2 Marks Questions

Question 512 Marks

If a $\triangle\text{ABC}, AD$ is the bisector of $\angle\text{A},$ Meeting side $BC$ at $D$.
If $AB = 10\ cm, AC = 14\ cm$ and $BC = 6\ cm,$ find $BD$ and $DC$

Answer

We have,
$AB = 10\ cm, AC = 14\ cm$ and $BC = 6\ cm$
Now, $\frac{\text{BD}}{\text{AB}}=\frac{\text{CD}}{\text{AC}}$
$⇒ BD × AC = CD × AB$
$⇒ (BC - CD) AC = CD × AB (\because BD = BC - CD)$
$⇒ (6 - CD) 14 = CD × 10$
$⇒ 84 - 14CD = 10CD$
$⇒ 24CD = 84$
$\Rightarrow \text{CD} =\frac{84}{24}$
$⇒ CD = 3.5$
$\because BD = BC - CD$
$⇒ BD = 6 - 3.5$
$⇒ BD = 2.5\ cm$

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Question 522 Marks

The area of two similar triangles are $36\ cm^2$ and $100\ cm^2$. If the length of a side of the smaller triangle in $3\ cm,$ find the length of the corresponding side of the larger triangle.

Answer

Area of smaller triangle $= 36\ cm^2$
and area of larger triangle $= 100\ cm^2$
One side of smaller triangle $= 3\ cm$
Let the corresponding side of larger triangle $= x$
$\triangle\text{s}$ are similar
$\therefore\frac{\text{area of smaller triangle}}{\text{area of larger triangle}}=\frac{(3)^2}{\text{x}^2}$
$\Rightarrow\frac{36}{100}=\frac{9}{\text{x}^2}\Rightarrow\text{x}^2=\frac{9\times100}{36}=25=(5)^2$
$\therefore\text{x}=5$
$\therefore$ Corresponding side of larger triangle $= 5\ cm.$

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Question 532 Marks

If $ABC$ and $DEF$ are similar triangles such that $\angle\text{A}=57^\circ$ and $\angle\text{E}=73^\circ,$ what is the measure of $\angle\text{C}?$

Answer

We have,

$\triangle\text{ABC}\sim\triangle\text{DEF}$
we know that corresponding angle of similar triangle are equal.
$\angle\text{A}=\angle\text{D}=57^\circ$ and $\angle\text{B}=\angle\text{E}=73^\circ$
$\angle\text{C}=\angle\text{F}=?$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle sum property)
$\Rightarrow57^\circ+73^\circ+\angle\text{C}=180^\circ$
$\Rightarrow130^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-130^\circ$
$\Rightarrow\angle\text{C}=50^\circ$
Thus, $\angle\text{C}=50^\circ$

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Question 542 Marks

In the given figure,

$AB || DC$ prove that
$DM × BV = BM × DU$

Answer

Since $\triangle\text{DMU}\sim\triangle\text{BMV}$
$\frac{\text{DM}}{\text{BM}}=\frac{\text{MU}}{\text{MV}}=\frac{\text{DU}}{\text{BV}}$
$\frac{\text{DM}}{\text{BM}}=\frac{\text{DU}}{\text{BV}}$
By cross multiplication, we get,
$DM × BV = DU × BM$
Hence proved that $DM × BV = DU × BM.$

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Question 552 Marks

In the given figure, $DE || BC$ and $\text{AD}=\frac{1}{2}\text{BD}.$ If $BC = 4.5\ cm,$ find $DE$.

Answer

Given: In $\triangle\text{ABC},\ \text{DE}||\text{BC}.\ \text{AD}=\frac{1}{2}\text{BD}$ and $BC = 4.5\ cm.$
To find: $DE$
In $\triangle\text{ABC}$ and $\triangle\text{ADE}$
$\angle\text{B}=\angle\text{ADE}$ (Corresponding angles)
$\angle\text{A}=\angle\text{A}$ (Common)
$\therefore\triangle\text{ABC}\sim\triangle\text{ADE}$ (AA Similarity)
$\frac{\text{AD}}{\text{AB}}=\frac{\text{DE}}{\text{BC}}$
$\frac{\text{AD}}{\text{AD}+\text{BD}}=\frac{\text{DE}}{\text{BC}}$
$\frac{\frac{1}{2}\text{BD}}{\frac{1}{2}\text{BD}+\text{BD}}=\frac{\text{DE}}{\text{BC}}$
$\frac{1}{3}=\frac{\text{DE}}{\text{BC}}$
$\frac{1}{3}=\frac{\text{DE}}{4.5}$
$\text{DE} = 1.5\text{cm.}$

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Question 562 Marks

In FIg. check whether $AD$ is the bisector of $\angle\text{A}$ of $\triangle\text{ABC}$ in the following.

$AB = 6\ cm, AC = 8\ cm, BD = 1.5\ cm$ and $CD = 2\ cm.$

Answer

It is given that, $AB = 6\ cm, AC = 8\ cm, BD = 1.5\ cm$ and $CD = 2\ cm.$
We have to check whether $AD$ is bisector of $\angle\text{A}$.
First we will check proportional retio between sides.
So $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\frac{6}{8}=\frac{1.5}{2}$
$\Rightarrow\frac{3}{4}=\frac{3}{4}$
$($It is proportional$)$
Hence, $AD$ is bisector of $\angle\text{A}$.

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Question 572 Marks

The lengths of the diagonals of a rhombus are $24\ cm$ and $10\ cm.$ Find each side of the rhombus.

Answer

We have,
$ABCD$ is a rhombus with diagonals $AC = 10\ cm$ and $BD = 24\ cm$
We know that diagonal of a rhombus bisect each other at $90^\circ $
$\therefore AO = OC = 5\ cm$ and $BO = OD = 12\ cm$
In $\triangle\text{AOB},$ by Pythagoras theorem
$ A B^2=A O^2+B O^2 $
$ \Rightarrow A B^2=5^2+12^2$
$\Rightarrow A B^2=25+144=169$
$\Rightarrow\text{AB}=\sqrt{169}=13\text{cm}$

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Question 582 Marks

In the following figures, you find who triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.

Answer

In two triangle, we observe that

In $\triangle\text{ABC}$ and $\triangle\text{CDE}$
$\frac{\text{CD}}{\text{CE}}=\frac{\text{CB}}{\text{CA}}$
$\angle\text{ACB}=\angle\text{DCE} ($Vertically opposite angles$)$
$\triangle\text{ABC}\sim\triangle\text{CDE} (SAS$ Similarity$)$
yes two triangles are similar.

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Question 592 Marks

In the adjoining figure, $DE$ is parallel to $BC$ and $AD = 1\ cm, BD = 2\ cm.$ What is the ratio of the area of $\triangle\text{ABC}$ to the area of $\triangle\text{ADE}?$

Answer

In $\triangle\text{ABC},\ \text{DE}||\text{BC}$
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}$

$\therefore\frac{\text{area}\triangle\text{ABC}}{\text{area}\triangle\text{ADE}}=\frac{\text{AB}^2}{\text{AD}^2}=\frac{(\text{AD}+\text{DB})^2}{\text{AD}^2}$
$=\frac{(1+2)^2}{(1)^2}=\frac{3^2}{1}=\frac{9}{1}$
$\therefore$ Ratio $= 9 : 1$

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Question 602 Marks

In the figure, $DE || BC$ in $\triangle\text{ABC}$ such that $BC = 8\ cm, AB = 6\ cm$ and $DA = 1.5\ cm.$ Find $DE.$

Answer

In the given figure,
$DE || BC$
$BC = 8\ cm, AB = 6\ cm$ and $DA = 1.5\ cm.$

In $\triangle\text{ABC}$ and $\triangle\text{ADE},$
$\angle\text{A}=\angle\text{A} ($Common$)$
$\angle\text{B}=\angle\text{ADE} ($Corresponding angles$)$
$\therefore\triangle\text{ABC}\sim\triangle\text{ADE} (AA$ axiom$)$
$\therefore\frac{\text{AB}}{\text{AD}}=\frac{\text{BC}}{\text{DE}}\Rightarrow\frac{6}{1.5}=\frac{8}{\text{DE}}$
$\Rightarrow\text{DE}=\frac{8\times1.5}{6}=\frac{12}{6}=2\text{cm}.$

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Question 612 Marks

In a $\triangle\text{ABC}, AB = BC = CA = 2a$ and $\text{AD}\perp\text{BC}.$ Prove that
$\text{AD}=\text{a}\sqrt{3}$

Answer

$\triangle\text{ABC}, AB = BC = AC = 2a$
$\triangle\text{AD}\perp\text{BC}$
$AD$ bisects $BC$ at $D$
$\text{BD}=\text{DC}=\frac{1}{2}\text{BC}=\text{a}$

Now in $\triangle\text{ABD}$
$ A B^2=A D^2+B D^2$
$ \Rightarrow(2 a)^2=A D^2+a^2 \Rightarrow 4 a^2=A D^2+a^2$
$ \Rightarrow A D^2=4 a^2-a^2=3 a^2=(\sqrt{3} a)^2$
$\therefore\text{AD}=\sqrt{3}\text{a}$

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Question 622 Marks

state $SAS$ similarity criterion.

Answer

If in two triangles, one pair of coresponding sides are proportional and the included angle are equal then the two triangles are similar.

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Maths 2 Marks Questions