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:
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.
Check out the RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms below:
RS Aggarwal Class 9 Chapter 9
Q.1: Three angles of a quadrilateral measure 56Â°, 115Â°, 84Â°. Find the measure of the fourth angle.
Let, the measure of the fourth angle be Xâ
Since the sum of the angle of a quadrilateral is 360â, we have:
56 + 115 + 84 + X = 360
â 255 + X = 360
â X = 105
Hence, the measure of the fourth angle is 105â
Q.2: The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the angles.
Let,Â â A = 2Xâ
Then â B=(4X)â
â C=(5X)â and
Since the sum of the angles of a quadrilateral is 360â, we have:
2X + 4X + 5X + 7X = 360â
â 18X = 360â
â X = 20â
Therefore, Angle A = 40â
Angle B = 80â
Angle C = 100â
Angle D = 140â
Q.3: In the adjoining figure, ABCD is a trapezium in which AB â¥ DC. If â A = 55Â°and â B = 70Â°, find â C and â D.
We have ABâ¥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 = 180â
ââ D = 180â â â A
Therefore, â D = 180â â 55â = 125â
Again, â B + â C = 180â
ââ C = 180â â â B
Therefore, â C = 180â â 70â = 110â
Q.4: In the adjoining figure, ABCD is a square and Â ÎEDC is an equilateral triangle. Prove that (i) AB = BE, (ii) â DAE = 15Â°.
ABCD is a square in which AB = BC = CD = DA. ÎEDC is an equilateral triangle in which ED = EC = DC
â EDC = â DEC = â DCE = 60â.
To prove: AE = BE and â DAE = 15â
Proof: In ÎADE and ÎBCE, we have:
AD = BC [Sides of a square]
DE = EC [Sides of an equilateral triangle]
â ADE = â BCE = 90â + 60â = 150â
i.e., AE = BE
Now, â ADE = 150â
DA = DC [Sides of a square]
DC = DE [Sides of an equilateral triangle]
So, DA = DE
ÎADE and ÎBCE are isosceles triangles.
i.e., â DAE = â DEA
Q.5: In the adjoining figure, BM â¥ AC and DN â¥ AC. If BM = DN, prove that AC bisects BD.
A quadrilateral ABCD, in which BM â¥ AC and DN â¥ AC and BM = DN
To prove: AC bisects BD, or DO = BO
Proof: Let, AC and BD intersect at O
Now, in ÎOND and ÎOMB, we have:
â OND = â OMB (90â each)
â DON = â BOM (Vertically opposite angles)
Also, DN = BM (Given)
i.e., ÎONDâ ÎOMB (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 â A and â C, Â (ii) BE = DE, (iii) Â â ABC = â ADC.
Given: ABCD is a quadrilateral in which AB = AD and BC = DC
(i) In ÎABC and ÎADC, we have:
AB = AD (Given)
BC = DC (Given)
AC is common
i.e., ÎABCâ ÎADC (SSS Congruence rule)
Therefore, â BAC = â DAC and
â BCA = â DCA (By CPCT)
Thus, AC bisects â A and â C
(ii) Now, in ÎABE and ÎADE, we have:
AB = AD (Given)
â BAE = â DAE (Proven above)
AE is common.
Therefore, ÎABEâ ÎADE (SAS congruence rule)
â BE = DE (By CPCT)
(iii) ÎABCâ ÎADC (Proven above)
Therefore, Angle ABC = Angle ADC (By CPCT)
Q.7: In the given figure, ABCD is a square and â PQR = 90Â°. If PB = QC = DR, prove that (i) QB = RC, (ii) PQ = QR (iii) â QPR = 45Â°
Let, ABCD be a parallelogram
â A = â C and â B = â D (Opposite angles)
Let â A be the smallest angle whose measure is xÂ°
â´ â B = (2x â 30)Â°
Now, â A + â B = 180Â° (Adjacent angles are supplementary)
â x + 2x â 30Â° = 180Â°
â 3x = 210Â°
â x = 70Â°
â´ â B = 2 x 70Â°- 30Â° = 110Â°
Hence, â A = â C = 70Â°; â B = â D = 110Â°
Q.8: If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.
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 ÎAOC, OA + OC > AC
Also, ÎBOD, OB + OD > BD
Adding these inequalities, we get:
(OA + OC) + (OB + OD) > (AC + BD)
â 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
Given: ABCD is a quadrilateral and AC is one of its diagonals
(i) In ÎABC, AB + BC > AC . . . . . . . . (1)
In ÎACD, CD + DA > AC . . . . . . . . . . (2)
Adding inequalities (1) and (2), we get:
AB + BC + CD + DA > 2AC
(ii) In ÎABC, 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 ÎACD, we have:
AC > |DAâCD| . . . . . . . (2)
From (1) and (2), we have:
AB + BC > |DAâCD|
â AB + BC + CD > DA
(iii) In ÎABC, AB + BC > AC
In ÎACD, CD + DA > AC
ÎBCD, BC + CD > BD
ÎABD, DA + AB > BD
Adding these inequalities, we get:
2 (AB + BC + CD + DA) > 2 (AC + BD)
â (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 â 1, â 2, â 3 and â 4 are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, ÎABD and ÎBCD.
In ÎABD, we have:
â 1 + â 2 + Â â A = 180Â° . . . . . . . . (i)
In ÎBCD, we have:
â 3 + â 4 + â C = 180Â° . . . . . . . (ii)
On adding (i) and (ii), we get:
(â 1 + â 3) + â A + â C + (â 4 + â 2) = 360Â°
â â A + â C + â B + â D = 360Â° Â Â ( Since,â 1 + â 3 = â B; â 4 + â 2 = â D)
â´Â â A + â C + â B + â D = 360Â°