Q: Prove that the sum of angles of a triangle is 180.

\$\textbf{Step -1: Prove that the sum of angles of a triangle is }\mathbf{180^\circ.}\$ \$\text{Let a triangle }ABC\text{ and draw a line }l\text{ through }A\text{ and parallel to the base }BC.\$ \$\because BC\parallel l\text{ and }AB \text{ and }AC\text{ are transversals on it.} \$ \$\therefore \angle 4=\angle 2\ldots(i)\$ \$\textbf{(Alternate angles)}\$ \$\text{and } \angle 5=\angle 3\ldots(ii)\$ \$\textbf{(Alternate angles)}\$ \$\text{Adding equations }(i)\text{ and }(ii),\text{ we get}\$ \$\angle 4+\angle 5=\angle 2+\angle 3\$ \$\text{Add }\angle 1\text{ on both sides.}\$ \$\Rightarrow \angle 1+\angle 4+\angle 5=\angle 1+\angle 2+\angle 3\ldots(iii)\$ \$\angle 1+\angle 4+\angle 5=180^\circ\$ \$\mathbf{(\because\textbf{l is a straight line})}\$ \$\text{From }(iii),\$ \$\angle 1+\angle 2+\angle 3=180^\circ\$ \$\text{Hence, sum of angles of a triangle is }180^\circ.\$ \$\textbf{ Hence Proved.}\$

Q: In Figure, the side \$QR\$ of \$\triangle PQR\$ is produced to a point \$S\$. If the bisectors of \$\angle PQR\$ and \$\angle PRS\$ meet at point \$T\$, then prove that \$\angle QTR=\dfrac { 1 }{ 2 } \angle QPR\$.

Given, Bisectors of \$\angle PQR \$and \$\angle PRS\$ meet at point \$T\$.To prove: \$\angle QTR = \dfrac{1}{2}\angle QPR\$.Proof,\$\angle TRS = \angle TQR + \angle QTR\$ (Exterior angle of a triangle equals to the sum of the two interior angles.)\$\Rightarrow \angle QTR = \angle TRS - \angle TQR\$ --- (i)Also \$\angle SRP = \angle QPR + \angle PQR\$\$ 2\angle TRS = \angle QPR + 2\angle TQR\$\$\angle QPR = 2\angle TRS - 2\angle TQR\$ \$\Rightarrow \dfrac{1}{2}\angle QPR = \angle TRS - \angle TQR\$ --- (ii)Equating (i) and (ii),\$\therefore \angle QTR =\dfrac{1}{2}\angle QPR\$ \$[hence \, proved]\$

Q: In a \$\Delta\$ABC, right angled at B, \$AB = 24\$ cm, \$BC = 7\$ cm. Determine(i) \$\sin A, \cos A\$(ii) \$\sin C, \cos C\$

R.E.F imageIn \$\Delta \$ ABC, B is at right angle.Given, AB=24cmBC=7cmusing Pythagoras theorem \$AB^{2}+BC^{2}=AC^{2}\$\$\Rightarrow (24)^{2}+(7)^{2}=AC^{2}\$\$\Rightarrow AC=\sqrt{(24^{2})+(7)^{2}}=\sqrt{576+49}=\sqrt{625}\$\$\Rightarrow AC=25 CM\$(i)sin A=\$\frac{BC}{AC}\$\$ = \frac{7}{25}\$cos A=\$\frac{AB}{AC}\$\$=\frac{24}{25}\$(ii)sin C=\$\frac{AB}{AC}\$\$=\frac{24}{25}\$cos c=\$\frac{BC}{AC}\$=\$\frac{7}{25}\$

Q: Prove that in a right triangle,the square of the hypotenuse is equal to the sum of the square of the other sides.

Given:- A right triangle \$ABC\$ right angled at \$B\$.To prove:- \${AC}^{2} = {AB}^{2} + {BC}^{2}\$Construction:- Draw \$BD \perp AC\$Proof:-In \$\triangle{ABC}\$ and \$\triangle{ABD}\$\$\angle{ABC} = \angle{ADB} \quad \left( \text{Each } 90° \right)\$\$\angle{A} = \angle{A} \quad \left( \text{Common} \right)\$\$\therefore \triangle{ABC} \sim \triangle{ABD} \quad \left( \text{By AA} \right)\$\$\cfrac{AB}{AC} = \cfrac{AD}{AB} \quad \left( \because \text{Sides of similar triangles are proportional} \right)\$\$\Rightarrow {AB}^{2} = AD \cdot AC ..... \left( 1 \right)\$Similarly, in \$\triangle{ABC}\$ and \$\triangle{BCD}\$\$\angle{ABC} = \angle{BDC} \quad \left( \text{Each } 90° \right)\$\$\angle{C} = \angle{C} \quad \left( \text{Common} \right)\$\$\therefore \triangle{ABC} \sim \triangle{BCD} \quad \left( \text{By AA} \right)\$\$\therefore \cfrac{DC}{BC} = \cfrac{BC}{AC}\$\$\Rightarrow {BC}^{2} = DC \cdot AC ..... \left( 2 \right)\$Adding equation \$\left( 1 \right) \& \left( 2 \right)\$, we have\${AB}^{2} + {BC}^{2} = AD \cdot AC + DC. AC\$\$\Rightarrow {AB}^{2} + {BC}^{2} = AC \left( AD + DC \right)\$\$\Rightarrow {AB}^{2} + {BC}^{2} = {AC}^{2}\$Hence proved.

Q: The sides AB and AC of \$\Delta ABC\$ are produced to points E and D respectively. If Bisectors BO and CO of, \$\angle CBE\,\,and\,\,\angle BCD\$ respectively meet at point O, then prove that \$\angle BOC = {90^0} - \frac{1}{2}\angle BAC.\$

∠CBE = 180 - ∠ABC∠CBO = 1/2 ∠CBE (BO is the bisector of ∠CBE)∠CBO = 1/2 ( 180 - ∠ABC) 1/2 x 180 = 90 ∠CBO = 90 - 1/2 ∠ABC .............(1) 1/2 x ∠ABC = 1/2∠ABC∠BCD = 180 - ∠ACD∠BCO = 1/2 ∠BCD ( CO is the bisector os ∠BCD)∠BCO = 1/2 (180 - ∠ACD)∠BCO = 90 - 1/2∠ACD .............(2)∠BOC = 180 - (∠CBO + ∠BCO)∠BOC = 180 - (90 - 1/2∠ABC + 90 - 1/2∠ACD)∠BOC = 180 - 180 + 1/2∠ABC + 1/2∠ACD∠BOC = 1/2 (∠ABC + ∠ACD)∠BOC = 1/2 ( 180 - ∠BAC) (180 -∠BAC = ∠ABC + ∠ACD)∠BOC = 90 - 1/2∠BACHence proved

Q: Find the diagonal of a square whose side is \$10\ cm\$

The side of square is \$10\ cm\$In right angled \$\triangle ABD\$\$AB^{2} + AD^{2} = BD^{2}\$ [according to Pythagoras theorem]\$(10)^{2} + (10)^{2} = BD^{2}\$ \$100 + 100 = BD^{2}\$\$BD^{2} = 200\$\$BD = 10\sqrt {2}cm\$\$\therefore\$ Diagonal of square \$= 10 \sqrt {2}cm\$.

Q: ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB & a, b, c have usual meaning, then prove that(i) pc = ab(ii) \$\frac{1}{{{p^2}}}\, = \,\,\frac{1}{{{a^2}}}\, + \,\frac{1}{{{b^2}}}\$

(i) Area of \$\Delta ABC = {1 \over 2}BC \times AC\$ \$ = {1 \over 2} \times a \times b = {{ab} \over 2}\$3 \$area = {1 \over 2} \times CD \times AB\$ \${1 \over 2} \times p \times c = {{pc} \over 2}\$ \${{ab} \over 2} = {{pc} \over 2}\$ \$ab = pc......(1)\$ (ii) In \$\Delta ABC\$ \$A{B^2} = A{C^2} + B{C^2}\$ \${\left( {{{ab} \over p}} \right)^2} = {b^2} + {a^2}\$ \${1 \over {{p^2}}} = {{{b^2} + {a^2}} \over {{a^2} + {b^2}}}\$ \${1 \over {{p^2}}} = {1 \over {{a^2}}} + {1 \over {{b^2}}}\$

Q: In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle. Prove it

Given:- \$ABC\$ is a triangle\${AC}^{2} = {AB}^{2} + {BC}^{2}\$To prove:- \$\angle{B} = 90°\$Construction:- Construct a triangle \$PQR\$ right angled at \$Q\$ such that, \$PQ = AB\$ and \$QR = BC\$Proof:-In \$\triangle{PQR}\$\${PR}^{2} = {PQ}^{2} + {QR}^{2} \quad \left( \text{By pythagoras theorem} \right)\$\$\Rightarrow {PR}^{2} = {AB}^{2} + {BC}^{2}..... \left( 1 \right) \quad \left( \because AB = PQ \text{ and } QR = BC \right)\$\${AC}^{2} = {AB}^{2} + {BC}^{2} ..... \left( 2 \right) \quad \left( \text{Given} \right)\$From equation \$\left( 1 \right) \& \left( 2 \right)\$, we have\${AC}^{2} = {PR}^{2}\$\$\Rightarrow AC = PR ..... \left( 3 \right)\$Now, in \$\triangle{ABC}\$ and \$\triangle{PQR}\$\$AB = PQ\$\$BC = QR\$\$AC = PR \quad \left( \text{From } \left( 3 \right) \right)\$\$\therefore \triangle{ABC} \cong \triangle{PQR} \quad \left( \text{By SSS congruency} \right)\$Therefore, by C.P.C.T.,\$\angle{B} = \angle{Q}\$\$\because \angle{Q} = 90°\$\$\therefore \angle{B} = 90°\$Hence proved.

Q: A \$15\$m long ladder reached a window \$12\$m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Let AC be the ladder and A be the window.Given: \$AC=15\$m, \$AB=12\$m, \$CB=a\$mIn right angled triangle ACB,\$(Hypotenuse)^2=(Base)^2+[Perpendicular)^2\$ [By Pythagoras theorem]\$\Rightarrow (AC)^2=(CB)^2+(AB)^2\$\$\Rightarrow (15)^2=(a)^2+(12)^2\$\$\Rightarrow 225=a^2+144\$\$\Rightarrow a^2=225-144=81\$\$\Rightarrow a=\sqrt{81}=9\$cmThus, the distance of the foot of the ladder from the wall is \$9\$m.

Question 1

Prove that the sum of angles of a triangle is 180.

Accepted Answer

$\textbf{Step -1:&#10;Prove that the sum of angles of a triangle is }\mathbf{180^\circ.}$&#10;&#10;     &#10;          $\text{Let a triangle }ABC\text{ and draw a&#10;line }l\text{ through }A\text{ and parallel to the base }BC.$                                $\because BC\parallel l\text{ and }AB&#10;\text{ and }AC\text{ are transversals on it.} $&#10;&#10;     &#10;          $\therefore \angle 4=\angle&#10;2\ldots(i)$                 &#10;   $\textbf{(Alternate angles)}$&#10;&#10;   &#10;            $\text{and } \angle 5=\angle&#10;3\ldots(ii)$             &#10; $\textbf{(Alternate angles)}$&#10;&#10;   &#10;            $\text{Adding equations }(i)\text{&#10;and }(ii),\text{ we get}$&#10;&#10;   &#10;            $\angle 4+\angle 5=\angle 2+\angle&#10;3$&#10;&#10;   &#10;            $\text{Add }\angle 1\text{ on both&#10;sides.}$&#10;&#10;   &#10;            $\Rightarrow \angle 1+\angle&#10;4+\angle 5=\angle 1+\angle 2+\angle 3\ldots(iii)$&#10;&#10;   &#10;            $\angle 1+\angle 4+\angle&#10;5=180^\circ$           &#10;$\mathbf{(\because\textbf{l is a straight line})}$&#10;&#10;   &#10;            $\text{From }(iii),$&#10;&#10;   &#10;            $\angle 1+\angle 2+\angle&#10;3=180^\circ$&#10;&#10;   &#10;            $\text{Hence, sum of angles of a&#10;triangle is }180^\circ.$&#10;&#10;$\textbf{ Hence Proved.}$

Question 2

In Figure, the side $QR$ of $\triangle PQR$ is produced to a point $S$. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point $T$, then prove that $\angle QTR=\dfrac { 1 }{ 2 } \angle QPR$.

Accepted Answer

Given, Bisectors of $\angle PQR $and $\angle PRS$ meet at point $T$.To prove: $\angle QTR = \dfrac{1}{2}\angle QPR$.Proof,$\angle TRS = \angle TQR + \angle QTR$ (Exterior angle of a triangle equals to the sum of the two interior angles.)$\Rightarrow \angle QTR = \angle TRS - \angle TQR$ --- (i)Also $\angle SRP =  \angle QPR + \angle PQR$$ 2\angle TRS = \angle QPR + 2\angle TQR$$\angle QPR =  2\angle TRS - 2\angle TQR$ $\Rightarrow \dfrac{1}{2}\angle QPR =  \angle TRS - \angle TQR$ --- (ii)Equating (i) and (ii),$\therefore \angle QTR  =\dfrac{1}{2}\angle QPR$   $[hence \, proved]$

Question 3

In a $\Delta$ABC, right angled at B, $AB = 24$ cm, $BC = 7$ cm. Determine(i) $\sin A, \cos A$(ii) $\sin C, \cos C$

Accepted Answer

R.E.F imageIn $\Delta $ ABC, B is at right angle.Given, AB=24cmBC=7cmusing Pythagoras theorem $AB^{2}+BC^{2}=AC^{2}$$\Rightarrow (24)^{2}+(7)^{2}=AC^{2}$$\Rightarrow AC=\sqrt{(24^{2})+(7)^{2}}=\sqrt{576+49}=\sqrt{625}$$\Rightarrow AC=25 CM$(i)sin A=$\frac{BC}{AC}$$ = \frac{7}{25}$cos A=$\frac{AB}{AC}$$=\frac{24}{25}$(ii)sin C=$\frac{AB}{AC}$$=\frac{24}{25}$cos c=$\frac{BC}{AC}$=$\frac{7}{25}$

Question 4

Prove that in a right triangle,the square of the hypotenuse is equal to the sum of the square of the other sides.

Accepted Answer

Given:- A right triangle $ABC$ right angled at $B$.To prove:- ${AC}^{2} = {AB}^{2} + {BC}^{2}$Construction:- Draw $BD \perp AC$Proof:-In $\triangle{ABC}$ and $\triangle{ABD}$$\angle{ABC} = \angle{ADB} \quad \left( \text{Each } 90&#176; \right)$$\angle{A} = \angle{A} \quad \left( \text{Common} \right)$$\therefore \triangle{ABC} \sim \triangle{ABD} \quad \left( \text{By AA} \right)$$\cfrac{AB}{AC} = \cfrac{AD}{AB} \quad \left( \because \text{Sides of similar triangles are proportional} \right)$$\Rightarrow {AB}^{2} = AD \cdot AC ..... \left( 1 \right)$Similarly, in $\triangle{ABC}$ and $\triangle{BCD}$$\angle{ABC} = \angle{BDC} \quad \left( \text{Each } 90&#176; \right)$$\angle{C} = \angle{C} \quad \left( \text{Common} \right)$$\therefore \triangle{ABC} \sim \triangle{BCD} \quad \left( \text{By AA} \right)$$\therefore \cfrac{DC}{BC} = \cfrac{BC}{AC}$$\Rightarrow {BC}^{2} = DC \cdot AC ..... \left( 2 \right)$Adding equation $\left( 1 \right) \& \left( 2 \right)$, we have${AB}^{2} + {BC}^{2} = AD \cdot AC + DC. AC$$\Rightarrow {AB}^{2} + {BC}^{2} = AC \left( AD + DC \right)$$\Rightarrow {AB}^{2} + {BC}^{2} = {AC}^{2}$Hence proved.

Question 5

A tree is broken at a height of $5$m from the ground and its top touches the ground at a distance of $12$m from the base of the tree. Find the original height of the tree.

Accepted Answer

Let A'CB represents the tree before it broken at the point C and let the top A' touches the ground at A after it broke. Then $\Delta$ABC is a right angled triangle, right angled at B.AB$=12$m and BC$=5$mUsing Pythagoras theorem, In $\Delta$ABC$(AC)^2+(AB)^2+(BC)^2$$\Rightarrow (AC)^2=(12)^2+(5)^2$$\Rightarrow (AC)^2=144+25$$\Rightarrow (AC)^2=169$$\Rightarrow AC=13$mHence, the total height of the tree$=AC+CB=13+5=18$m.

Question 6

The sides AB and AC of $\Delta ABC$ are produced to points E and D respectively. If Bisectors BO and CO of, $\angle CBE\,\,and\,\,\angle BCD$ respectively meet at point O, then prove that $\angle BOC = {90^0} - \frac{1}{2}\angle BAC.$

Accepted Answer

&#8736;CBE = 180 - &#8736;ABC&#8736;CBO = 1/2 &#8736;CBE (BO is the bisector of &#8736;CBE)&#8736;CBO = 1/2 ( 180 - &#8736;ABC)                                                   1/2 x 180 = 90 &#8736;CBO = 90 - 1/2 &#8736;ABC    .............(1)                                   1/2 x &#8736;ABC = 1/2&#8736;ABC&#8736;BCD = 180 - &#8736;ACD&#8736;BCO = 1/2 &#8736;BCD     ( CO is the bisector os &#8736;BCD)&#8736;BCO = 1/2 (180 - &#8736;ACD)&#8736;BCO = 90 - 1/2&#8736;ACD    .............(2)&#8736;BOC = 180 - (&#8736;CBO + &#8736;BCO)&#8736;BOC = 180 - (90 - 1/2&#8736;ABC + 90 - 1/2&#8736;ACD)&#8736;BOC = 180 - 180 + 1/2&#8736;ABC + 1/2&#8736;ACD&#8736;BOC = 1/2 (&#8736;ABC + &#8736;ACD)&#8736;BOC = 1/2 ( 180 - &#8736;BAC)      (180 -&#8736;BAC = &#8736;ABC + &#8736;ACD)&#8736;BOC = 90 - 1/2&#8736;BACHence proved

Question 7

Find the diagonal of a square whose side is $10\ cm$

Accepted Answer

The side of square is $10\ cm$In right angled $\triangle ABD$$AB^{2} + AD^{2} = BD^{2}$ [according to Pythagoras theorem]$(10)^{2} + (10)^{2} = BD^{2}$ $100 + 100 = BD^{2}$$BD^{2} = 200$$BD = 10\sqrt {2}cm$$\therefore$ Diagonal of square $= 10 \sqrt {2}cm$.

Question 8

ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB & a, b, c have usual meaning, then prove that(i) pc = ab(ii) $\frac{1}{{{p^2}}}\, = \,\,\frac{1}{{{a^2}}}\, + \,\frac{1}{{{b^2}}}$

Accepted Answer

(i) Area of $\Delta ABC = {1 \over 2}BC&#10;\times AC$&#10;&#10;$ = {1 \over 2} \times a \times b = {{ab}&#10;\over 2}$3&#10;&#10;$area = {1 \over 2} \times CD \times AB$&#10;&#10;${1 \over 2} \times p \times c = {{pc}&#10;\over 2}$&#10;&#10;${{ab} \over 2} = {{pc} \over 2}$&#10;&#10;$ab = pc......(1)$&#10;&#10;(ii) In $\Delta ABC$&#10;&#10;$A{B^2} = A{C^2} + B{C^2}$&#10;&#10;${\left( {{{ab} \over p}} \right)^2} =&#10;{b^2} + {a^2}$&#10;&#10;${1 \over {{p^2}}} = {{{b^2} + {a^2}}&#10;\over {{a^2} + {b^2}}}$&#10;&#10;${1 \over {{p^2}}} = {1 \over {{a^2}}} +&#10;{1 \over {{b^2}}}$

Question 9

In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle. Prove it

Accepted Answer

Given:- $ABC$ is a triangle${AC}^{2} = {AB}^{2} + {BC}^{2}$To prove:- $\angle{B} = 90&#176;$Construction:- Construct a triangle $PQR$ right angled at $Q$ such that, $PQ = AB$ and $QR = BC$Proof:-In $\triangle{PQR}$${PR}^{2} = {PQ}^{2} + {QR}^{2} \quad \left( \text{By pythagoras theorem} \right)$$\Rightarrow {PR}^{2} = {AB}^{2} + {BC}^{2}..... \left( 1 \right) \quad \left( \because AB = PQ \text{ and } QR = BC \right)$${AC}^{2} = {AB}^{2} + {BC}^{2} ..... \left( 2 \right) \quad \left( \text{Given} \right)$From equation $\left( 1 \right) \& \left( 2 \right)$, we have${AC}^{2} = {PR}^{2}$$\Rightarrow AC = PR ..... \left( 3 \right)$Now, in $\triangle{ABC}$ and $\triangle{PQR}$$AB = PQ$$BC = QR$$AC = PR \quad \left( \text{From } \left( 3 \right) \right)$$\therefore \triangle{ABC} \cong \triangle{PQR} \quad \left( \text{By SSS congruency} \right)$Therefore, by C.P.C.T.,$\angle{B} = \angle{Q}$$\because \angle{Q} = 90&#176;$$\therefore \angle{B} = 90&#176;$Hence proved.

Question 10

A $15$m long ladder reached a window $12$m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Accepted Answer

Let AC be the ladder and A be the window.Given: $AC=15$m, $AB=12$m, $CB=a$mIn right angled triangle ACB,$(Hypotenuse)^2=(Base)^2+[Perpendicular)^2$ [By Pythagoras theorem]$\Rightarrow (AC)^2=(CB)^2+(AB)^2$$\Rightarrow (15)^2=(a)^2+(12)^2$$\Rightarrow 225=a^2+144$$\Rightarrow a^2=225-144=81$$\Rightarrow a=\sqrt{81}=9$cmThus, the distance of the foot of the ladder from the wall is $9$m.