AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that<br/><br/>(i) $\triangle \mathrm{DAP} \cong \triangle \mathrm{EBP}$<br/><br/>(ii) $A D=B E$
Solution: It is given that $\angle E P A=\angle D P B$ $\Rightarrow \angle E P A+\angle D P E=\angle D P B+\angle D P E$ $\Rightarrow \angle \mathrm{DPA}=\angle \mathrm{EPB}$ In $\Delta$ DAP and $\Delta$ EBP, $\angle \mathrm{DAP}=\angle \mathrm{EBP}($ Given $)$ $\mathrm{AP}=\mathrm{BP}(\mathrm{P}$ is mid-point of $\mathrm{AB})$ $\angle \mathrm{DPA}=\angle \mathrm{EPB}($ From above $)$ $\therefore \triangle \mathrm{DAP} \cong \triangle \mathrm{EBP}(\mathrm{ASA}$ congruence rule $)$ $\therefore \m...
Read More →In the given figure, $A C=A E, A B=A D$ and $\angle B A D=\angle E A C$. Show that $B C=D E$.
Solution: It is given that $\angle B A D=\angle E A C$ $\angle B A D+\angle D A C=\angle E A C+\angle D A C$ $\angle B A C=\angle D A E$ In $\triangle B A C$ and $\triangle D A E$, $\mathrm{AB}=\mathrm{AD}$ (Given) $\angle B A C=\angle D A E($ Proved above $)$ $\mathrm{AC}=\mathrm{AE}$ (Given) $\therefore \triangle \mathrm{BAC} \cong \triangle \mathrm{DAE}($ By $S A S$ congruence rule $)$ $\therefore \mathrm{BC}=\mathrm{DE}(\mathrm{By} \mathrm{CPCT})$...
Read More →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...
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