Detailed Solutions and Diagrams for Practice Problems (Chapter 8: Quadrilaterals)
Problem 1: Diagonals of a Parallelogram
Question:
In a parallelogram ABCDABCD, the diagonals ACAC and BDBD intersect at OO. Prove that △AOB≅△COD\triangle AOB \cong \triangle COD.
Solution:
Given: ABCDABCD is a parallelogram, and diagonals ACAC and BDBD intersect at OO.
To Prove: △AOB≅△COD\triangle AOB \cong \triangle COD.
Proof:
In parallelogram ABCDABCD, diagonals bisect each other. AO=COandBO=DOAO = CO \quad \text{and} \quad BO = DO
∠AOB=∠COD\angle AOB = \angle COD (vertically opposite angles).
△AOB≅△COD\triangle AOB \cong \triangle COD (by SAS criterion).
Thus, the two triangles are congruent.
Problem 2: Sum of Angles in a Quadrilateral
Question:
A quadrilateral ABCDABCD has the following angles:
∠A=90∘, ∠B=110∘, ∠C=85∘.\angle A = 90^\circ, \, \angle B = 110^\circ, \, \angle C = 85^\circ.
Find ∠D\angle D.
Solution:
The sum of the angles of a quadrilateral is 360∘360^\circ.
∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^\circ
Substitute the given values:
90∘+110∘+85∘+∠D=360∘90^\circ + 110^\circ + 85^\circ + \angle D = 360^\circ ∠D=360∘−285∘=75∘\angle D = 360^\circ – 285^\circ = 75^\circ
Answer:
∠D=75∘\angle D = 75^\circ
Problem 3: Cyclic Quadrilateral
Question:
Prove that in a cyclic quadrilateral ABCDABCD, ∠A+∠C=180∘\angle A + \angle C = 180^\circ.
Solution:
Given: ABCDABCD is a cyclic quadrilateral (all vertices lie on a circle).
To Prove: ∠A+∠C=180∘\angle A + \angle C = 180^\circ.
Proof:
In a cyclic quadrilateral, opposite angles are supplementary. ∠A+∠C=180∘and∠B+∠D=180∘\angle A + \angle C = 180^\circ \quad \text{and} \quad \angle B + \angle D = 180^\circ
This follows directly from the property of cyclic quadrilaterals.
Thus, ∠A+∠C=180∘\angle A + \angle C = 180^\circ.
Problem 4: Diagonals in a Rhombus
Question:
Prove that the diagonals of a rhombus bisect each other at right angles.
Solution:
Given: ABCDABCD is a rhombus. Diagonals ACAC and BDBD intersect at OO.
To Prove: AC⊥BDAC \perp BD and AO=OC,BO=ODAO = OC, BO = OD.
Proof:
A rhombus is a parallelogram with all sides equal.
The diagonals of a parallelogram bisect each other. AO=OCandBO=ODAO = OC \quad \text{and} \quad BO = OD
In a rhombus, diagonals bisect each other at 90∘90^\circ.
Thus, AC⊥BDAC \perp BD.
Conclusion: The diagonals of a rhombus bisect each other at right angles.
Problem 5: Finding a Missing Side
Question:
In a parallelogram, one side measures 12 cm12 \, \text{cm}, and the other measures 9 cm9 \, \text{cm}. Find the perimeter.
Solution:
The perimeter of a parallelogram is given by:
P=2×(Length+Breadth)P = 2 \times (\text{Length} + \text{Breadth})
Substitute Length=12 cm\text{Length} = 12 \, \text{cm} and Breadth=9 cm\text{Breadth} = 9 \, \text{cm}:
P=2×(12+9)=2×21=42 cmP = 2 \times (12 + 9) = 2 \times 21 = 42 \, \text{cm}
Answer:
The perimeter is 42 cm42 \, \text{cm}.
Additional Practice Problems
Prove that opposite sides of a parallelogram are equal.
In a cyclic quadrilateral, one angle is 80∘80^\circ. Find the opposite angle.
If the diagonals of a rhombus are 10 cm10 \, \text{cm} and 24 cm24 \, \text{cm}, find its area.
A parallelogram has sides 5 cm5 \, \text{cm} and 12 cm12 \, \text{cm}. If one of its diagonals is 13 cm13 \, \text{cm}, find the length of the other diagonal.
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