Solution

\( AB + BC > \color{red} AC \)

The sum of the lengths of any two sides of a triangle is greater than the length of third side.

Answer \( {\color{orange} (iii)} \ \color{red} AC \)

Answer \( {\color{orange} (i)} \ \color{red} 2:1 \)

The centroid divides each median in the ratio \( \color{red} 2:1 \)

Solution

\begin{align*} \text{Pythagoras theorem} \\ \color{magenta} \text{(Side 1)}^2 + \text{(Side 2)}^2 &= \color{magenta} \text{(Hypotenuse)}^2 \\ 8^2 + x^2 &= 17^2 \\ 64 + x^2 &= 289 \\ x^2 &= 225 \\ x &= 15 \\ \end{align*}

Answer \( {\color{orange} (d)} \ \color{red} 15 \text{ cm} \)

Answer \( {\color{orange} (i)} \ \color{red} \text{Orthocentre} \)

The point of concurrence of the altitudes of a triangle is called the \(\color{red} orthocentre \).

Solution

The adjacent interior angle to an exterior angle is the opposite interior angle.

Answer \( {\color{orange} (iii)} \ \color{red} \angle PRQ \)

Solution

The point of concurrence of the angle bisectors of a triangle is called the incentre of the triangle. Since the angle bisectors of a triangle will all lie in the interior of the triangle, the incentre of a triangle always lies in its interior.

Solution

Let the base angles be \( x \) each. \begin{align*} 80 + x + x &= 180 \text{ (Angle sum property)}\\ 80 + 2x &= 180^\circ \\ 2x &= 180 - 80^\circ \\ 2x& = 100^\circ \\ x& = \frac{100^\circ}{2} \\ x &= 50^\circ \end{align*}

Answer Base angles are \( \color{red} 50^\circ,50^\circ \)

Solution

\begin{align*} x + 50^\circ &= 120^\circ \text{ (Exterior angle property)} \\ x &= 120^\circ - 50^\circ \\ x &= 70^\circ \\ \\ y + 120^\circ &= 180^\circ \text{ (Linear pair)} \\ y &= 180^\circ - 120^\circ \\ y &= 60^\circ \end{align*}

Answer \( \color{red} x = 70^\circ, \ y = 60^\circ \)

Solution

\begin{align*} \text{Longest side} & = 2.5 \ cm \\ \text{Side 1} & = 2 \ cm \\ \text{Side 2} & = 1.5 \ cm \\ \text{Pythagoras theorem} \\ \color{magenta} \text{(Side 1)}^2 + \text{(Side 2)}^2 &= \color{magenta} \text{(Hypotenuse)}^2 \\ 2^2 + (1.5)^2 &= (2.5)^2 \\ 4 + 2.25 &= 6.25 \\ 6.25 &= 6.25 \end{align*}

Answer \( \color{red} \text{Yes} \), the given are sides of a right-angled triangle.

Answer In a right angled triangle, the circumcentre is at the mid-point of the hypotenuse.

Solution

\begin{align*} AC & = x \\ AB & = 5 \ m \\ BC & = 12 \ m \\ &\text{Pythagoras theorem} \\ \color{magenta} (AC)^2 &= \color{magenta} (AB)^2 + (BC)^2 \\ x^2 &= (5)^2 + (12)^2 \\ x^2 &= 25 + 144 \\ x^2 &= 169 \\ x &= 13 \ m \\ \end{align*}

Answer Distance from the starting point \( \color{red} 13 \ m \).

Solution

\begin{align*} In \ & \triangle \ ABC \\ AC & = 13 \ cm \\ BC & = 5 \ cm \\ AB & = x \\ \text{Pythagoras theorem} \\ \color{magenta} (AB)^2 + (BC)^2 & = \color{magenta} (AC)^2 \\ x^2 + (5)^2 &= (13)^2 \\ x^2 + 25 &= 169 \\ x^2 &= 169 - 25 \\ x^2 &= 144 \\ x &= 12 \ m \\ AB &= 12 \ m \\ \\ Perimeter & = 2(L+B) \\ & = 2(12+5) \\ & = 2 \times 17 \\ Perimeter & = 34 \ cm \end{align*}

Answer Perimeter \( \color{red} 34 \ cm \).

Solution

\begin{align*} In \ & \triangle \ PQR \\ PQ &= PR \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle PQR &= \angle PRQ \\ \text{By the angle sum }& \text{property}\\ \angle P + \angle PQR + \angle PRQ &= 180^\circ \\ 30^\circ + \angle PQR + \angle PQR &= 180^\circ \\ 2 \angle PQR &= 180^\circ \\ 2 \angle PQR &= 180^\circ - 30^\circ \\ 2 \angle PQR &= 150^\circ \\ \angle PQR &= \frac{150^\circ}{2} \\ \angle PQR &= 75^\circ \\ \angle PRQ &= 75^\circ \end{align*}

Answer \( \angle PQR = {\color{red} 75^\circ} , \angle PRQ = {\color{red} 75^\circ}\).

Solution

\begin{align*} In \ & \triangle \ SQR \\ SQ &= SR \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle SQR &= \angle SRQ \\ \text{By the angle sum }& \text{property}\\ \angle S + \angle SQR + \angle SRQ &= 180^\circ \\ 70^\circ + \angle SQR + \angle SQR &= 180^\circ \\ 2 \angle SQR &= 180^\circ \\ 2 \angle SQR &= 180^\circ - 70^\circ \\ 2 \angle SQR &= 110^\circ \\ \angle SQR &= \frac{110^\circ}{2} \\ \angle SQR &= 55^\circ \\ \angle SRQ &= 55^\circ \end{align*}

Answer \( \angle SQR = {\color{red} 55^\circ} , \angle SRQ = {\color{red} 55^\circ}\).

Solution

\begin{align*} x & = \angle PQR - \angle SRQ \\ x & = 75^\circ - 55^\circ \\ x & = 20^\circ \\ \end{align*}

Answer \( x = {\color{red} 20^\circ}\)

Answer Angles opposite to equal sides of a triangle are equal

\begin{align*} & In \ \triangle \ PQR \\ & PQ = PR \ (Given) \\ & \color{red} \text{Angles opposite to equal sides of a triangle are equal}\\ & \implies \angle Q = \angle R \end{align*}

Answer Corresponding angles

\begin{align*} & In \ \triangle \ PQR \\ & LM \parallel QR \ (Given) \\ & PQ \ \text{(Transversal line)} \\ & \implies \angle PLM = \angle Q \color{red} \text{ (Corresponding angles)} \end{align*}

Answer Corresponding angles

\begin{align*} & In \ \triangle \ PQR \\ & LM \parallel QR \ (Given) \\ & PR \ \text{(Transversal line)} \\ & \implies \angle PML = \angle R \color{red} \text{ (Corresponding angles)} \end{align*}

Answer From (i) , (ii) and (iii)

\begin{align*} \angle Q &= \angle R \quad (i) \\ \angle PLM &= \angle Q \quad (ii) \\ \angle PML &= \angle R \quad (iii) \\ \implies \angle PLM &= \angle PML \end{align*}

Answer In a triangle, if two angles are equal, then the sides opposite to these angles are also equal.

\begin{align*} & In \ \triangle \ PLM \\ & \angle PLM = \angle PML \implies \text{from (iv)} \\ \end{align*}

In a triangle, if two angles are equal, then the sides opposite to these angles are also equal.

\begin{align*} & PL = PM \\ & \triangle PLM \color{red} \text{ is isosceles} \end{align*}

Solution

\begin{align*} \text{Let the vertex angle be } & = x \\ \text{Let base angles be } & = 2x \\ \text{(Angle sum } & \text{property)} \\ \color{magenta} \angle A + \angle B + \angle C & = \color{magenta} 180^\circ \\ 2x + 2x + x &= 180^\circ \\ 5x &= 180^\circ \\ x &= 36^\circ \\ \color{green} \text{Vertex angle} & = \color{green} 36^\circ \\ \text{Base angles} & = 2x\\ & = 2 \times 36^\circ\\ \color{green} \text{Base angles} & = \color{green} 72^\circ\\ \end{align*}

Answer Angles of the triangle are \( \color{red} 36^\circ , 72^\circ, 72^\circ \).

Solution

\begin{align*} AC & = x \\ AB & = 5 \ m \\ BC & = 12 \ m \\ &\text{Pythagoras theorem} \\ \color{magenta} (AC)^2 &= \color{magenta} (AB)^2 + (BC)^2 \\ x^2 &= (5)^2 + (12)^2 \\ x^2 &= 25 + 144 \\ x^2 &= 169 \\ x &= 13 \ m \\ \\ &\text{Total height of the pole} \\ & = BC + CA \\ & = 12 \ m + 13 \ m \\ & = 25 \ m \\ \end{align*}

Answer Total height of the pole \( \color{red} 25 \ m \)

Answer Yes, it is the vertex C.

Answer All the three medians are equal 5.2cm

Solution

\begin{align*} \text{Given :} \\ \angle ACD &= 120^\circ \\ \angle A &= \angle B \\ \\ \text{To find} & = \angle A \ , \angle B \ , \angle ACB \\ \\ \color{magenta} \angle ACB + \angle ACD & = \color{magenta} 180^\circ \text{ (Linear pair)} \\ \angle ACB + 120^\circ & = 180^\circ \\ \angle ACB & = 180^\circ - 120^\circ \\ \color{green} \angle ACB & = \color{green} 60^\circ \\ \\ \color{magenta}\angle A + \angle B &= \color{magenta} \angle ACD \text{ (Exterior angle property)} \\ \angle A + \angle A &= 120^\circ \\ 2 \ \angle A &= 120^\circ \\ \angle A &= \frac{120^\circ}{2} \\ \\ \angle A &= 60^\circ \\ \therefore \angle B &= 60^\circ \\ \angle ACB & = 60^\circ \\ \triangle ABC & \text{ is an equilateral triangle} \end{align*}

Answer The angles of the triangle are \(\color{red}{60^\circ,\, 60^\circ,\, 60^\circ.}\) It is an \(\color{red}{\text{equilateral triangle.}}\)

Solution

\begin{align*} \angle 3 + 136^\circ & = 180^\circ \color{magenta} \text{ (Linear pair)} \\ \angle 3 & = 180^\circ - 136^\circ \\ \color{green}\angle 3 & = \color{green}44^\circ \\ \\ \angle 2 + 104^\circ & = 180^\circ \color{magenta} \text{ (Linear pair)} \\ \angle 2 & = 180^\circ - 104^\circ \\ \color{green}\angle 2 & = \color{green}76^\circ \\ \\ \angle 1 + \angle 2 + \angle 3 & = 180^\circ \color{magenta} \text{ (Angle sum property)} \\ \angle 1 + 76^\circ + 44^\circ & = 180^\circ \\ \angle 1 + 120^\circ & = 180^\circ \\ \angle 1 & = 180^\circ - 120^\circ \\ \color{green} \angle 1 & = \color{green} 60^\circ \end{align*}

Answer \( \angle 1 = {\color{red} 60^\circ} , \angle 2 = {\color{red} 76^\circ} ,\angle 3 = {\color{red} 44^\circ} \)

Solution

\begin{align*} \angle CAB & = \angle EAF \color{magenta} \text{ (Vertically opposite angles)} \\ \color{green}\angle CAB & = \color{green} 45^\circ \\ \\ \angle ABC + \angle CAB & = \angle ACD \color{magenta} \text{ (Exterior angle property)} \\ \angle ABC + 45^\circ & = 105^\circ \\ \angle ABC & = 105^\circ - 45^\circ \\ \color{green} \angle ABC & = \color{green} 60^\circ \\ \\ \angle ACB + \angle ACD & = 180^\circ \color{magenta} \text{ (Linear pair)} \\ \angle ACB + 105^\circ & = 180^\circ \\ \angle ACB & = 180^\circ - 105^\circ \\ \color{green}\angle ACB & = \color{green} 75^\circ \end{align*}

Answer \( \angle CAB = {\color{red} 45^\circ} , \angle ABC = {\color{red} 60^\circ} ,\angle ACB = {\color{red} 75^\circ} \)

Solution

\begin{align*} \angle x &= \angle A + \angle B \color{magenta} \text{ (Exterior angle property)} \\ x &= 30^\circ + 30^\circ \\ \color{green} x &= \color{green} 60^\circ \\ \\ \angle y &= \angle D + \angle DCE \\ y &= 30^\circ + 60^\circ \\ \color{green} y &= \color{green} 90^\circ \\ \end{align*}

Answer \( x = {\color{red} 60^\circ} , y = {\color{red} 90^\circ} \)

Solution

The sum of any two sides of a triangle is always greater than the third side.

Length of the third side is greater than 15 - 12 = 3 cm.

Length of the third side is less than 15 + 12 = 27 cm.

Answer \( 3 cm < {\color{red} \text{Length of the third side}} < 27cm \)

Answer

In △ABC and △DEF.
\(\color{red} \angle B = \angle E \), is required to make the two triangles congruent by SAS congruence condition.

Answer

\[ \begin{align*} \text{In } \triangle ADB & \text{ and } \triangle ADC \\ \text{AB} &= \text{AC} \text{ (Given)} \color{red} (S)\\ \angle BAD &= \angle CAD \text{ ( AD bisects } \angle A \,) \color{red} (A)\\ \text{AD} &= \text{AD} \text{ (Common side)} \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ADB \cong \triangle ADC \\ \\ \text{Third pair } & \implies AD =AD \\ \text{Yes, } &BD = CD \text{ (by CPCT)} \end{align*} \]

Answer \( \angle \text{BAC} = \angle \text{DCA}\) (Alternate angles)

Answer Yes, \( \triangle \text{ABC} \cong \triangle \text{CDA} \) by SAS congruence condition.

Answer

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle CDA \\ \text{AB} &= \text{CD} \text{ (Given) } \color{red} (S) \\ \angle \text{BAC} &= \angle \text{DCA} \text{ (Alternate angles) } \color{red} (A) \\ \text{AC} &= \text{CA} \text{ (Common side) } \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABC \cong \triangle CDA \end{align*} \]

Solution

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle EDF \\ \text{AB} &= \text{ED} \implies (3 \,cm) \color{red} (S) \\ \angle \text{B} &= \angle \text{D} \implies (70^\circ ) \color{red} (A) \\ \text{BC} &= \text{DF} \implies (4 \,cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABC \cong \triangle EDF \end{align*} \]

Answer Yes, \( \color{red} \triangle \text{ABC} \cong \triangle \text{EDF} \)

Solution

\[ \begin{align*} \text{In } \triangle PQR & \text{ and } \triangle ZXY \\ \text{PQ} &= \text{ZX} \implies (3 \,cm) \color{red} (S) \\ \angle \text{Q} &= \angle \text{X} \implies (90^\circ ) \color{red} (A) \\ \text{QR} &= \text{XY} \implies (2 \,cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle PQR \cong \triangle ZXY \end{align*} \]

Answer Yes, \( \color{red} \triangle PQR \cong \triangle ZXY \)

Solution

\[ \begin{align*} \text{In } \triangle AOB & \text{ and } \triangle COD \\ \text{AO} &= \text{CO} \implies (Given) \color{red} (S) \\ \angle \text{AOB} &= \angle \text{COD} \implies \text{(Vertically opposite angles )} \color{red} (A) \\ \text{BO} &= \text{DO} \implies (Given) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle AOB \cong \triangle COD \end{align*} \]

Answer Yes, \( \color{red} \triangle \text{AOB} \cong \triangle \text{COD} \)

Solution

\[ \begin{align*} \text{In } \triangle PQS & \text{ and } \triangle RSQ \\ \text{PS} &= \text{RQ} \implies (4 \, cm) \color{red} (S) \\ \angle \text{PSQ} &= \angle \text{RQS} \implies \text{(Alternate angles )} \color{red} (A) \\ \text{QS} &= \text{SQ} \implies (5.8 \, cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle PQS \cong \triangle RSQ \end{align*} \]

Answer Yes, \( \color{red} \triangle \text{PSQ} \cong \triangle \text{RQS} \)

Solution

\[ \begin{align*} \text{In } \triangle PRQ & \text{ and } \triangle XYZ \\ \text{PR} &= \text{XY} \implies (4 \,cm) \color{red} (S) \\ \angle \text{PRQ} &= \angle \text{XYZ} \implies (55^\circ ) \color{red} (A) \\ \text{RQ} &= \text{YZ} \implies (2 \,cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle PQR \cong \triangle XZY \end{align*} \]

Answer Yes, \( \color{red} \triangle PRQ \cong \triangle XYZ \)

Solution

\[ \begin{align*} \text{In } \triangle ABD & \text{ and } \triangle BAC \\ \text{AD} &= \text{BC} \implies (Given) \color{red} (S) \\ \text{BD} &= \text{AC} \implies \text{(Given )} \color{red} (S) \\ \text{AB} &= \text{BA} \implies \text{(Common Side)} \color{red} (S) \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \\ \\ \text{In } \triangle ABD & \text{ and } \triangle BAC \\ \text{AD} &= \text{BC} \implies (Given) \color{red} (S) \\ \angle \text{D} &= \angle \text{C} \implies \text{(By CPCT)} \color{red} (A) \\ \text{BD} &= \text{AC} \implies (Given) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \\ \\ \end{align*} \]

Answer Yes, \( \color{red} \triangle ABD \cong \triangle BAC \)

Answer

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle ADC \\ \text{AB} &= \text{AD} \implies (Given) \color{red} (S) \\ \angle BAC &= \angle DAC \implies (Given) \color{red} (A) \\ \text{AC} &= \text{AC} \implies \text{(Common side)} \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \end{align*} \]

Solution

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle PQR \\ \text{AB} &= \text{QR} \implies 2 \, cm \\ \text{BC} &= \text{RP} \implies 3 \, cm \\ \text{AC} &= \text{QP} \implies 2.5 \, cm \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle ABC \cong \triangle QRP \end{align*} \]

Answer \( \color{red} \triangle ABC \cong \triangle QRP \)

Solution

\[ \begin{align*} \text{In } \triangle PQS & \text{ and } \triangle PRS \\ \text{PQ} &= \text{PR} \implies 3.5 \, cm \\ \text{QS} &= \text{RS} \implies 2 \, cm \\ \text{SP} &= \text{SP} \implies 2.5 \, cm \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle PQS \cong \triangle PRS \end{align*} \]

Answer \( \color{red} \triangle PQS \cong \triangle PRS \)

Solution

\[ \begin{align*} \text{In } \triangle ABD & \text{ and } \triangle FEC \\ \text{AB} &= \text{FE} \implies 3 \, cm \\ \text{BD} &= \text{EC} \implies 3 \, cm \\ \text{DA} &= \text{CF} \implies 5.5 \, cm \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle FEC \end{align*} \]

Answer \( \color{red} \triangle ABD \cong \triangle FEC \)

Solution

\[ \begin{align*} \text{In } \triangle POQ & \text{ and } \triangle ROS \\ \text{PO} &= \text{RO} \implies 3 \, cm \\ \text{OQ} &= \text{OS} \implies 4 \, cm \\ \text{Third side} & \text{ not given.} \end{align*} \]

Answer Triangles are \( \color{red} \text{Not congruent} \) by SSS condition.

Answer \( \color{red} Yes \) , by SSS congruence condition.

Answer

\[ \begin{align*} \text{In } \triangle PQS & \text{ and } \triangle RQS \\ \text{PS} &= \text{RS} \text{ (Given)} \\ \text{PQ} &= \text{RQ} \text{ (Given)} \\ \text{SQ} &= \text{SQ} \text{ (Common side)} \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle PQS \cong \triangle RQS \end{align*} \]

Answer \( \color{red} Yes \), by SSS congruence condition.

Answer

\[ \begin{align*} \text{In } \triangle ADB & \text{ and } \triangle ADC \\ \text{AB} &= \text{AC} \text{ (Given)} \\ \text{BD} &= \text{CD} \text{ ( D is the mid point of BC)} \\ \text{AD} &= \text{AD} \text{ (Common side)} \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle ADB \cong \triangle ADC \end{align*} \]

Answer

In △PQR and △XYZ.
\(\color{red} PR = XY \), is required to make the two triangles congruent by SSS congruence condition.

Answer

Answer To save electricity, switch off all the electronic gadgets when not in use or any other.

Answer \( (iii) \ \color{red} Triangle \)

Answer \( (ii) \ \color{red} 1 m, 2 m, 2 m \)

The sum of any two sides of a triangle is greater than the third side. \[ \begin{align*} 1 + 2 = 3 > 2 \\ 2 + 2 = 4 > 2 \\ 1 + 2 = 3 > 2 \\ \end{align*} \]

\[ \begin{align*} SSS , SAS , ASA , RHS \end{align*} \]

Answer \( \color{red} Three \)

Answer

\[ \begin{align*} 20 + 12 = 32 > 10 \\ 12 + 10 = 22 > 20 \\ 20 + 10 = 30 > 12 \\ \end{align*} \]

Yes it is possible to construct a triangle with dimensions \( 20 m, 12 m, 10 m \), because the sum of any two sides of a triangle is always greater than the third side.

Answer

\[ \begin{align*} 20 + 13 = 33 > 7 \\ 20 + 7 = 27 > 13 \\ 13 + 7 = 20 \cancel> 20 \\ \end{align*} \]

Triangle is not possible because the sum of two sides is not greater than the third side \( \color{red} 13 + 7 = 20 \cancel> 20 \).

Answer We can find the third angle using the angle sum property and construct a triangle.

Answer \( \color{red} 6 \)

Answer

\[ \begin{align*} In \ \triangle ABC \\ \angle A + \angle B + \angle C & = 180^\circ \\ 65^\circ + 75^\circ + \angle C & = 180^\circ \\ 140^\circ + \angle C & = 180^\circ \\ \angle C & = 180^\circ - 140^\circ \\ \angle C & = 40^\circ \end{align*} \]

Answer

\[ \begin{align*} In \ \triangle ABC \\ \angle A + \angle B + \angle C & = 180^\circ \\ 90^\circ + \angle B + 45^\circ & = 180^\circ \\ \angle B + 135^\circ & = 180^\circ \\ \angle B & = 180^\circ - 135^\circ \\ \angle B & = 45^\circ \end{align*} \]

Answer

Answer

XY is perpendicular to side AB.

PQ is perpendicular to side AC.

Answer

Answer

Hypotenuse = 5cm

Answer

\[ \begin{align*} AB &= 4.6cm \\ AC &= 2.3cm \\ 2AC &= 4.6cm \\ \therefore AB &= 2AC \end{align*} \]

Answer

Answer

Answer

Answer

Answer

Answer

Answer

\( \angle E = {\color{red} 40^\circ} , \angle F = {\color{red} 40^\circ} \)

Answer

\( \angle E = {\color{red} 40^\circ} , \angle F = {\color{red} 40^\circ} \)

Answer

XY is perpendicular to side PQ.

Answer

The sum of the lengths of any two sides of a triangle is greater than the length of third side.

\begin{align*} \text{Sum of two sides} && \quad \text{Third Side} \\ \end{align*} \begin{align*} 3 + 6.1 &= 9.1 & > && 2.6 \\ 2.6 + 6.1 &= 8.7 & > && 3 \\ 3 + 2.6 &= 5.6 & \color{red}{<} && 6.1 \\ \end{align*}

Triangle is not possible.

Answer

No, the construction is not possible.

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

\begin{align*} \text{Sum of two sides} && \quad \text{Third Side} \\ \end{align*} \begin{align*} 7 + 3 &= 10 & > && 4 \\ 7 + 4 &= 11 & > && 3 \\ 3 + 4 &= 7 & \color{red}{=} && 7 \\ \end{align*}

\[ \begin{align*} \text{Average age of 35 students} & = 14 \,years \\ \text{Sum of ages of 35 students} & = 14 \times 35 \\ & = 490 \\ \\ \text{Average age of 35 students and 1 teacher} & = 15 \,years \\ \text{Sum of ages of 35 students and 1 teacher} & = 15 \times 36 \\ & = 540 \\ \\ \text{Age of the teacher} & = 540 - 490 \\ & = 50 \,years \end{align*} \]

Answer Age of the teacher \( \boxed{\color{red} 50\, years}\)

\[ \begin{align*} \text{Average marks of 10 students} & = 60 \\ \text{Sum of marks of 10 students} & = 60 \times 10 \\ & = 600 \\ \\ \text{Average marks of 9 students} & = 59 \\ \text{Sum of marks of 9 students} & = 59 \times 9 \\ & = 531 \\ \\ \text{Highest mark} & = 600 - 531 \\ & = 69 \, Marks \end{align*} \]

Answer Highest mark \( \boxed{\color{red} 69 \, Marks}\)

\[ \begin{align*} \text{Let the 2 numbers be } & x \text{ and } y \\ \\ \text{Mean of 2 numbers} & = 10 \\ \frac{x + y}{2} & = 10 \\ \\ x + y & = 2\times 10 \\ x + y & = 20 \\ \\ \text{Difference between 2 numbers} &=0 \\ \\ x - y &=0 \\ x &=0 + y \\ x &= y \\ x + y & = 20 \\ x + x & = 20 \\ 2x & = 20 \\ \\ x & = \frac{20}{2} \\ \\ x & = 10 \\ y & = 10 \\ \end{align*} \]

Answer The numbers are \( \boxed{\color{red} 10 , 10}\)

Solution

\begin{align*} & 75\%, 3.2 , 450\%, 0.25, 3\frac{1}{5},1.54 \\ \\ 75\% &= \frac{75}{100} \implies 0.75 \\ \\ 450\% &= \frac{450}{100} \implies 4.5 \\\\ 3\frac{1}{5} &= \frac{16}{5} \implies 3.2 \\ \\ \color{orange}Observations & = 0.75, 3.2 , 4.5, 0.25, 3.2,1.54 \\ \\ & \color{magenta} \boxed{Mean} \\ \text{Mean} &= \frac{\text{Sum of all observations}}{\text{No. of observations}} \\ \\ &= \frac{0.25 + 0.75 + 1.54 + 3.2 + 3.2 + 4.5}{6} \\ &= \frac{13.44}{6} \\ \\ \color{green} Mean &=\color{green} 2.24 \\ \\ & \color{magenta} \boxed{Median} \\ \text{Ascending order} & = \cancel{0.25}, \cancel{ 0.75}, 1.54, 3.2, \cancel{3.2}, \cancel{4.5} \\ \\ \text{Median} &= \frac{\text{Sum of two middle most observations}}{2} \\ \\ &= \frac{1.54 + 3.2}{2} \\ \\ &= \frac{4.74}{2} \\ \\ \color{green} Median &= \color{green} 2.37 \\ \\ & \color{magenta} \boxed{Mode} \\ \text{Mode} &= \text{The most frequently occurring value} \\ \text{Ascending order} & = 0.25, 0.75, 1.54, {\color{magenta}3.2, 3.2}, 4.5 \\ &= 3.2 \quad \text{(appears 2 times)} \\ \color{green} Mode &= \color{green} 3.2 \\ \\ \end{align*}

Answer Mean \( = \color{red} 2.24 \) , Median \( = \color{red} 2.37 \), Mode \( = \color{red} 3.2 \)