l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ΔABC ≅ ΔCDA.
Solution: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{CDA}$, $\angle \mathrm{BAC}=\angle \mathrm{DCA}$ (Alternate interior angles, as $p \| q$ ) $\mathrm{AC}=\mathrm{CA}$ (Common) $\angle \mathrm{BCA}=\angle \mathrm{DAC}$ (Alternate interior angles, as $/ \| m$ ) $\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{CDA}(\mathrm{By} \mathrm{ASA}$ congruence rule $)$...
Read More →$A D$ and $B C$ are equal perpendiculars to a line segment $A B$ (See the given figure). Show that $C D$ bisects $A B$.
Solution: In $\triangle \mathrm{BOC}$ and $\triangle \mathrm{AOD}$, $\angle B O C=\angle A O D$ (Vertically opposite angles) $\angle C B O=\angle D A O\left(\right.$ Each $\left.90^{\circ}\right)$ $\mathrm{BC}=\mathrm{AD}$ (Given) $\therefore \triangle \mathrm{BOC} \cong \triangle \mathrm{AOD}(\mathrm{AAS}$ congruence rule $)$' $\therefore \mathrm{BO}=\mathrm{AO}(\mathrm{By} \mathrm{CPCT})$ $\Rightarrow \mathrm{CD}$ bisects $\mathrm{AB}$....
Read More →In quadrilateral $A C B D, A C=A D$ and $A B$ bisects $\angle A$ (See the given figure). Show that $\triangle A B C \cong \triangle A B D$. What can you say about $B C$ and $B D$ ?
Solution: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ABD}$ $A C=A D$ (Given) $\angle \mathrm{CAB}=\angle \mathrm{DAB}(\mathrm{AB}$ bisects $\angle \mathrm{A})$ $A B=A B$ (Common) $\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{ABD}$ (By $S A S$ congruence rule) $\therefore \mathrm{BC}=\mathrm{BD}(\mathrm{By} \mathrm{CPCT})$ Therefore, $B C$ and $B D$ are of equal lengths....
Read More →In the given figure, the side $Q R$ of $\triangle P Q R$ is produced to a point $S$. If the bisectors of $\angle P Q R$ and $\angle P R S$ meet at point $T$, then prove that $\angle Q T R=$ $\frac{1}{2} \angle \mathrm{QPR}$
Solution: In $\triangle Q T R, \angle T R S$ is an exterior angle. $\therefore \angle \mathrm{QTR}+\angle \mathrm{TQR}=\angle \mathrm{TRS}$ $\angle Q T R=\angle T R S-\angle T Q R(1)$ For $\triangle P Q R, \angle P R S$ is an external angle. $\therefore \angle \mathrm{QPR}+\angle \mathrm{PQR}=\angle \mathrm{PRS}$ $\angle Q P R+2 \angle T Q R=2 \angle T R S($ As $Q T$ and $R T$ are angle bisectors $)$ $\angle Q P R=2(\angle T R S-\angle T Q R)$ $\angle Q P R=2 \angle Q T R$ [By using equation (1)...
Read More →In the given figure, if $P Q \perp P S, P Q \| S R, \angle S Q R=28^{\circ}$ and $\angle Q R T=65^{\circ}$, then find the values of $x$ and $y$.
Solution: It is given that $P Q \| S R$ and $Q R$ is a transversal line. $\angle P Q R=\angle Q R T$ (Alternate interior angles) $x+28^{\circ}=65^{\circ}$ $x=65^{\circ}-28^{\circ}$ $x=37^{\circ}$ By using the angle sum property for $\triangle S P Q$, we obtain $\angle S P Q+x+y=180^{\circ}$ $90^{\circ}+37^{\circ}+y=180^{\circ}$ $y=180^{\circ}-127^{\circ}$ $y=53^{\circ}$ $\therefore x=37^{\circ}$ and $y=53^{\circ}$...
Read More →In the given figure, if lines $P Q$ and $R S$ intersect at point $T$, such that $\angle P R T=40^{\circ}, \angle R P T=95^{\circ}$ and $\angle T S Q=75^{\circ}$, find $\angle S Q T$.
Solution: Using angle sum property for $\triangle \mathrm{PRT}$, we obtain $\angle \mathrm{PRT}+\angle \mathrm{RPT}+\angle \mathrm{PTR}=180^{\circ}$ $40^{\circ}+95^{\circ}+\angle \mathrm{PTR}=180^{\circ}$ $\angle P T R=180^{\circ}-135^{\circ}$ $\angle \mathrm{PTR}=45^{\circ}$ $\angle S T Q=\angle P T R=45^{\circ}$ (Vertically opposite angles) $\angle S T Q=45^{\circ}$ By using angle sum property for $\triangle S T Q$, we obtain $\angle S T Q+\angle S Q T+\angle Q S T=180^{\circ}$ $45^{\circ}+\an...
Read More →In the given figure, if $A B \| D E, \angle B A C=35^{\circ}$ and $\angle C D E=53^{\circ}$, find $\angle D C E$.
Solution: $A B \| D E$ and $A E$ is a transversal. $\angle B A C=\angle C E D$ (Alternate interior angles) $\therefore \angle C E D=35^{\circ}$ In $\triangle \mathrm{CDE}$ $\angle C D E+\angle C E D+\angle D C E=180^{\circ}$ (Angle sum property of a triangle) $53^{\circ}+35^{\circ}+\angle D C E=180^{\circ}$ $\angle \mathrm{DCE}=180^{\circ}-88^{\circ}$ $\angle \mathrm{DCE}=92^{\circ}$...
Read More →In the given figure, $\angle X=62^{\circ}, \angle X Y Z=54^{\circ} .$ If $Y O$ and $Z O$ are the bisectors of $\angle X Y Z$ and $\angle X Z Y$ respectively of $\triangle X Y Z$, find $\angle O Z Y$ and $\angle Y O Z$.
Solution: As the sum of all interior angles of a triangle is $180^{\circ}$, therefore, for $\triangle X Y Z$, $\angle X+\angle X Y Z+\angle X Z Y=180^{\circ}$ $62^{\circ}+54^{\circ}+\angle X Z Y=180^{\circ}$ $\angle X Z Y=180^{\circ}-116^{\circ}$ $\angle X Z Y=64^{\circ}$ $\angle O Z Y=\frac{64}{2}=32^{\circ}(O Z$ is the angle bisector of $\angle X Z Y)$ Similarly, $\angle O Y Z=\frac{54}{2}=27^{\circ}$ Using angle sum property for $\triangle O Y Z$, we obtain $\angle O Y Z+\angle Y O Z+\angle O...
Read More →In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.
Solution: It is given that, $\angle S P R=135^{\circ}$ and $\angle P Q T=110^{\circ}$ $\angle S P R+\angle Q P R=180^{\circ}($ Linear pair angles $)$ $\Rightarrow 135^{\circ}+\angle Q P R=180^{\circ}$ $\Rightarrow \angle Q P R=45^{\circ}$ Also, $\angle \mathrm{PQT}+\angle \mathrm{PQR}=180^{\circ}$ (Linear pair angles) $\Rightarrow 110^{\circ}+\angle \mathrm{PQR}=180^{\circ}$ $\Rightarrow \angle \mathrm{PQR}=70^{\circ}$ As the sum of all interior angles of a triangle is $180^{\circ}$, therefore, ...
Read More →In the given figure, if $A B \| C D, \angle A P Q=50^{\circ}$ and $\angle P R D=127^{\circ}$, find $x$ and $y$.
Solution: $\angle \mathrm{APR}=\angle \mathrm{PRD}$ (Alternate interior angles) $50^{\circ}+y=127^{\circ}$ $y=127^{\circ}-50^{\circ}$ $y=77^{\circ}$ Also, $\angle \mathrm{APQ}=\angle \mathrm{PQR}$ (Alternate interior angles) $50^{\circ}=x$ $\therefore x=50^{\circ}$ and $y=77^{\circ}$...
Read More →In the given figure, if $P Q \| S T, \angle P Q R=110^{\circ}$ and $\angle R S T=130^{\circ}$, find $\angle Q R S$.
Solution: Let us draw a line $X Y$ parallel to $S T$ and passing through point $R$. $\angle \mathrm{PQR}+\angle \mathrm{QRX}=180^{\circ}$ (Co-interior angles on the same side of transversal $\mathrm{QR}$ ) $\Rightarrow 110^{\circ}+\angle Q R X=180^{\circ}$ $\Rightarrow \angle Q R X=70^{\circ}$ Also, $\angle \mathrm{RST}+\angle S R Y=180^{\circ}$ (Co-interior angles on the same side of transversal SR) $130^{\circ}+\angle S R Y=180^{\circ}$ $\angle S R Y=50^{\circ}$ $X Y$ is a straight line. RQ an...
Read More →In the given figure, If $A B \| C D, E F \perp C D$ and $\angle G E D=126^{\circ}$, find $\angle A G E, \angle G E F$ and $\angle F G E$.
Solution: It is given that, $A B \| C D$ $\mathrm{EF} \perp \mathrm{CD}$ $\angle G E D=126^{\circ}$ $\Rightarrow \angle G E F+\angle F E D=126^{\circ}$ $\Rightarrow \angle G E F+90^{\circ}=126^{\circ}$ $\Rightarrow \angle G E F=36^{\circ}$ $\angle A G E$ and $\angle G E D$ are alternate interior angles. $\Rightarrow \angle A G E=\angle G E D=126^{\circ}$ However, $\angle A G E+\angle F G E=180^{\circ}$ (Linear pair) $\Rightarrow 126^{\circ}+\angle \mathrm{FGE}=180^{\circ}$ $\Rightarrow \angle \m...
Read More →In the given figure, if $A B\|C D, C D\| E F$ and $y: z=3: 7$, find $x$.
Solution: It is given that $A B \| C D$ and $C D \| E F$ $\therefore \mathrm{AB}\|\mathrm{CD}\| \mathrm{EF}$ (Lines parallel to the same line are parallel to each other) It can be observed that $x=z$ (Alternate interior angles) ...(1) It is given that $y: z=3: 7$ Let the common ratio between $y$ and $z$ be $a$. $\therefore y=3 a$ and $z=7 a$ Also, $x+y=180^{\circ}$ (Co-interior angles on the same side of the transversal) $z+y=180^{\circ}[U$ sing equation (1) $]$ $7 a+3 a=180^{\circ}$ $10 a=180^{...
Read More →(a) Tabulate the differences in the characteristics of states of matter. <br/><br/>(b) Comment upon the following: rigidity, compressibility, fluidity, fliling a gas container, shape, kinetic energy and density
Solution (a) Difference in the characteristics of 3 states of matter (b) Comment on: Rigidity. The tendency of a substance to retain/maintain their shape when subjected to outside force. Compressibility: The matter has intermolecular space. The external force applied on the matter can bring these particles closer. This property is called compressibility. Gases and liquids are compressible. Fluidity: The tendency of particles to flow is called fluidity. Liquids and gases flow. Filling of a gas co...
Read More →In the given figure, find the values of x and y and then show that AB || CD.
Solution: It can be observed that, $50^{\circ}+x=180^{\circ}($ Linear pair $)$ $x=130^{\circ} \ldots$(1) Also, $y=130^{\circ}$ (Vertically opposite angles) As $x$ and $y$ are alternate interior angles for lines $A B$ and $C D$ and also measures of these angles are equal to each other, therefore, line $A B \|$ CD....
Read More →It is given that $\angle X Y Z=64^{\circ}$ and $X Y$ is produced to point $P$. Draw a figure from the given information. If ray $Y Q$ bisects $\angle Z Y P$, find $\angle X Y Q$ and reflex $\angle Q Y P$.
Solution: It is given that line $Y Q$ bisects $\angle P Y Z$. Hence, $\angle Q Y P=\angle Z Y Q$ It can be observed that $P X$ is a line. Rays $Y Q$ and $Y Z$ stand on it. $\therefore \angle X Y Z+\angle Z Y Q+\angle Q Y P=180^{\circ}$ $\Rightarrow 64^{\circ}+2 \angle Q Y P=180^{\circ}$ $\Rightarrow 2 \angle Q Y P=180^{\circ}-64^{\circ}=116^{\circ}$ $\Rightarrow \angle Q Y P=58^{\circ}$ Also, $\angle Z Y Q=\angle Q Y P=58^{\circ}$ Reflex $\angle Q Y P=360^{\circ}-58^{\circ}=302^{\circ}$ $\angle ...
Read More →In the given figure, if $x+y=w+z$, then prove that $\mathrm{AOB}$ is a line.
Solution: It can be observed that, $x+y+z+w=360^{\circ}$ (Complete angle) It is given that, $x+y=z+w$ $\therefore x+y+x+y=360^{\circ}$ $2(x+y)=360^{\circ}$ $x+y=180^{\circ}$ Since $x$ and $y$ form a linear pair, $A O B$ is a line....
Read More →The mass per unit volume of a substance is called density. <br/><br/>(density = mass/volume). <br/><br/>Arrange the following in order of increasing density: air, exhaust from chimneys, honey, water, chalk, cotton and iron.
Solution Increasing density: air exhaust from chimneys cotton water honey chalk iron...
Read More →In the given figure, $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$, then prove that $\angle \mathrm{PQS}=\angle \mathrm{PRT}$.
Solution: In the given figure, ST is a straight line and ray QP stands on it. $\therefore \angle \mathrm{PQS}+\angle \mathrm{PQR}=180^{\circ}$ (Linear Pair) $\angle P Q R=180^{\circ}-\angle P Q S$(1) $\angle \mathrm{PRT}+\angle \mathrm{PRQ}=180^{\circ}($ Linear Pair $)$ $\angle P R Q=180^{\circ}-\angle P R T(2)$ It is given that $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$. Equating equations (1) and (2), we obtain $180^{\circ}-\angle \mathrm{PQS}=180^{\circ}-\angle \mathrm{PRT}$ $\angle \mathrm{PQ...
Read More →What are the characteristics of the particles of matter?
Solution The characteristics of the particles of matter are: Particles have intermolecular space. Particles have intermolecular force. Particles of matter are moving continuously....
Read More →In the given figure, lines $\mathrm{XY}$ and $\mathrm{MN}$ intersect at $\mathrm{O}$. If $\angle \mathrm{POY}=90^{\circ}$ and $a: b=2: 3$, find $c$.
Solution: Let the common ratio between $a$ and $b$ be $x$. $\therefore a=2 x$, and $b=3 x$ $X Y$ is a straight line, rays $O M$ and $O P$ stand on it. $\therefore \angle \mathrm{XOM}+\angle \mathrm{MOP}+\angle \mathrm{POY}=180^{\circ}$ $b+a+\angle \mathrm{POY}=180^{\circ}$ $3 x+2 x+90^{\circ}=180^{\circ}$ $5 x=90^{\circ}$ $x=18^{\circ}$ $a=2 x=2 \times 18=36^{\circ}$ $b=3 x=3 \times 18=54^{\circ}$ MN is a straight line. Ray OX stands on it. $\therefore b+c=180^{\circ}$ (Linear Pair) $54^{\circ}+...
Read More →A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?
Solution A diver is able to cut through water in a swimming pool. This shows that the particles of water have intermolecular space and has less force of attraction....
Read More →In the given figure, lines $\mathrm{AB}$ and $\mathrm{CD}$ intersect at $\mathrm{O}$. If $\angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ}$ and $\angle \mathrm{BOD}=40^{\circ}$, find $\angle \mathrm{BOE}$ and reflex $\angle \mathrm{COE}$.
Solution: $\mathrm{AB}$ is a straight line, rays $\mathrm{OC}$ and $\mathrm{OE}$ stand on it. $\therefore \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180^{\circ}$ $\Rightarrow(\angle \mathrm{AOC}+\angle \mathrm{BOE})+\angle \mathrm{COE}=180^{\circ}$ $\Rightarrow 70^{\circ}+\angle \mathrm{COE}=180^{\circ}$ $\Rightarrow \angle \mathrm{COE}=180^{\circ}-70^{\circ}=110^{\circ}$ Reflex $\angle \mathrm{COE}=360^{\circ}-110^{\circ}=250^{\circ}$ $\mathrm{CD}$ is a straight line, rays $\ma...
Read More →Give reasons for the following observation: <br/><br/>The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food you have to go close.
Solution The smell of hot sizzling food reaches severed meters away, as the particles of hot food have more kinetic energy and hence the rate of diffusion is more than the particles of cold food....
Read More →Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution: Yes. According to Euclid’s 5th postulate, when n line falls on l and m and if $\angle \mathrm{l}+\angle 2180^{\circ}$, then $\angle 3+\angle 4180^{\circ}$, producing line $/$ and $m$ further will meet in the side of $\angle 1$ and $\angle 2$ which is less than $180^{\circ}$. If $\angle \mathrm{l}+\angle 2=180^{\circ}$, then $\angle 3+\angle 4=180^{\circ}$ The lines / and $m$ neither meet at the side of $\angle 1$ and $\angle 2$ nor at the side of $\angle 3$ and $\angle 4$. This means t...
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