Problems on Parallelogram

These are the various types of solved problems on parallelogram.

1. Prove that any two adjacent angles of a parallelogram are supplementary.

Solution:

Let ABCD be a parallelogram

Problems on Parallelogram

Then, AD ∥ BC and AB is a transversal. 

Therefore, A + B = 180° [Since, sum of the interior angles on the same side of the transversal is 180°] 

Similarly, ∠B + ∠C = 180°, ∠C + ∠D = 180° and ∠D + ∠A = 180°. 

Thus, the sum of any two adjacent angles of a parallelogram is 180°. 

Hence, any two adjacent angles of a parallelogram are supplementary. 



2. Two adjacent angles of a parallelogram are as 2 : 3. Find the measure of each of its angles.

Solution:

Let ABCD be a given parallelogram

Problems on Parallelogram

Then, ∠A and ∠B are its adjacent angles.

Let ∠A = (2x)° and ∠B = (3x)°.

Then, ∠A + ∠B = 180° [Since, sum of adjacent angles of a ∥gm is 180°]

⇒ 2x + 3x = 180

⇒ 5x = 180

⇒ x = 36.

Therefore, ∠A = (2 × 36)° = 72° and ∠B = (3 × 36°) = 108°.

Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]

= 108° + ∠C = 180° [Since, ∠B = 108°]

∠C = (180° - 108°) = 72°.

Also, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]

⇒ 72° + ∠D = 180°

⇒ ∠D = (180° - 72°) 108°.

Therefore, ∠A = 72°, ∠B = 108°, ∠C = 72°and ∠D = 108°.



3. In the adjoining figure, ABCD is a parallelogram in which ∠A = 75°. Find the measure of each of the angles ∠B, ∠C and ∠D.

Solution:

It is given that ABCD is a parallelogram in which ∠A = 75°.

Problems on Parallelogram

Since the sum of any two adjacent angles of a parallelogram is 180°,

∠A + ∠B = 180°

⇒ 75° + ∠B = 180°

⇒∠B = (180° - 75°) = 105°

Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]

⇒ 105° + ∠C = 180°

⇒ ∠C = (180° - 105°) = 75°.

Further, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]

⇒ 75° + ∠D = 180°

⇒ ∠D = (180° - 75°) = 105°.

Therefore, ∠B = 105°, ∠C = 75° and ∠D = 105°.



4. In the adjoining figure, ABCD is a parallelogram in which

∠BAD = 75° and ∠DBC = 60°. Calculate:

(i) ∠CDB and (ii) ∠ADB.

Problems on Parallelogram

Solution:

We know that the opposite angles of a parallelogram are equal.

Therefore, ∠BCD = ∠BAD = 75°.

(i) Now, in ∆ BCD, we have

∠CDB + ∠DBC + ∠BCD = 180° [Since, sum of the angles of a triangle is 180°]

⇒ ∠CDB + 60° + 75° = 180°

⇒ ∠CDB + 135° = 180°

⇒ ∠CDB = (180° - 135°) = 45°.

(ii) AD ∥ BC and BD is the transversal.

Therefore, ∠ADB = ∠DBC = 60° [alternate interior angles]

Hence, ∠ADB = 60°.


5. In the adjoining figure, ABCD is a parallelogram in which

∠CAD = 40°, ∠BAC = 35° and ∠COD = 65°.

Calculate: (i) ∠ABD (ii) ∠BDC (iii) ∠ACB (iv) ∠CBD.

Problems on Parallelogram

Solution:

(i) ∠AOB = ∠COD = 65° (vertically opposite angles)

Now, in ∆OAB, we have:

∠OAB + ∠ABO + ∠AOB =180° [Since, sum of the angles of a triangle is 180°]

⇒ 35°+ ∠ABO + 65° = 180°

⇒ ∠ABO + 100° = 180°

⇒ ∠ABO = (180° - 100°) = 80°

⇒ ∠ABD = ∠ABO = 80°.

(ii) AB ∥ DC and BD is a transversal.

Therefore, ∠BDC = ∠ABD = 80° [alternate interior angles]

Hence, ∠BDC = 80°.

(iii) AD ∥ BC and AC is a transversal.

Therefore, ∠ACB = ∠CAD = 40° [alternate interior angles]

Hence, ∠ACB = 40°.

(iv) ∠BCD = ∠BAD = (35° + 40°) = 75° [opposite angles of a parallelogram]

Now, in ∆CBD, we have

∠BDC + ∠BCD + ∠CBD = 180° [sum of the angles of a triangle is 180°]

⇒ 80° + 75° + ∠CBD = 180°

⇒ 155° + ∠CBD = 180°

⇒ ∠CBD = (180° - 155°) = 25°.

Hence, ∠CBD = 25°.



6. In the adjoining figure, ABCD is a parallelogram, AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB = 90°.

Problems on Parallelogram

Solution:

We know that the sum of two adjacent angles of a parallelogram is 180°

Therefore, ∠A + ∠B = 180° ……………. (i)

Since AO and BO are the bisectors of ∠A and ∠B, respectively, we have

∠OAB = 1/2∠A and ∠ABO = 1/2∠B.

From ∆OAB, we have

∠OAB + ∠AOB + ∠ABO = 180° [Since, sum of the angles of a triangle is 180°]

⇒ ¹/₂∠A + ∠ABO + ¹/₂∠B = 180°

⇒ ¹/₂(∠A + ∠B) + ∠AOB = 180°

⇒ (¹/₂ × 180°) + ∠AOB = 180° [using (i)]

⇒ 90° + ∠AOB = 180°

⇒ ∠AOB = (180° - 90°) = 90°.

Hence, ∠AOB = 90°.



7. The ratio of two sides of a parallelogram is 4 : 3. If its perimeter is 56 cm, find the lengths of its sides.

Solution:

Let the lengths of two sides of the parallelogram be 4x cm and 3x cm respectively.

Then, its perimeter = 2(4x + 3x) cm = 8x + 6x = 14x cm.

Therefore, 14x = 56 ⇔ x = ⁵⁶/₁₄ = 4.

Therefore, one side = (4 × 4) cm = 16 cm and other side = (3 × 4) cm = 12 cm.



8. The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. Find its breadth.

Solution:

Let ABCD be the given rectangle in which length AB = 8 cm and diagonal AC = 10 cm.

Problems on Parallelogram

Since each angle of a rectangle is a right angle, we have

∠ABC = 90°.

From right ∆ABC, we have

AB² + BC² = AC² [Pythagoras’ Theorem]

⇒ BC² = (AC² - AB²) = {(1O)² - (8)²} = (100 - 64) = 36

⇒ BC = √36 = 6cm.

Hence, breadth = 6 cm.



9. In the adjacent figure, ABCD is a rhombus whose diagonals AC and BD intersect at a point O. If side AB = 10cm and diagonal BD = 16 cm, find the length of diagonal AC.

Problems on Parallelogram

Solution:

We know that the diagonals of a rhombus bisect each other at right angles

Therefore, BO = ¹/₂BD = (¹/₂ × 16) cm = 8 cm, AB = 10 cm and ∠AOB = 90°.

From right ∆OAB, we have

AB² = AO² + BO²

⇒ AO² = (AB² – BO²) = {(10) ² - (8)²} cm²

                             = (100 - 64) cm²

                             = 36 cm²

     ⇒ AO = √36 cm = 6 cm.

Therefore, AC = 2 × AO = (2 × 6) cm = 12 cm.



Parallelogram

Parallelogram

Properties of a Rectangle Rhombus and Square

Problems on Parallelogram

Practice Test on Parallelogram


Parallelogram - Worksheet

Worksheet on Parallelogram







8th Grade Math Practice

From Problems on Parallelogram to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Roman Numerals | System of Numbers | Symbol of Roman Numerals |Numbers

    Feb 22, 24 04:21 PM

    List of Roman Numerals Chart
    How to read and write roman numerals? Hundreds of year ago, the Romans had a system of numbers which had only seven symbols. Each symbol had a different value and there was no symbol for 0. The symbol…

    Read More

  2. Worksheet on Roman Numerals |Roman Numerals|Symbols for Roman Numerals

    Feb 22, 24 04:15 PM

    Roman Numbers Table
    Practice the worksheet on roman numerals or numbers. This sheet will encourage the students to practice about the symbols for roman numerals and their values. Write the number for the following: (a) V…

    Read More

  3. Roman Symbols | What are Roman Numbers? | Roman Numeration System

    Feb 22, 24 02:30 PM

    Roman Numbers
    Do we know from where Roman symbols came? In Rome, people wanted to use their own symbols to express various numbers. These symbols, used by Romans, are known as Roman symbols, Romans used only seven…

    Read More

  4. Place Value | Place, Place Value and Face Value | Grouping the Digits

    Feb 19, 24 11:57 PM

    Place-value of a Digit
    The place value of a digit in a number is the value it holds to be at the place in the number. We know about the place value and face value of a digit and we will learn about it in details. We know th…

    Read More

  5. Math Questions Answers | Solved Math Questions and Answers | Free Math

    Feb 19, 24 11:14 PM

    Math Questions Answers
    In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students

    Read More