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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,...

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]...

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...

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) $\Rightarrow \frac{A L}{D M}=\frac{A O}{D O}$ ...

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...

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...

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. ...

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) $\Rightarrow \frac{A N}{N D}=\frac{A L}{L C}$ .....

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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