Class 9 Maths Chapter 8 Previous Year Questions - Quadrilaterals

Sol: Given: one of the angles = 140°

 Remaining three angles = 360° - 140° = 220°
Ratio is 3 : 3 : 2
∴ The smallest angle 

Sol:

AB || DC we have
∠A + ∠D = 180°
55°+ ∠D = 180°
∠D = 180°-55°
= 125°
also,
∠B + ∠C = 180°
70° + ∠C = 180 °
∠C = 180° -70°
= 110°

 Sol: ∵ Opposite sides of a parallelogram are equal.
∴ AB = CD = 4.5 cm, and BC = AD

Now, AB + CD + BC + AD = 21 cm
⇒ AB + AB + BC + BC = 21 cm
⇒2[AB + BC] = 21 cm
⇒ 2[4.5 cm + BC] = 21 cm
⇒ [4.5 cm + BC] = (21/2)
= 11.5 cm ⇒ BC = 11.5 ∠ 4.5 = 7 cm
Thus, BC = 7 cm, CD = 4.5 cm and AD = 7 cm.

Sol: Since, the adjacent angles of a parallelogram are supplementary.

∴ ∠ B + ∠ C = 180° ( adjacent angles )
⇒ (3x - 10)° + (2x + 10)° = 180°
⇒ 3x + 2x - 10° + 10° = 180°
⇒ 5x = 180°
⇒ x= (180°/5)= 36°
Thus, the required value of x is 36°.

Sol: Since DE || BC and D is the mid-point of AB.
∴ E must be the mid-point of AC.
∴ AE = EC ⇒ x = 5 cm

Also, DE || BC ⇒ DE = (1/2)BC
∴ 2DE = 2(1/2)BC
⇒ 2DE = BC
⇒ 2 x 6 cm = BC or BC = 12 cm
⇒ y = 12 cm
Thus, x = 5 cm and y = 12 cm

 Sol: Let us join BE and extend it meet CD produced at P.

In ΔAEB and ΔDEP, we get AB || PC and BP is a transversal,
∴ ∠ABE = ∠EPD             [Alternate angles]
AE = ED            [∵ E is midpoint of AB]
∠AEB = ∠PED             [Vertically opp. angles]
⇒ ΔAEB ≌ ΔDEP
⇒ BE = PE and AB = DP [SAS]
⇒ BE = PE and AB = DP
Now, in ΔEPC, E is a mid point of BP and F is mid point of BC
∴ EF || PC and EF =(1/2)PC            [Mid point theorem]
i.e., EF || AB and EF = (1/2) (PD + DC)
= (1/2) (AB + DC)
Thus, EF || AB and EF = (1/2) (AB + DC)

 Sol:

 Since ∠ A : ∠ B : ∠ C : ∠ D = 1 : 2 : 3 : 4

∴ If ∠ A = x, then ∠ B = 2x, ∠ C = 3x and ∠ D = 4x. ∴ ∠ A + ∠ B + ∠ C + ∠ D = 360°
⇒ x + 2x + 3x + 4x = 360° ⇒ 10x = 36°
⇒ x= (360°/10)= 36°
Therefore meausre of all the angles are :
∠ A = x = 36° 
∠ B = 2x = 2 x 36° = 72° 
∠ C = 3x = 3 x 36° = 108° 
∠ D = 4x = 4 x 36° = 144°

Sol:
Here, in ΔABC, AB = 8 cm, BC = 9 cm, AC = 10 cm.
In ΔAOB, X and Y are the mid-points of AO and BO.
∴ By using mid-point theorem, we have
XY = 1/2 AB = 1/2 x 8 cm = 4 cm
Similarly, in Δ𝜏BOC, Y and Z are the mid-points of BO and CO.
∴ By using mid-point theorem, we have
YZ = 1/2 BC = 1/2 x 9cm = 4.5 cm
And, in Δ𝜏COA, Z and X are the mid-points of CO and AO.
∴ ZX = 1/2 AC = 1/2 x 10 cm = 5 cm
Hence, the lengths of the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.

Sol:
Here, E and F are the mid-points of AD and BC respectively.
In ΔBFG and ΔCFD
BF = CF [given]
∠BFG = ∠CFD (vert. opp. ∠s]
∠BGF = ∠CDF (alt. int. ∠s, as AB || DC)
So, by using AAS congruence axiom, we have
ΔBFG ≅ ΔCFD
⇒ DF = FG [c.p.c.t.)
Now, in ΔAGD, E and F are the mid-points of AD and GD.
∴ By mid-point theorem, we have
EF || AG
or EO || AB
Also, in ΔADC, EO || DC
∴ EO is a line segment from mid-point of one side parallel to another side.
Thus, it bisects the third side.
Hence, AO = CO

Sol:
Since AE = DE
∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ]
Again, BC || AD
∠EBC = ∠A …. (ii) (corresponding ∠s]
From (i) and (ii), we have
∠D = ∠EBC …. (iii)
But ∠EBC + ∠ABC = 180° (a linear pair]
∠D + ∠ABC = 180° (using (iii)]
Now, a pair of opposite angles of quadrilateral ABCD is supplementary
Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA
∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD]
∠BAD = ∠CDA [using (i)]
AD = AD (common]
So, by using AAS congruence axiom, we have
ΔABD ≅ ΔDCA
Hence, BD = CA [c.p.c.t.]