2 Marks Questions - Maths STD 10 Questions - Vidyadip

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Question 12 Marks

$\triangle\text{ABC}\sim\triangle\text{DEF}$ and their areas are respectively $64cm^2$ and $121cm^2$. If $EF = 15.4\ cm$, find $BC$.

Answer

It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}$
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
Let $BC$ be $x\ cm$.
$\Rightarrow\frac{64}{121}=\frac{\text{x}^2}{(15.4)^2}$
$\Rightarrow\text{x}^2=\frac{64\times15.4\times15.4}{121}$
$\Rightarrow\text{x}=\sqrt{\frac{64\times15.4\times15.4}{121}}$
$=\frac{8\times15.4}{11}$
$=11.2$
Hence, $BC = 11.2\ cm$

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Question 22 Marks

$ABC$ is an isosceles triangle, right-angled at $B$. Similar triangle $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$. Find the ratio between the areas of $\triangle\text{ABE}$ and $\triangle\text{ACD}.$

Answer

$ABC$ is an isosceles triangle right angled at $B,$
Let $AB = BC = x\ cm$

By Pythagoras theorem,
$ \mathrm{AC}^2=A B^2+B C^2$
$=x^2+x^2 $
$A C^2=2 x^2 $
$\text{AC}=\sqrt{2}\text{x}$
$\triangle\text{ACD}\approx\triangle\text{ABE}$ (Given)
$\therefore\frac{\text{ar}\triangle\text{ABE}}{\text{ar}\triangle\text{ACD}}=\frac{\text{AB}^2}{\text{AC}^2}=\frac{\text{x}^2}{\big(\sqrt{2}\text{x}\big)^2}$
$=\frac{\text{x}^2}{2\text{x}^2}=\frac{1}{2}=1:2$

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Question 32 Marks

$\triangle\text{ABC}$ and $\triangle\text{DBC}$ lie on the same side of $BC$, as shown in the figure. From a point $P$ on $BC, PQ || AB$ and $PR || BD$ are drawn, meeting $AC$ at $Q$ and $CD$ at $R$ respectively. Prove that $QR || AD$.

Answer

In $\triangle\text{CAB},\text{PQ }||\text{ AB}.$
Applying Thales' theorem, we get:
$\frac{\text{CP}}{\text{PB}}=\frac{\text{CQ}}{\text{QA}}\dots(1)$
Similarly, applying Thales' theorem $\triangle\text{BDC},$ where $PR || BD$, we get:
$\frac{\text{CP}}{\text{PB}}=\frac{\text{CR}}{\text{RD}}\dots(2)$
Hence, from $(1)$ and $(2),$ we have:
$\frac{\text{CQ}}{\text{QA}}=\frac{\text{CR}}{\text{RD}}$
Applying the converse of Thales' theorem, we conclude that $QR || AD$ in $\triangle\text{ADC}.$
This completes the proof.

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Question 42 Marks

In a $\triangle\text{ABC},\text{AD}$ is a median and $\text{AL}\perp\text{BC}.$
Prove that:

$\text{AC}^2=\text{AD}^2+\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
$\text{AC}^2+\text{AB}^2=2\text{AD}^2+\frac{\text{1}}{2}\text{BC}^2$

Answer

In right $\triangle\text{ALD},$
By Pythagoras theorem,
$ \mathrm{AD}^2=\mathrm{AL}^2+\mathrm{DL}^2$
$ \Rightarrow \mathrm{AL}^2=\mathrm{AD}^2-\mathrm{DL}^2 ....(i)$
In right $\triangle\text{ACL},$
By Pythagoras theorem,
$\text{AC}^2 =\text{ AL}^2 + \text{LC}^2$
$\Rightarrow \text{AC}^2 = (\text{AD}^2 -\text{ DL}^2) + (\text{DL} +\text{ DC)}^2\dots(\text{from (i))}$
$\Rightarrow\text{AC}^2=\text{AD}^2-\text{DL}^2+\Big(\text{DL}+\frac{\text{BC}}{2}\Big)^2$$.....($Since $AD$ is the median$)$
$\Rightarrow\text{AC}^2=\text{AD}^2-\text{DL}^2+\text{DL}^2+\Big(\frac{\text{BC}}{2}\Big)^2+\text{BC}.\text{DL}$
$\Rightarrow\text{AC}^2=\text{AD}^2+\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
Hence proved.

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Question 52 Marks

The areas of two similar triangles $ABC$ and $PQR$ are in the ratio $9 : 16$. If $BC = 4.5cm$, find the length of $QR$.

Answer

It is given that $\triangle\text{ABC}\sim\triangle\text{PQR}.$
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\Rightarrow\frac{9}{16}=\frac{\text{4.5}^2}{\text{QR}^2}$
$\Rightarrow\text{QR}^2=\frac{4.5\times4.5\times16}{9}$
$\Rightarrow\text{QR}=\sqrt{\frac{4.5\times4.5\times16}{9}}$
$=\frac{4.5\times4}{3}$
$=6\text{cm}$
Hence, $QR = 6\ cm$

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Question 62 Marks

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\triangle\text{ABC}$ such that $DE || BC$:
If $AB = 13.3\ cm, AC = 11.9\ cm$ and $EC = 5.1\ cm$, find $AD$.

Answer

In $\triangle\text{ABC},$ it is given that $DE || BC$.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Adding $1$ to both sides, we get:
$\frac{\text{AD}}{\text{DB}}+1=\frac{\text{AE}}{\text{EC}}+1$
$\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}}$
$\Rightarrow\frac{\text{13.3}}{\text{DB}}=\frac{\text{11.9}}{\text{5.1}}$
$\text{DB}=\frac{13.3\times5.1}{11.9}=5.7\text{cm}$
Therefore, $\text{AD}=\text{AB}-\text{BD}=13.5-5.7=7.6\text{cm}$

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Question 72 Marks

In the given figure, $LM || CD$.
Prove that $\frac{\text{AM}}{\text{AB}}=\frac{\text{AN}}{\text{AD}}.$

Answer

$LM || CB$ and $LN || CD$
Therefore, Applying Thale's theorem, we have:
$\frac{\text{AB}}{\text{AM}}=\frac{\text{AC}}{\text{AL}}$ and $\frac{\text{AD}}{\text{AN}}=\frac{\text{AC}}{\text{AL}}$
$\Rightarrow\frac{\text{AB}}{\text{AM}}=\frac{\text{AD}}{\text{AN}}$
$\therefore\frac{\text{AM}}{\text{AB}}=\frac{\text{AN}}{\text{AD}}$
This completes the proof.

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Question 82 Marks

In the given pairs of triangles, find which pair of triangles are similar. State the similarity criterior and write the similarity relation in symbolic from.

Answer

In $\triangle\text{ABC}$ and $\triangle\text{PQR}$
$\angle\text{A}=\angle\text{Q}=50^\circ$
$\angle\text{B}=\angle\text{P}=60^\circ$
$\angle\text{C}=\angle\text{R}=70^\circ$
$\therefore\triangle\text{ABC}\sim\triangle\text{QPR}$ (by $AAA$ similarity)

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Question 92 Marks

A ladder $10m$ long reaches the window of a house $8m$ above the ground. Find the distance of the foot of the ladder from the base of the wall.

Answer

Let the ladder be $AB$ and $BC$ be the height of the window from the ground.

We have:
$AB = 10m$ and $BC = 8m$
Applying Pythagoras theorem in right-angled triangle $ACB$, we have:
$ A B^2=A C^2+B C^2 $
$ \Rightarrow A C^2=A B^2-B C^2=10^2-8^2=100-64=36$
$\Rightarrow AC = 6m$
Hence, the foot of the ladder is $6m$ away from the base of the wall.

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Question 102 Marks

In the given pairs of triangles, find which pair of triangles are similar. State the similarity criterior and write the similarity relation in symbolic from.

Answer

In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle Sum Property)
$\Rightarrow80^\circ+\angle\text{B}+70^\circ=180^\circ$
$\Rightarrow\angle\text{B}=30^\circ$
$\angle\text{A}=\angle\text{M}$ and $\angle\text{B}=\angle\text{N}$
Therefore, by $AA$ similarity theorem, $\triangle\text{ABC}\sim\triangle\text{MNR}$

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Question 112 Marks

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\triangle\text{ABC}$ such that $DE || BC:$
$AD = 4cm, DB = (x - 4)cm, AE = 8cm$ and $EC = (3x - 19)cm$.

Answer

In $\triangle\text{ABC},$ it is given that $DE || BC$.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{4}}{\text{x}-4}=\frac{\text{8}}{3\text{x}-19}$
$\Rightarrow\text{4}(3\text{x}-19)=8(\text{x}-4)$
$\Rightarrow12\text{x}-76=8\text{x}-32$
$\Rightarrow4\text{x}=44$
$\Rightarrow\text{x}=11\text{cm}$

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Question 122 Marks

In the given figure, $DE || BC$ such that $AD = x\ cm$, $DB = (3x + 4)cm, AE = (x + 3)cm$ and $EC = (3x + 19)cm$. Find the value of $x$.

Answer

$\therefore\text{DE }||\text{ BC}$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$ (Basic proportionality theorem)
$\frac{\text{x}}{3\text{x}+4}=\frac{\text{x}+3}{3\text{x}+19}$
$\Rightarrow\text{x}(3\text{x}+19)=(\text{x}+3)(3\text{x}+4)$
$\Rightarrow3\text{x}^2+19\text{x}=3\text{x}^2+4\text{x}+9\text{x}+12$
$\Rightarrow19\text{x}-13\text{x}=12$
$\Rightarrow6\text{x}=12$
$\Rightarrow\text{x}=2$

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Question 132 Marks

Find the length of the altitude of an equilateral triangle of side $2a\ cm.$

Answer

Let the triangle be $ABC$ with $AD$ as its altitude. Then, $D$ is the midpoint of $BC$.
In right-angled triangle $ABD$, we have:

$A B^2=A D^2+D B^2$
$\Rightarrow A D^2=A B^2-D B^2=4 a^2-a^2$ $\big(\therefore\text{BD}=\frac{1}{2}\text{BC}\big)$
$=3 a^2$
$\text{AD}=\sqrt{3\text{a}}$
Hence, the length of the altitude of an equilateral triangle of side $2a\ cm$ is $\sqrt{3}\text{a}\text{ cm}.$

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Question 142 Marks

$M$ is a point on the side $BC$ of a parallelogram $ABCD$. $DM$ when produced meet $AB$ produced at $N$. Prove that.

$\frac{\text{DM}}{\text{MN}}=\frac{\text{DC}}{\text{BN}}$
$\frac{\text{DN}}{\text{DM}}=\frac{\text{AN}}{\text{DC}}$

Answer

Given: $ABCD$ is a parallelogram
To prove:

$\frac{\text{DM}}{\text{MN}}=\frac{\text{DC}}{\text{BN}}$
$\frac{\text{DN}}{\text{DM}}=\frac{\text{AN}}{\text{DC}}$

Proof: In $\triangle\text{DMC}$ and $\triangle\text{NMB}$
$\angle\text{DMC}=\angle\text{NMB}$ (Vertically opposite angle)
$\angle\text{DCM}=\angle\text{NBM}$ (Altrnate angles)
By AAA-similarity
$\triangle\text{DMC}\sim\triangle\text{NMB}$
$\therefore\frac{\text{DM}}{\text{MN}}=\frac{\text{DC}}{\text{BN}}$
Now, $\frac{\text{MN}}{\text{DM}}=\frac{\text{BN}}{\text{DC}}$
Adding 1 to both sides, we get
$\frac{\text{MN}}{\text{DM}}+1=\frac{\text{BN}}{\text{DC}}+1$
$\Rightarrow\frac{\text{MN}+\text{DM}}{\text{DM}}=\frac{\text{BN}+\text{DC}}{\text{DC}}$
$\Rightarrow\frac{\text{MN}+\text{DM}}{\text{DM}}=\frac{\text{BN}+\text{AB}}{\text{DC}}$ $[\therefore\text{ABCD}$ is a parallelogram$]$
$\Rightarrow\frac{\text{DN}}{\text{DM}}=\frac{\text{AN}}{\text{DC}}$

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Question 152 Marks

In a $\triangle\text{ABC},\text{AD}$ is the bisector of $\angle\text{A}.$
If $AB = 6.4\ cm, AC = 8\ cm$ and $BD = 5.6\ cm$, find $DC$.

Answer

It is given that $AD$ bisects $\angle\text{A}.$
Applying angle-bisector theorem in $\triangle\text{ABC},$ we get:
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{5.6}{\text{DC}}=\frac{6.4}{8}$
$\Rightarrow\text{DC}=\frac{8\times5.6}{6.4}=7\text{cm}$

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Question 162 Marks

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\triangle\text{ABC}$ such that $DE || BC$:
$AD = x\ cm, DB = (x - 2)cm, AE = (x + 2)cm$ and $EC = (x - 1)cm$.

Answer

In $\triangle\text{ABC},$ it is given that $DE || BC$.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}}{\text{x}-2}=\frac{\text{x}+2}{\text{x}-1}$
$\Rightarrow\text{x}(\text{x}-1)=(\text{x}-2)(\text{x}+2)$
$\Rightarrow\text{x}^2-\text{x}=\text{x}^2-4$
$\Rightarrow\text{x}=4\text{cm}$

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2 Marks Questions - Maths STD 10 Questions - Vidyadip