# RS Aggarwal Class 9 Solutions Chapter 9 - Quadrilaterals And Parallelograms

## RS Aggarwal Class 9 Chapter 9 - Quadrilaterals And Parallelograms Solutions Free PDF

A quadrilateral is a polygon which contains four side whereas, a parallelogram is a type of quadrilateral, whose pairs of opposite sides are parallel. One of the most common properties of all quadrilaterals is that sum of all the angles is 360 degrees. Some of the different types of quadrilaterals are:

1. Rhombus
2. Trapezium
3. Square
4. Parallelogram
5. Rectangle

Parallelograms have some unique properties such as:

• Opposite sides are congruent
• Opposite angles are congruent
• It has adjacent angles which are supplementary
• It has diagonals which bisect each other.

Students must practice these solutions to gain more marks in the exam. We have provided here all the exercise questions of RS Aggarwal Maths solution to help you to revise the complete syllabus. It is an important tool to prepare effectively for the Class 9 Maths examination. All the solutions are explained in a detailed step-by-step manner for the better understanding of the students.

Q.1: Three angles of a quadrilateral measure 56Â°, 115Â°, 84Â°. Find the measure of the fourth angle.

Sol:

Let, the measure of the fourth angle be Xâˆ˜$X^{\circ}$

Since the sum of the angle of a quadrilateral is 360âˆ˜$360^{\circ}$, we have:

56 + 115 + 84 + X = 360

â‡’$\Rightarrow$ 255 + X = 360

â‡’$\Rightarrow$ X = 105

Hence, the measure of the fourth angle is 105âˆ˜$105^{\circ}$

Q.2: The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the angles.

Sol:

Let,Â âˆ A$\angle A$ = 2Xâˆ˜$2X^{\circ}$

Then âˆ B=(4X)âˆ˜$\angle B = (4X)^{\circ}$

âˆ C=(5X)âˆ˜$\angle C = (5X)^{\circ}$ and

âˆ D=(7X)âˆ˜$\angle D = (7X)^{\circ}$

Since the sum of the angles of a quadrilateral is 360âˆ˜$360^{\circ}$, we have:

2X + 4X + 5X + 7X = 360âˆ˜$360^{\circ}$

â‡’$\Rightarrow$ 18X = 360âˆ˜$360^{\circ}$

â‡’$\Rightarrow$ X = 20âˆ˜$20^{\circ}$

Therefore, Angle A = 40âˆ˜$40^{\circ}$

Angle B = 80âˆ˜$80^{\circ}$

Angle C = 100âˆ˜$100^{\circ}$

Angle D = 140âˆ˜$140^{\circ}$

Q.3: In the adjoining figure, ABCD is a trapezium in which AB âˆ¥$\parallel$ DC. If âˆ  A = 55Â°and âˆ  B = 70Â°, find âˆ  C and âˆ  D.

Sol:

We have ABâˆ¥DC$AB\parallel DC$

Angle A and Angle D are the interior angles on the same side of the transversal line AD, whereas angle B and Angle C are the same interior angles on the same side of transversal line BC.

Now, Angle A + Angle D = 180âˆ˜$180^{\circ}$

â‡’$\Rightarrow$ âˆ D$\angle D$ = 180âˆ˜$180^{\circ}$ â€“ âˆ A$\angle A$

Therefore, âˆ D$\angle D$ = 180âˆ˜$180^{\circ}$ â€“ 55âˆ˜$55^{\circ}$ = 125âˆ˜$125^{\circ}$

Again, âˆ B$\angle B$ + âˆ C$\angle C$ = 180âˆ˜$180^{\circ}$

â‡’$\Rightarrow$ âˆ C$\angle C$ = 180âˆ˜$180^{\circ}$ â€“ âˆ B$\angle B$

Therefore, âˆ C$\angle C$ = 180âˆ˜$180^{\circ}$ â€“ 70âˆ˜$70^{\circ}$ = 110âˆ˜$110^{\circ}$

Q.4: In the adjoining figure, ABCD is a square and Â Î”EDC$\Delta EDC$ is an equilateral triangle. Prove that (i) AB = BE, (ii) âˆ  DAE = 15Â°.

Sol:

ABCD is a square in which AB = BC = CD = DA. Î”EDC$\Delta EDC$ is an equilateral triangle in which ED = EC = DC

âˆ EDC$\angle EDC$ = âˆ DEC$\angle DEC$ = âˆ DCE$\angle DCE$ = 60âˆ˜$60^{\circ}$.

To prove: AE = BE and âˆ DAE$\angle DAE$ = 15âˆ˜$15^{\circ}$

Proof: In Î”ADE$\Delta ADE$ and Î”BCE$\Delta BCE$, we have:

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

âˆ ADE$\angle ADE$ = âˆ BCE$\angle BCE$ = 90âˆ˜$90^{\circ}$ + 60âˆ˜$60^{\circ}$ = 150âˆ˜$150^{\circ}$

Î”ADE$\Delta ADE$ â‰…$\cong$ Î”BCE$\Delta BCE$

i.e., AE = BE

Now, âˆ ADE$\angle ADE$ = 150âˆ˜$150^{\circ}$

DA = DC [Sides of a square]

DC = DE [Sides of an equilateral triangle]

So, DA = DE

Î”ADE$\Delta ADE$ and Î”BCE$\Delta BCE$ are isosceles triangles.

i.e., âˆ DAE$\angle DAE$ = âˆ DEA$\angle DEA$

= 12(180âˆ˜âˆ’150âˆ˜)$\frac{1}{2}(180^{\circ}-150^{\circ})$

= 30âˆ˜2$\frac{30^{\circ}}{2}$

= 15âˆ˜$15^{\circ}$

Q.5: In the adjoining figure, BM âŠ¥ AC and DN âŠ¥ AC. If BM = DN, prove that AC bisects BD.

Sol:

A quadrilateral ABCD, in which BM âŠ¥$\perp$ AC and DN âŠ¥$\perp$ AC and BM = DN

To prove: AC bisects BD, or DO = BO

Proof: Let, AC and BD intersect at O

Now, in Î”OND$\Delta OND$ and Î”OMB$\Delta OMB$, we have:

âˆ OND$\angle OND$ = âˆ OMB$\angle OMB$ (90âˆ˜$90^{\circ}$ each)

âˆ DON$\angle DON$ = âˆ BOM$\angle BOM$ (Vertically opposite angles)

Also, DN = BM (Given)

i.e., Î”OND$\Delta OND$ â‰…$\cong$ Î”OMB$\Delta OMB$ (AAS congruence rule)

Therefore, OD = OB (CPCT)

Hence, AC bisects BD

Q.6: In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that:

(i) AC bisects âˆ  A and âˆ  C, Â (ii) BE = DE, (iii) Â âˆ  ABC = âˆ ADC.

Sol:

Given: ABCD is a quadrilateral in which AB = AD and BC = DC

(i) In Î”ABC$\Delta ABC$ and Î”ADC$\Delta ADC$, we have:

BC = DC (Given)

AC is common

i.e., Î”ABC$\Delta ABC$ â‰…$\cong$ Î”ADC$\Delta ADC$ (SSS Congruence rule)

Therefore, âˆ BAC$\angle BAC$ = âˆ DAC$\angle DAC$ and

âˆ BCA$\angle BCA$ = âˆ DCA$\angle DCA$ (By CPCT)

Thus, AC bisects âˆ A$\angle A$ and âˆ C$\angle C$

(ii) Now, in Î”ABE$\Delta ABE$ and Î”ADE$\Delta ADE$, we have:

âˆ BAE$\angle BAE$ = âˆ DAE$\angle DAE$ (Proven above)

AE is common.

Therefore, Î”ABE$\Delta ABE$ â‰…$\cong$ Î”ADE$\Delta ADE$ (SAS congruence rule)

â‡’$\Rightarrow$ BE = DE (By CPCT)

(iii) Î”ABC$\Delta ABC$ â‰…$\cong$ Î”ADC$\Delta ADC$ (Proven above)

Therefore, Angle ABC = Angle ADC (By CPCT)

Q.7: In the given figure, ABCD is a square and âˆ PQR = 90Â°. If PB = QC = DR, prove that (i) QB = RC, (ii) PQ = QR (iii) âˆ  QPR = 45Â°

Sol:

Let, ABCD be a parallelogram

âˆ  A = âˆ  C and âˆ  B = âˆ  D (Opposite angles)

Let âˆ  A be the smallest angle whose measure is xÂ°

âˆ´ âˆ B = (2x â€“ 30)Â°

Now, âˆ  A + âˆ  B = 180Â° (Adjacent angles are supplementary)

â‡’$\Rightarrow$ x + 2x â€“ 30Â° = 180Â°

â‡’$\Rightarrow$ 3x = 210Â°

â‡’$\Rightarrow$ x = 70Â°

âˆ´ âˆ B = 2 x 70Â°- 30Â° = 110Â°

Hence, âˆ  A = âˆ  C = 70Â°; âˆ  B = âˆ  D = 110Â°

Q.8: If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.

Sol:

Let, ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral

Join O with A, B, C, and D

We know that the sum of any sides of a triangle is greater than the third side

So, in Î”AOC$\Delta AOC$, OA + OC > AC

Also, Î”BOD$\Delta BOD$, OB + OD > BD

(OA + OC) + (OB + OD) > (AC + BD)

â‡’$\Rightarrow$ OA + OB + OC + OD > AC + BD

Q.9: In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that: (i) AB + BC + CD + DA > 2AC (ii) AB + BC + CD > DA (iii) AB + BC + CD + DA > AC + BD

Sol:

Given: ABCD is a quadrilateral and AC is one of its diagonals

(i) In Î”ABC$\Delta ABC$, AB + BC > AC . . . . . . . . (1)

In Î”ACD$\Delta ACD$, CD + DA > AC . . . . . . . . . . (2)

Adding inequalities (1) and (2), we get:

AB + BC + CD + DA > 2AC

(ii) In Î”ABC$\Delta ABC$, we have :

AB + BC > AC . . . . . . . . (1)

We also know that the length of each side of a triangle is greater than the positive difference of the the other two sides.

In Î”ACD$\Delta ACD$, we have:

AC > |DAâ€“CD|$\left | DA â€“ CD \right |$ . . . . . . . (2)

From (1) and (2), we have:

AB + BC > |DAâ€“CD|$\left | DA â€“ CD \right |$

â‡’$\Rightarrow$ AB + BC + CD > DA

(iii) In Î”ABC$\Delta ABC$, AB + BC > AC

In Î”ACD$\Delta ACD$, CD + DA > AC

Î”BCD$\Delta BCD$, BC + CD > BD

Î”ABD$\Delta ABD$, DA + AB > BD

2 (AB + BC + CD + DA) > 2 (AC + BD)

â‡’$\Rightarrow$ (AB + BC + CD + DA) > (AC + BD)

Q.10: Prove that the sum of all angles of a quadrilateral is 360Â°.

Sol: Let ABCD be a quadrilateral and âˆ 1, âˆ 2, âˆ 3 and âˆ 4 are its four angles as shown in the figure.

Join BD which divides ABCD in two triangles, Î”ABD$\Delta ABD$ and Î”BCD$\Delta BCD$.

In Î”ABD$\Delta ABD$, we have:

âˆ  1 + âˆ  2 + Â âˆ  A = 180Â° . . . . . . . . (i)

In Î”BCD$\Delta BCD$, we have:

âˆ  3 + âˆ  4 + âˆ  C = 180Â° . . . . . . . (ii)

On adding (i) and (ii), we get:

(âˆ 1 + âˆ 3) + âˆ A + âˆ C + (âˆ  4 + âˆ 2) = 360Â°

â‡’$\Rightarrow$ âˆ  A + âˆ  C + âˆ  B + âˆ  D = 360Â° Â Â ( Since,âˆ  1 + âˆ  3 = âˆ  B; âˆ  4 + âˆ  2 = âˆ  D)

âˆ´Â âˆ  A + âˆ  C + âˆ  B + âˆ  D = 360Â°

### RS Aggarwal Class 9 Solutions Chapter 9 – Quadrilaterals And Parallelograms

The RS Aggarwal Class 9 Solutions provides all the answers in a proper stepwise method which makes you well versed with all the solutions including the tougher ones. If you want to score good marks in your Class 9 examination, then, practicing RS Aggarwal Class 9 Maths solutions is an utmost necessity. It is one of the best study material when it comes a providing a question bank to practice from and also test the students understanding of concepts.