.$ If $\angle\text{B}=70^\circ$ and $\angle\text{C}=50^\circ$ then $\angle\text{BAD}=?$

✓
30º
B
40º
C
45º
D
50º

Answer

Correct option: A.

30º

in $\triangle\text{ABC}$ it is given that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}.$
⇒ AD bisects $\angle\text{BAC}$
In $\triangle\text{ABC},$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{BAC}+70^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=60^\circ$
$\Rightarrow\angle\text{BAD}=\angle\text{DAC}=30^\circ$

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MCQ 541 Mark

The line segments joining the midpoint of the sides of a triangle form four triangles, each of which is:

A
Congruent to the original triangle.
✓
Similar to the original triangle.
C
An isosceles triangle.
D
An equilateral triangle.

Answer

Correct option: B.

Similar to the original triangle.

The line segments joining the midpoint of the side of a triangle form four triangles, each of which is similar to the original triangle.

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MCQ 551 Mark

In a $\triangle\text{ABC}$ it is given that AD is the internal bisector of $\angle\text{A}.$ If BD = 4cm, DC = 5cm and AB = 6cm, then AC =?

A
4.5cm
B
8cm
C
9cm
✓
7.5cm

Answer

Correct option: D.

7.5cm

since AD is the bisector of $\angle\text{A},$
by the angle bisector theorem,
$\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{6}{\text{x}}=\frac{4}{5}$
$\Rightarrow\text{x}=7.5\text{cm}$
So, AC = 7.5cm.

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MCQ 561 Mark

In the given figure, DE || BC. If DE = 5cm, BC = 8cm and AD = 3.5cm then AB =?

✓
5.6cm
B
4.8cm
C
5.2cm
D
6.4cm

Answer

Correct option: A.

5.6cm

$\therefore\text{DE }||\text{ BC}$
$\therefore\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{DE}}{\text{BC}}$ (Thales' theorem)
$\Rightarrow\frac{3.5}{\text{AB}}=\frac{5}{\text{8}}$
$\Rightarrow\text{AB}=\frac{3.5\times8}{\text{5}}=5.6\text{cm}$

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MCQ 571 Mark

In $\triangle\text{ABC}$ and $\triangle\text{DEF},$ it is given that $\angle\text{B}=\angle\text{E},\angle\text{F}=\angle\text{C}$ and $\text{AB}=3\text{DE},$ then the two triangles are:

A
Congruent but not similar
✓
Similar but not congruent
C
Neither congruent not similar
D
Similar as well as congruent

Answer

Correct option: B.

Similar but not congruent

In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
It is given that $\angle\text{B}=\angle\text{E},\angle\text{F}=\angle\text{C},$ and hence $\angle\text{A}=\angle\text{D}$
So, the two triangles are similar.
Since AB = 3DE
$\Rightarrow\text{AB}\not=\text{DE}$
So, the triangles are not congruent.
Thus, the two triangles are similar, but not cogruent.

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MCQ 581 Mark

In an equilateral $\triangle\text{ABC},\text{D}$ is the midpoint of AB and E is the midpoint of AC. Then, $\text{ar}(\triangle\text{ABC}):\text{ar}(\triangle\text{ADE})=?$

A
2 : 1
✓
4 : 1
C
1 : 2
D
1 : 4

Answer

Correct option: B.

4 : 1

Since D and E are the mid-point of AB and AC respectively.
$\frac{\text{AB}}{\text{AD}}=\frac{\text{AC}}{\text{AE}}=\frac{2}{1}$ and $\angle\text{CAD}=\angle\text{EAD}$ ....(common angle)
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{ADE}$ ....(SAS criterion for Similarity)
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{ADE})}=\frac{\text{AB}^2}{\text{AD}^2}=\frac{2^2}{1^2}=\frac{4}{1}$
So, the ratio is 4 : 1.

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Maths M.C.Q (1 Marks)