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Tick the correct answer and justify :

[question] Question. Tick the correct answer and justify : In $\Delta \mathrm{ABC}, \mathrm{AB}=6 \sqrt{3} \mathrm{~cm}, \mathrm{AC}=12 \mathrm{~cm}$ and $\mathrm{BC}=6 \mathrm{~cm}$. The angle $\mathrm{B}$ is : (1) 120° (2) 60° (3) 90° (4) 45° [/question] [solution] Solution: $\mathrm{AB}^{2}=(6 \sqrt{3})^{2}=108$ $\mathrm{BC}^{2}=6^{2}=36$ $A C^{2}=12^{2}=144$ So, $\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$ $\Delta \mathrm{ABC}$ is right $\Delta$, right angled at $\mathrm{B}$ $\angle \ma...

In an equailateral triangle,

[question] Question. In an equailateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. [/question] [solution] Solution: Altitude of equilateral $\Delta=\frac{\sqrt{3}}{2}$ side $h=\frac{\sqrt{3}}{2} a$ $h^{2}=\frac{3}{4} a^{2}$ $4 h^{2}=3 a^{2}$ [/solution]...

In an equailateral triangle ABC,

[question] Question. In an equailateral triangle $\mathrm{ABC}, \mathrm{D}$ is a point on side $\mathrm{BC}$ such that $\mathrm{BD}=\frac{1}{3} \mathrm{BC}$. Prove that $9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}$. [/question] [solution] Solution: AB = BC = CA = a (say) $B D=\frac{1}{3} B C=\frac{1}{3} a$ $\Rightarrow C D=\frac{2}{3} B C=\frac{2}{3} a$ $\mathrm{AE} \perp \mathrm{BC}$ $\Rightarrow B E=E C=\frac{1}{2} a$ $\mathrm{DE}=\frac{1}{2} \mathrm{a}-\frac{1}{3} \mathrm{a}=\frac{1}{6} \mathrm{a}$ $...

The perpendicular from $A$ on side $B C$ of a $\triangle A B C$ intersects $B C$ at $D$ such that $D B=3 C D(s e e$ figure).

[question] Question. The perpendicular from $A$ on side $B C$ of a $\Delta A B C$ intersects $B C$ at $D$ such that $D B=3 C D(s e e$ figure). Prove that $2 \mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}$. [/question] [solution] Solution: DB = 3 CD $\Rightarrow \mathrm{CD}=\frac{1}{4} \mathrm{BC}$ ...(1) and $D B=\frac{3}{4} B C$ In $\triangle \mathrm{ABD}, \quad \mathrm{AB}^{2}=\mathrm{DB}^{2}+\mathrm{AD}^{2}$ In $\Delta \mathrm{ACD}, \quad \mathrm{AC}^{2}=\mathrm{CD}^{2}+\mathrm{AD}^{2}$ Su...

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

[question] Question. $\mathrm{D}$ and $\mathrm{E}$ are points on the sides $\mathrm{CA}$ and $\mathrm{CB}$ respectively of a triangle $\mathrm{ABC}$ right angled at $\mathrm{C}$. Prove that $\mathrm{AE}^{2}+\mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{DE}^{2}$. [/question] [solution] Solution: In right angled $\triangle \mathrm{ACE}$, $\mathrm{AE}^{2}=\mathrm{CA}^{2}+\mathrm{CE}^{2}$ ...(1) and in right angled $\triangle B C D$, $\mathrm{BD}^{2}=\mathrm{BC}^{2}+\mathrm{CD}^{2}$ ...(2) Adding (1) and ...

Two poles of height 6 m and 11 m stand on a plane ground.

[question] Question. Two poles of height 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. [/question] [solution] Solution: Let AD and BE be two poles of height 6 m and 11 m and AB = 12 m In $\triangle \mathrm{DEC}$, by pythagoras theorem $\mathrm{DE}^{2}=\mathrm{CD}^{2}+\mathrm{CE}^{2}$ $\mathrm{DE}^{2}=12^{2}+5^{2}(\mathrm{DC}=\mathrm{AB}=12 \mathrm{~m})$ $\mathrm{DE}=\sqrt{144+25}=\sqrt{169}=13 \mathrm{~m}$ Thus,...

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.

[question] Question. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attachedto the other end. How far from the base of the pole should the stake be driven so that the wirewill be taut? [/question] [solution] Solution: Let AB be the vertical pole of 18 m and AC be the wire of 24 m. The $\Delta \mathrm{ABC}$, by pythagoras theorem $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$ $24^{2}=18^{2}+\mathrm{BC}^{2}$ $\mathrm{BC}^{2}=252$ $\mathrm{BC}=6 \sqrt{7} \math...

A ladder 10 m long reaches a window 8 m above the ground.

[question] Question. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall [/question] [solution] Solution: Let AC = x metres be the distance of the foot of the ladder from the base of the wall. AB = 8 m (Height of window) BC = 10 m (length of ladder) Now, $\quad x^{2}+(8)^{2}=(10)^{2}$ $\Rightarrow x^{2}=100-64=36 \Rightarrow x=6$, i.e., $A C=6 m$ [/solution]...

In figure, O is a point in the interior of a triangle ABC,

[question] Question. In figure, $\mathrm{O}$ is a point in the interior of a triangle $\mathrm{ABC}, \mathrm{OD} \perp \mathrm{BC}, \mathrm{OE} \perp \mathrm{AC}$ and $\mathrm{OF} \perp \mathrm{AB}$. Show that (i) $\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}$ $=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}$ (ii)$\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}=\mathrm{AE}^{2}+\mathrm{CD}^{2}+\mathrm{BF}^{2}$. [/question] [solution] Solution: (...

Prove that the sum of the squares of the sides

[question] Question. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [/question] [solution] Solution: ABCD is a rhombus in which AB = BC = CD = DA = a (say). Its diagonals AC and BD are right bisectors of each other at O. In $\Delta \mathrm{OAB}, \angle \mathrm{AOB}=90^{\circ}$ $\mathrm{OA}=\frac{1}{2} \mathrm{AC}$ and $\mathrm{OB}=\frac{1}{2} \mathrm{BD}$ By pythagoras theorem, we have $\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^...

ABC is an equilateral triangle of side 2a.

[question] Question. ABC is an equilateral triangle of side 2a. Find each of its altitudes. [/question] [solution] Solution: Altitude of equilateral triangle $=\frac{\sqrt{3}}{2} \times$ Side $=\frac{\sqrt{3}}{2} \times 2 a=\sqrt{3} a$ [/solution]...

ABC is an isosceles triangle with AC = BC.

[question] Question. $\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AC}=\mathrm{BC}$. If $\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$, prove that $\mathrm{ABC}$ is a right triangle. [/question] [solution] Solution: $\mathrm{As}, \mathrm{AB}^{2}=2 \mathrm{AC}^{2}$ $\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}$ $=\mathrm{AC}^{2}+\mathrm{BC}^{2} \quad[\because \mathrm{AC}=\mathrm{BC}]$ As it satisfy the pythagoran triplet So, $\triangle \mathrm{ABC}$ is right triangle, right angled at $\angle \mat...

ABC is an isosceles triangle right angled at C.

[question] Question. $\mathrm{ABC}$ is an isosceles triangle right angled at $\mathrm{C}$. Prove that $\mathrm{AB}^{2}=2 \mathrm{AC}^{2} .$ [/question] [solution] Solution: In $\triangle \mathrm{ABC}, \angle \mathrm{ACB}=90^{\circ}$. We are given that $\triangle \mathrm{ABC}$ is an isosceles triangle. $\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}=45^{\circ}$ $\Rightarrow \mathrm{AC}=\mathrm{BC}$ ... (1) By pythagoras theorem, we have $\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}$ $=\mathrm...

In figure, ABD is a right triangle right angled at A and AC

[question] Question. In figure, $A B D$ is a right triangle right angled at $A$ and $A C \perp B D$. Show that (i) $\mathrm{AB}^{2}=\mathrm{BC} \cdot \mathrm{BD}$ (ii) $\mathrm{AC}^{2}=\mathrm{BC} . \mathrm{DC}$ (iii) $\mathrm{AD}^{2}=\mathrm{BD} \cdot \mathrm{CD}$ [/question] [solution] Solution: In the given figure, we have $\Delta \mathrm{ABC} \sim \Delta \mathrm{DAC} \sim \Delta \mathrm{DBA}$ (i) $\triangle \mathrm{ABC} \sim \Delta \mathrm{DBA}$ $\Rightarrow \frac{\operatorname{ar}(\Delta A ...

PQR is a triangle right angled at P and M is a point on QR

[question] Question. $\mathrm{PQR}$ is a triangle right angled at $\mathrm{P}$ and $\mathrm{M}$ is a point on $\mathrm{QR}$ such that $\mathrm{PM} \perp \mathrm{QR}$. Show that $\mathrm{PM}^{2}=\mathrm{QM} \times \mathrm{MR}$. [/question] [solution] Solution: $\angle 1+\angle 2=\angle 2+\angle 4$ $\left(\mathrm{Each}=90^{\circ}\right)$ $\Rightarrow \angle 1=\angle 4$ Similarly, $\angle 2=\angle 3$. Now, this gives $\Delta \mathrm{QPM} \sim \Delta \mathrm{PRM}$ (AA similarity) $\Rightarrow \frac{...

Sides of two similar triangles are in the ratio 4 : 9.

[question] Question. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (1) 2 : 3 (2) 4 : 9 (3) 81 : 16 (4) 16 : 81 [/question] [solution] Solution: $\frac{\text { area of } 1^{\text {st }} \Delta}{\text { area of } 2^{\text {nd }} \Delta}=\left(\frac{4}{9}\right)^{2}=\frac{16}{81}$ [/solution]...

ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

[question] Question. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is (1) 2 : 1 (2) 1 : 2 (3) 4 : 1 (4) 1 : 4 [/question] [solution] Solution: $(\because B C=2 B D)$ Since, both are equilateral triangles $\Delta \mathrm{ABC} \sim \Delta \mathrm{EBD}$ $\frac{\operatorname{ar} \Delta \mathrm{ABC}}{\operatorname{ar} \Delta \mathrm{BDE}}=\left(\frac{\mathrm{BC}}{\mathrm{BD}}\right)^{2}=\left(\frac{2}{1}\right)^{2}=4: 1$ [/so...

Prove that the area of an equilateral triangle described on one side

[question] Question. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. [/question] [solution] Solution: ABCD is a square having sides of length = a. Then the diagonal $\mathrm{BD}=\mathrm{a} \sqrt{2}$. We construct equilateral $\Delta \mathrm{s}$ PAB and QBD $\Rightarrow \Delta \mathrm{PAB} \sim \Delta \mathrm{QBD}$ (Equilateral triangles are similar) $\Rightarrow \frac{\operat...

Prove that the ratio of the areas of two similar triangles is equal

[question] Question. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. [/question] [solution] Solution: In figure, $\mathrm{AD}$ is a median of $\triangle \mathrm{ABC}$ and $\mathrm{PM}$ is a median of $\triangle \mathrm{PQR}$. Here, $\mathrm{D}$ is mid-point of $\mathrm{BC}$ and $M$ is mid-point of $Q R$. Now, we have $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$. $\Rightarrow \angle \mathrm{B}=\angle \mathrm{Q}...

D, E and F are respectively the mid-points of sides AB,

[question] Question. $\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ of $\triangle \mathrm{ABC}$. Find the ratio of the areas of $\Delta \mathrm{DEF}$ and $\triangle \mathrm{ABC}$. [/question] [solution] Solution: $\mathrm{DF}=\frac{1}{2} \mathrm{BC}, \mathrm{DE}=\frac{1}{2} \mathrm{AC}, \mathrm{EF}=\frac{1}{2} \mathrm{AB}$ [By midpoint theorem] So, $\frac{D F}{B C}=\frac{D E}{A C}=\frac{E F}{A B}=\frac{1}{2}$ $\ther...

If the areas of two similar triangles are equal,

[question] Question. If the areas of two similar triangles are equal, prove that they are congruent. [/question] [solution] Solution: Let $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ and area $(\Delta \mathrm{ABC}) \quad=$ area $(\Delta \mathrm{PQR}) \quad($ Given $)$ i.e., $\frac{\operatorname{area}(\Delta A B C)}{\operatorname{area}(\Delta P O R)}=1$ $\Rightarrow \frac{A B^{2}}{P Q^{2}}=\frac{B C^{2}}{Q R^{2}}=\frac{C A^{2}}{P R^{2}}=1$ $\Rightarrow \mathrm{AB}=\mathrm{PQ}, \mathrm{BC}=\math...

In figure, ABC and DBC are two triangles on the same base BC.

[question] Question. In figure, $\mathrm{ABC}$ and $\mathrm{DBC}$ are two triangles on the same base $\mathrm{BC}$. If $\mathrm{AD}$ intersects $\mathrm{BC}$ at $\mathrm{O}$, show that $\frac{\operatorname{ar}(A B C)}{\operatorname{ar}(D B C)}=\frac{A O}{D O}$.' [/question] [solution] Solution: Draw $\mathrm{AL} \perp \mathrm{BC}$ and $\mathrm{DM} \perp \mathrm{BC}$ (see figure) $\Delta \mathrm{OLA} \sim \Delta \mathrm{OMD}$(AA similarity criterion) $\Rightarrow \frac{A L}{D M}=\frac{A O}{D O}$ ...

Diagonals of trapezium ABCD with AB

[question] Question. Diagonals of trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at the point $\mathrm{O}$. If $\mathrm{AB}=2 \mathrm{CD}$, find the ratio of the areas of triangles $\mathrm{AOB}$ and COD. [/question] [solution] Solution: In $\Delta \mathrm{AOB}$ and $\Delta \mathrm{COD}$, $\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate interior angles) $\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate interior angles) $\therefore$ By AA, similarity $\Del...

Let $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ and their areas be $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively.

[question] Question. Let $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ and their areas be $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. If $\mathrm{EF}=15.4 \mathrm{~cm}$, find BC. [/question] [solution] Solution: $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}($ Given $)$ $\Rightarrow \frac{\operatorname{ar}(A B C)}{\operatorname{ar}(D E F)}=\frac{B C^{2}}{E F^{2}}$ (By theorem 6.7) $\Rightarrow \frac{64}{121}=\frac{B C^{2}}{E F^{2}} \quad \Rightarrow\left\{\frac{B C}{E F}...

A vertical stick of length 6 m casts a shadow 4 m long

[question] Question. A vertical stick of length $6 \mathrm{~m}$ casts a shadow $4 \mathrm{~m}$ long on the ground and at the same time a tower casts a shadow $28 \mathrm{~m}$ long. Find the height of the tower. [/question] [solution] Solution: $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ $\therefore \quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ $\frac{6}{x}=\frac{4}{28}$ $\Rightarrow x=42 \mathrm{~m}$ [/solution]...

$\mathrm{D}$ is a point on the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$.

[question] Question. $\mathrm{D}$ is a point on the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$. Show that $\mathrm{CA}^{2}=$ CB. CD. [/question] [solution] Solution: For $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DAC}$, We have $\angle \mathrm{BAC}=\angle \mathrm{ADC} \quad($ Given $)$ and $\angle \mathrm{ACB}=\angle \mathrm{DCA} \quad($ Each $=\angle \mathrm{C})$ $\Rightarrow \Delta \mathrm{ABC} \sim \Delta \mathrm{DAC} \quad$ (AA si...

In figure, $\mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles,

[question] Question. In figure, $\mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively. Prove that: (i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$ (ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$ [/question] [solution] Solution: (i) In $\Delta \mathrm{ABC}$ and $\Delta \mathrm{AMP}$ $\angle \mathrm{CAB}=\angle \mathrm{PAM}($ common $)$ $\angle \mathrm{ABC}=\angle \mathrm{AMP}=90^{\circ}$ $\therefore$ B...

$\mathrm{E}$ is a point on the side $\mathrm{AD}$ produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$.

[question] Question. $\mathrm{E}$ is a point on the side $\mathrm{AD}$ produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$. Show that $\triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$ [/question] [solution] Solution: In $\Delta \mathrm{ABE}$ and $\Delta \mathrm{CFB}$, $\angle \mathrm{EAB}=\angle \mathrm{BCF}$ (opp. angles of parallelogram) $\angle \mathrm{AEB}=\angle \mathrm{CBF}$ (Alternate interior angles, $\mathrm{As} \mathrm{AE} \| \mathr...

In figure, altitudes $\mathrm{AD}$ and $\mathrm{CE}$ of $\triangle \mathrm{ABC}$ intersect each other at the point $\mathrm{P}$.

[question] Question. In figure, altitudes $\mathrm{AD}$ and $\mathrm{CE}$ of $\triangle \mathrm{ABC}$ intersect each other at the point $\mathrm{P}$. Show that : (i) $\triangle \mathrm{AEP} \sim \Delta \mathrm{CDP}$ (ii) $\triangle \mathrm{ABD} \sim \Delta \mathrm{CBE}$ (iii) $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$ (iv) $\Delta \mathrm{PDC} \sim \Delta \mathrm{BEC}$ [/question] [solution] Solution: (i) In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{CDP}$, $\angle \mathrm{APE}=\a...

In figure, if $\triangle \mathrm{ABE}$

[question] Question. In figure, if $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$, show that $\triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}$. [/question] [solution] Solution: In figure, $\Delta \mathrm{ABE} \cong \triangle \mathrm{ACD}$(Given) $\Rightarrow \mathrm{AB}=\mathrm{AC}$ and $\mathrm{AE}=\mathrm{AD} \quad(\mathrm{CPCT})$ $\Rightarrow \frac{A B}{A C}=1$ and $\frac{A D}{A E}=1$ $\Rightarrow \frac{A B}{A C}=\frac{A D}{A E} \quad(E a c h=1)$ Now, in $\triangle \mathrm{ADE}$ and...

$\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$.

[question] Question. $\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$. Show that $\triangle \mathrm{RPQ} \sim$ $\triangle \mathrm{RTS}$. [/question] [solution] Solution: In figure, We have RPQ and RTS in which $\angle \mathrm{RPQ}=\angle \mathrm{RTS}$ (Given) $\angle \mathrm{PRQ}=\angle \mathrm{SRT}(\mathrm{Each}=\angle \mathrm{R})$ Then by AA similarity criterion, we have $\Delta \ma...

In figure, $\frac{Q R}{O S}=\frac{Q T}{P R}$ and $\angle 1=\angle 2$.

[question] Question. In figure, $\frac{Q R}{O S}=\frac{Q T}{P R}$ and $\angle 1=\angle 2$. Show that $\Delta P Q S \sim \Delta T Q R .$ [/question] [solution] Solution: In figure, $\angle 1=\angle 2$ (Given) $\Rightarrow \mathrm{PQ}=\mathrm{PR}$ (Sides opposite to equal angles of $\Delta \mathrm{PQR}$ ) We are given that $\frac{Q R}{O S}=\frac{Q T}{P R}$ $\Rightarrow \frac{\mathrm{QR}}{\mathrm{OS}}=\frac{\mathrm{QT}}{\mathrm{PQ}} \quad(\because \mathrm{PQ}=\mathrm{PR}$ proved $)$ $\Rightarrow \f...

Diagonals AC and BD of a trapezium ABCD with AB

[question] Question. Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{O A}{O C}=\frac{O B}{O D}$. [/question] [solution] Solution: In figure, $\mathrm{AB} \| \mathrm{DC}$ $\Rightarrow \angle 1=\angle 3, \angle 2=\angle 4$ (Alternate interior angles) Also $\angle \mathrm{DOC}=\angle \mathrm{BOA}$ (Vertically opposite angles) $\Rightarrow \...

In figure, $\Delta \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$

[question] Question. In figure, $\triangle \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$ and $\angle \mathrm{CDO}=70^{\circ}$. Find $\angle \mathrm{DOC}, \angle \mathrm{DCO}$ and $\angle \mathrm{OAB}$. [/question] [solution] Solution: From figure, $\angle D O C+125^{\circ}=180^{\circ}$ $\Rightarrow \angle \mathrm{DOC}=180^{\circ}-125^{\circ}=55^{\circ}$ $\angle \mathrm{DCO}+\angle \mathrm{CDO}+\angle \mathrm{DOC}=180^{\circ}$ (Sum of three angles of $\Delta \mathrm{ODC}...

State which pairs of triangles in figure,

[question] Question. State which pairs of triangles in figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form: [/question] [solution] Solution: (i) Yes. $\angle \mathrm{A}=\angle \mathrm{P}=60^{\circ}, \angle \mathrm{B}=\angle \mathrm{Q}=80^{\circ}$, $\angle \mathrm{C}=\angle \mathrm{R}=40^{\circ}$ Therefore, $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$. By AAA similarity criterion (ii) Yes. ...

The diagonals of a quadrilateral ABCD intersect

[question] Question. The diagonals of a quadrilateral $\mathrm{ABCD}$ intersect each other at the point $\mathrm{O}$ such that $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$. Show that $\mathrm{ABCD}$ is a trapezium. [/question] [solution] Solution: In figure $\frac{A O}{B O}=\frac{C O}{D O}$ $\Rightarrow \frac{A O}{O C}=\frac{B O}{O D} \quad \ldots$ (1) (given) Through $\mathrm{O}$, we draw $\mathrm{OE} \| \mathrm{BA}$ OE meets AD at E. From $\Delta \mathrm{DAB}$, $\mathrm{EO...

ABCD is a trapezium in which AB

[question] Question. $\mathrm{ABCD}$ is a trapezium in which $\mathrm{AB} \| \mathrm{DC}$ and its diagonals intersect each other at the point $\mathrm{O}$. Show that $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$. [/question] [solution] Solution: We draw EOF $\| \mathrm{AB}($ also $\| \mathrm{CD})$ (see figure) In $\Delta \mathrm{ACD}, \quad \mathrm{OE} \| \mathrm{CD}$ $\Rightarrow \frac{A E}{E D}=\frac{A O}{O C} \ldots(1)$ In $\triangle \mathrm{ABD}, \mathrm{OE} \| \mathrm{BA...

Using Theorem 6.1,

[question] Question. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. [/question] [solution] Solution: In $\triangle \mathrm{ABC}, \mathrm{D}$ is mid point of $\mathrm{AB}$ (see figure) i.e., $\frac{A D}{D B}=1$ Straight line $\ell \| \mathrm{BC}$. Line $\ell$ is drawn through $\mathrm{D}$ and it meets $\mathrm{AC}$ at $\mathrm{E}$. By Basic Proportionality Theorem $\frac{A D}{D B}=\frac{A E}{E C} \Rightar...

In figure, A, B and C are points on OP,

[question] Question. In figure, $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are points on $\mathrm{OP}, \mathrm{OQ}$ and $\mathrm{OR}$ respectively such that $\mathrm{AB} \| \mathrm{PQ}$ and $\mathrm{AC}$ $\| \mathrm{PR}$. Show that $\mathrm{BC} \| \mathrm{QR}$. [/question] [solution] Solution: In $\triangle \mathrm{POO}$ AB $\| P Q$ (given) $\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OA}}{\mathrm{AP}} \ldots$ (i) (Basic Proportionality Theorem) In $\triangle \mathrm{POR}$ $\mathrm{AC} \| \math...

In figure, DE

[question] Question. In figure, $D E \| O Q$ and $D F \| O R$. Show that $E F \| Q R$. [/question] [solution] Solution: In figure, $D E \| O Q$ and DF $\| O R$, then by Basic Proportionality Theorem, We have $\quad \frac{P E}{E Q}=\frac{P D}{D O}....(1) and $\quad \frac{P F}{F R}=\frac{P D}{D O}$\ldots(2)$ From $(1)$ and $(2), \quad \frac{P E}{E Q}=\frac{P F}{F R}$ Now, in $\triangle \mathrm{PQR}$, we have proved that $\Rightarrow \frac{P E}{E Q}=\frac{P F}{F R}$ $\mathrm{EF} \| \mathrm{QR}$ (By...

In figure, $\mathrm{DE} \| \mathrm{AC}$ and $\mathrm{DF} \| \mathrm{AE}$. Prove that $\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$.

[question] Question. In figure, $D E \| A C$ and $D F \| A E$. Prove that $\frac{B F}{F E}=\frac{B E}{E C}$. [/question] [solution] Solution: In $\triangle \mathrm{ABE}$ $\mathrm{DF} \| \mathrm{AE}$ (Given) $\frac{B D}{D A}=\frac{B F}{F E} \ldots$ (i) (Basic Proportionality Theorem) In $\triangle \mathrm{ABC}$, $D E \| A C$(Given) $\frac{B D}{D A}=\frac{B E}{E C}$ (ii) (Basic Proportionality Theorem) From (i) and (ii), we get $\frac{B F}{F E}=\frac{B E}{E C} \quad$ Hence proved. [/solution]...

In figure, if $\mathrm{LM} \| \mathrm{CB}$ and $\mathrm{LN} \| \mathrm{CD}$, prove that $\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$.

[question] Question. In figure, if $\mathrm{LM} \| \mathrm{CB}$ and $\mathrm{LN} \| \mathrm{CD}$, prove that $\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$. [/question] [solution] Solution: In $\triangle \mathrm{ACB}$ (see figure), $\mathrm{LM} \| \mathrm{CB}$ (Given) $\Rightarrow \frac{A M}{M B}=\frac{A L}{L C}$ ...(1) (Basic Proportionality Theorem) In $\triangle \mathrm{ACD}$ (see figure), $\mathrm{LN} \| \mathrm{CD}$ (Given) $\Rightarrow \frac{A N}{N D}=\frac{A L}{L C}$ .....

$E$ and $F$ are points on the sides $P Q$ and $P R$ respectively of a $\triangle P Q R$.

[question] Question. $\mathrm{E}$ and $\mathrm{F}$ are points on the sides $\mathrm{PQ}$ and $\mathrm{PR}$ respectively of a $\Delta \mathrm{PQR}$. For each of the following cases, State whether EF $\| \mathrm{QR}$ : (i) $\mathrm{PE}=3.9 \mathrm{~cm}, \mathrm{EQ}=3 \mathrm{~cm}, \mathrm{PF}=3.6 \mathrm{~cm}$ and $\mathrm{FR}=2.4 \mathrm{~cm}$. (ii) $\mathrm{PE}=4 \mathrm{~cm}, \mathrm{QE}=4.5 \mathrm{~cm}, \mathrm{PF}=8 \mathrm{~cm}$ and $\mathrm{RF}=9 \mathrm{~cm}$. (iii) $\mathrm{PQ}=1.28 \mat...

In figure,

[question] Question. In figure, (i) and (ii), $\mathrm{DE} \| \mathrm{BC}$. Find $\mathrm{EC}$ in (i) and $\mathrm{AD}$ in (ii). [/question] [solution] Solution: (i) In figure, (i) DE $\| \mathrm{BC}$ (Given) $\Rightarrow \frac{A D}{D B}=\frac{A E}{E C}$ (By Basic Proportionality Theorem) $\Rightarrow \frac{1.5}{3}=\frac{1}{E C}$ $\{\because \mathrm{AD}=1.5 \mathrm{~cm}, \mathrm{DB}=3 \mathrm{~cm}$ and $\mathrm{AE}=1 \mathrm{~cm}\}$ $\Rightarrow \mathrm{EC}=\frac{3}{1.5}=2 \mathrm{~cm}$ (ii) In ...

State whether the following quadrilaterals are similar or not :

[question] Question. State whether the following quadrilaterals are similar or not : [/question] [solution] Solution: The two quadrilateral in figure are not similar because their corresponding angles are not equal. [/solution]...

Give two different examples of pair of

[question] Question. Give two different examples of pair of (i) Similar figures. (ii) Non-similar figures. [/question] [solution] Solution: (i) 1. Pair of equilateral triangles are similar figures. 2. Pair of squares are similar figures. (ii) 1. One equilateral triangle and one isosceles triangle are non-similar. 2. Square and rectangle are non-similar. [/solution]...

Fill in the blanks using the correct word given in brackets :

[question] Question. Fill in the blanks using the correct word given in brackets : (i) All circles are ______. (congruent, similar) (ii) All squares are _______. (similar, congruent) (iii) All _______ triangles are similar.(isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if(a) their corresponding angles are _______ and (b) their corresponding sides are _______. (equal, proportional) [/question] [solution] Solution: (i) All circles are similar. (ii) All squares ...

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