त्रिभुज और चतुर्भुज (Triangles and Quadrilaterals)

Step 1: Apply the Angle Sum Property of a triangle In $\Delta PQR$, the sum of all three interior angles is strictly $180^\circ$:

$$\angle P + \angle Q + \angle R = 180^\circ$$

Step 2: Substitute the known values and solve for $\angle R$

$$48^\circ + 73^\circ + \angle R = 180^\circ$$

$$121^\circ + \angle R = 180^\circ$$

$$\angle R = 180^\circ - 121^\circ = 59^\circ$$

Step 3: Classify the triangle by its angles The measures of the three interior angles are $48^\circ$, $73^\circ$, and $59^\circ$. Since every single angle is strictly less than $90^\circ$ ($48^\circ < 90^\circ$, $73^\circ < 90^\circ$, and $59^\circ < 90^\circ$), all three angles are acute. Therefore, $\Delta PQR$ is an Acute-Angled Triangle (and because all three angles are unequal, it is also a Scalene Triangle).

Final Answer: (i) $\angle R = 59^\circ$ (ii) $\Delta PQR$ is an Acute-Angled Triangle.

Governing Theorem: According to the Triangle Inequality Theorem, a triangle can physically exist if and only if the sum of the lengths of any two sides is strictly greater than the third side:

$$\text{Shorter Side } 1 + \text{Shorter Side } 2 > \text{Longest Side}$$

Case (a): Side lengths 5 cm, 8 cm, 14 cm

• The two shorter sides are $5\text{ cm}$ and $8\text{ cm}$; the longest side is $14\text{ cm}$.
• Sum of the two shorter sides:

$$5 + 8 = 13\text{ cm}$$

• Comparison with longest side:

$$13\text{ cm} < 14\text{ cm}$$

• Conclusion: Since the sum of the two shorter sides is less than the third side ($13 < 14$), the two shorter segments cannot meet to enclose an interior region.

$$\mathbf{\text{No, a triangle CANNOT be formed.}}$$

Case (b): Side lengths 7 cm, 11 cm, 18 cm

• The two shorter sides are $7\text{ cm}$ and $11\text{ cm}$; the longest side is $18\text{ cm}$.
• Sum of the two shorter sides:

$$7 + 11 = 18\text{ cm}$$

• Comparison with longest side:

$$18\text{ cm} = 18\text{ cm}$$

• Conclusion: The sum is equal to the third side, not strictly greater. The segments would collapse completely flat onto a single straight line with zero interior area.

$$\mathbf{\text{No, a triangle CANNOT be formed.}}$$

Case (c): Side lengths 6 cm, 9 cm, 13 cm

• The two shorter sides are $6\text{ cm}$ and $9\text{ cm}$; the longest side is $13\text{ cm}$.
• Sum of the two shorter sides:

$$6 + 9 = 15\text{ cm}$$

• Comparison with longest side:

$$15\text{ cm} > 13\text{ cm}$$

• Since $6 + 9 = 15 > 13$, $6 + 13 = 19 > 9$, and $9 + 13 = 22 > 6$, the condition holds for all three pairings.
• Conclusion:

$$\mathbf{\text{Yes, a valid triangle CAN be formed.}}$$

Step 1: Set up the algebraic equation using Exterior Angle Theorem Let the common ratio multiplier for the two interior opposite angles be $x$.

• First interior opposite angle $= 2x$
• Second interior opposite angle $= 3x$

By the Exterior Angle Theorem, the exterior angle equals the sum of its two interior opposite angles:

$$\text{Exterior } \angle = 2x + 3x$$

$$120^\circ = 5x$$

$$x = \frac{120^\circ}{5} = 24^\circ$$

Step 2: Calculate the two interior opposite angles

• First interior opposite angle $= 2x = 2 \times 24^\circ = \mathbf{48^\circ}$
• Second interior opposite angle $= 3x = 3 \times 24^\circ = \mathbf{72^\circ}$

Step 3: Calculate the adjacent interior angle The exterior angle and its adjacent interior angle form a Linear Pair on a straight line:

$$\text{Adjacent Interior } \angle = 180^\circ - \text{Exterior } \angle$$

$$\text{Adjacent Interior } \angle = 180^\circ - 120^\circ = \mathbf{60^\circ}$$

Step 4: Verification of Angle Sum Property

$$\text{Sum of all three interior angles} = 48^\circ + 72^\circ + 60^\circ = 180^\circ$$

The sum strictly equals $180^\circ$, confirming the solution is 100% correct.

Final Answer: (i) Interior opposite angles are 48° and 72°. (ii) Adjacent interior angle is 60°. (iii) Verified: $48^\circ + 72^\circ + 60^\circ = 180^\circ$.

Step 1: Formulate the equation using the Quadrilateral Angle Sum Property The sum of all four interior angles of any planar quadrilateral is strictly $360^\circ$:

$$\angle A + \angle B + \angle C + \angle D = 360^\circ$$

Substitute the given expressions:

$$(3x + 10) + (2x + 20) + (4x - 15) + (x + 25) = 360^\circ$$

Step 2: Combine like terms and solve for x

$$(3x + 2x + 4x + x) + (10 + 20 - 15 + 25) = 360$$

$$10x + 40 = 360$$

$$10x = 360 - 40$$

$$10x = 320$$

$$x = \frac{320}{10} = \mathbf{32}$$

Step 3: Calculate the degree measure of each angle

• $\angle A = 3x + 10 = 3(32) + 10 = 96 + 10 = \mathbf{106^\circ}$
• $\angle B = 2x + 20 = 2(32) + 20 = 64 + 20 = \mathbf{84^\circ}$
• $\angle C = 4x - 15 = 4(32) - 15 = 128 - 15 = \mathbf{113^\circ}$
• $\angle D = x + 25 = 32 + 25 = \mathbf{57^\circ}$

Check: $106^\circ + 84^\circ + 113^\circ + 57^\circ = 360^\circ$. (Correct!)

Step 4: Check if any pair of opposite angles is supplementary

• Pair 1 ($\angle A$ and $\angle C$):

$$\angle A + \angle C = 106^\circ + 113^\circ = 219^\circ \neq 180^\circ$$

• Pair 2 ($\angle B$ and $\angle D$):

$$\angle B + \angle D = 84^\circ + 57^\circ = 141^\circ \neq 180^\circ$$

Final Answer: (i) $x = 32$ (ii) $\angle A = 106^\circ$, $\angle B = 84^\circ$, $\angle C = 113^\circ$, $\angle D = 57^\circ$ (iii) Neither pair of opposite angles is supplementary ($219^\circ \neq 180^\circ$ and $141^\circ \neq 180^\circ$).

Step 1: Calculate the four angles of the parallelogram In any parallelogram, consecutive (adjacent) angles are supplementary (they sum to $180^\circ$ as co-interior angles between parallel lines):

$$\angle A + \angle B = 180^\circ$$

Given the ratio $\angle A : \angle B = 4 : 5$, let $\angle A = 4k$ and $\angle B = 5k$:

$$4k + 5k = 180^\circ \implies 9k = 180^\circ \implies k = 20^\circ$$

• $\angle A = 4k = 4 \times 20^\circ = \mathbf{80^\circ}$
• $\angle B = 5k = 5 \times 20^\circ = \mathbf{100^\circ}$

In a parallelogram, opposite angles are equal:

• $\angle C = \angle A = \mathbf{80^\circ}$
• $\angle D = \angle B = \mathbf{100^\circ}$

Step 2: Find the value of y and length of side AD In a parallelogram, opposite sides are equal in length:

• $AB = CD$:

$$3y - 2 = 16$$

$$3y = 16 + 2 = 18$$

$$y = \frac{18}{3} = \mathbf{6}$$

• $AD = BC$:

$$AD = \mathbf{11\text{ cm}}$$

Step 3: Calculate the perimeter of parallelogram ABCD The perimeter is the sum of all four sides:

$$\text{Perimeter } (P) = 2(AB + BC) = 2(16 + 11) = 2(27) = \mathbf{54\text{ cm}}$$

Final Answer: (i) Angles are $\angle A = 80^\circ$, $\angle B = 100^\circ$, $\angle C = 80^\circ$, and $\angle D = 100^\circ$. (ii) $y = 6$ and $AD = 11\text{ cm}$. (iii) Perimeter $= 54\text{ cm}$.

Part (a): Isosceles Triangle Analysis [2.5 Marks]

Step 1: Calculate the base angles $\angle Y$ and $\angle XZY$ In $\Delta XYZ$, side $XY = XZ$, which means the angles opposite to these equal sides are strictly equal:

$$\angle Y = \angle XZY$$

By the Angle Sum Property of a triangle:

$$\angle X + \angle Y + \angle XZY = 180^\circ$$

$$52^\circ + 2\angle Y = 180^\circ$$

$$2\angle Y = 180^\circ - 52^\circ = 128^\circ$$

$$\angle Y = \frac{128^\circ}{2} = \mathbf{64^\circ}$$

Therefore, base angle $\angle Y = 64^\circ$ and base angle $\angle XZY = \mathbf{64^\circ}$.

Step 2: Calculate the exterior angle $\angle XZW$

• Method 1 (Exterior Angle Theorem):

$$\text{Ext } \angle XZW = \text{Int Opp } \angle X + \text{Int Opp } \angle Y = 52^\circ + 64^\circ = \mathbf{116^\circ}$$

• Method 2 (Linear Pair Verification):

$$\angle XZW = 180^\circ - \angle XZY = 180^\circ - 64^\circ = \mathbf{116^\circ}$$

Both methods confirm $\angle XZW = 116^\circ$.

Part (b): Rhombus Diagonals & Angle Determination [2.5 Marks]

Step 1: Determine $\angle AOB$ In any rhombus, the diagonals bisect each other perpendicularly at right angles ($90^\circ$):

$$AC \perp BD \implies \angle AOB = \mathbf{90^\circ}$$

Step 2: Determine $\angle ABO$ Consider right-angled triangle $\Delta AOB$:

• $\angle AOB = 90^\circ$
• Given $\angle OAB = 38^\circ$

By Angle Sum Property in $\Delta AOB$:

$$\angle OAB + \angle AOB + \angle ABO = 180^\circ$$

$$38^\circ + 90^\circ + \angle ABO = 180^\circ$$

$$128^\circ + \angle ABO = 180^\circ$$

$$\angle ABO = 180^\circ - 128^\circ = \mathbf{52^\circ}$$

Step 3: Determine $\angle ABC$ In a rhombus, the diagonal bisects the interior vertex angle. Thus, diagonal $BD$ bisects $\angle ABC$:

$$\angle ABC = 2 \times \angle ABO = 2 \times 52^\circ = \mathbf{104^\circ}$$

Step 4: Determine $\angle BCD$ Consecutive angles of a rhombus (being a parallelogram) are supplementary:

$$\angle ABC + \angle BCD = 180^\circ$$

$$104^\circ + \angle BCD = 180^\circ$$

$$\angle BCD = 180^\circ - 104^\circ = \mathbf{76^\circ}$$

(Alternatively, opposite angles are equal: $\angle BCD = \angle BAD = 2 \times \angle OAB = 2 \times 38^\circ = 76^\circ$.)

Final Summary of Answers:

• Part (a): Base angles $\angle Y = \angle XZY = \mathbf{64^\circ}$; Exterior angle $\angle XZW = \mathbf{116^\circ}$.
• Part (b): (i) $\angle AOB = \mathbf{90^\circ}$, (ii) $\angle ABO = \mathbf{52^\circ}$, (iii) $\angle ABC = \mathbf{104^\circ}$, (iv) $\angle BCD = \mathbf{76^\circ}$.