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,
$AC^2 = AC^2 + AD^2$
$\Rightarrow DC^2 = AC^2 - AD^2$
$\Rightarrow DC^2 = 12^2 - 6^2$
$\Rightarrow DC^2 = 144 - 36$
$\Rightarrow DC^2= 108$
$\Rightarrow\text{DC}=\sqrt{3\times4\times9}$
$\Rightarrow\text{DC}=6\sqrt{3}\text{cm}$
So, the height is $6\sqrt3\text{cm}.$

View full question & answer→

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,
$\frac{25}{36}=\frac{(3.5)^2}{\text{h}^2}$
$\Rightarrow\text{h}^2=\frac{(3.5)^2\times36}{25}$
$\Rightarrow\text{h}^2=17.64$
$\Rightarrow\text{h}=4.2\ \text{cm}$

View full question & answer→

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 $\angle\text{BAD}=?$

✓
$30^\circ$
B
$40^\circ$
C
$45^\circ$
D
$50^\circ$

Answer

Correct option: A.

$30^\circ$

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

View full question & answer→

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.

View full question & answer→

MCQ 551 Mark

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

A
$4.5\ cm$
B
$8\ cm$
C
$9\ cm$
✓
$7.5\ cm$

Answer

Correct option: D.

$7.5\ cm$

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, $AC = 7.5\ cm.$

View full question & answer→

MCQ 561 Mark

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

✓
$5.6\ cm$
B
$4.8\ cm$
C
$5.2\ cm$
D
$6.4\ cm$

Answer

Correct option: A.

$5.6\ cm$

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

View full question & answer→

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 $\text{AB = 3DE}$
$\Rightarrow\text{AB}\not=\text{DE}$
So, the triangles are not congruent.
Thus, the two triangles are similar, but not cogruent.

View full question & answer→

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$)$
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{ADE} ....(\text{SAS}$ 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 $4 : 1.$

View full question & answer→

Maths M.C.Q (1 Marks)