CBSE Class 9 Mathematics Quadrilaterals Worksheet Set B

Very Short Answer Type Questions:

Question: Two consecutive angles of a parallelogram are (x + 60°) and (2x + 30°). What special name can you give to this parallelogram?
Answer: We know that consecutive interior angles of a parallelogram are supplementary.
∴ (x + 60°) + (2x + 30°) = 180°
∴ 3x + 90° = 180° ⇒ 3x = 90° ⇒ x = 30°
Thus, two consecutive angles are (30° + 60°), (2 × 30° + 30°) i.e., 90° and 90°.
Hence, the special name of the given parallelogram is rectangle.

Question: ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°. Find angles C and D of the trapezium.
Answer:  We have given, a trapezium ABCD, whose parallel sides are AB and DC.

Since, AB || CD and AD is a transversal.
∴ ∠A + ∠D = 180° [Angles on same side of transversal]
⇒ ∠D = 180° – ∠A = 180° – 45° = 135°
Similarly, ∠C = 135°

Question: In the figure, it is given that QLMN and NLRM are parallelograms. Can you say that QL = LR? Why or why not?

Answer: Yes, QL = LR
As, opposite sides of a parallelogram are equal.
∴ In parallelogram QLMN, QL = NM …(i)
In parallelogram NLRM, NM = LR …(ii)
From (i) and (ii), QL = LR

Question: In the given figure, PQRS is a parallelogram in which ∠PSR = 125°. Find the measure of ∠RQT.

Answer:  ∠PQR = ∠PSR = 125°
(∵Opposite angles of a parallelogram are equal) 
Now, ∠PQR + ∠RQT = 180° (Linear pair)
⇒ 125° + ∠RQT = 180° ⇒ ∠RQT = 55°

Question: If one angle of a rhombus is a right angle, then it is necessarily a _______ .
Answer: If one angle of a rhombus is a right angle, then it is necessarily a square.

Short Answer Type Questions:

Question: PQRS is a rhombus with ∠QPS = 50°. Find ∠RQS.
Answer: Since a rhombus satisfies all the properties of a parallelogram.

∴ ∠QPS = ∠QRS [Opposite angles of a parallelogram]
⇒ ∠QRS = 50° [∠QPS = 50° (Given)]
 Diagonals of a rhombus bisect the opposite angles. 
∴ ∠ORQ = 1/2 ∠QRS ⇒ ∠ORQ = 25°
Now, in ΔORQ, we have
∠OQR + ∠ORQ + ∠ROQ = 180°
⇒ ∠OQR + 25° + 90° = 180° [Diagonals of a rhombus are perpendicular to each other ⇒ ∠ROQ = 90°]
⇒ ∠OQR = 180° – 115° = 65°
∴ ∠RQS = 65°

Question: In ΔABC, AD is the median and DE || AB, such that E is a point on AC. Prove that BE is another median.
Answer: In ΔABC, DE || AB and AD is the median.

So, D is the mid-point of BC.
By converse of mid-point theorem,
E is the mid-point of AC.
Hence, BE is median.

Question: In ΔABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
Answer: Let D, E and F be the mid-points of sides BC, CA and AB respectively.
In ΔABC, F and E are mid-points of AB and AC.

∴ FE || BC and FE = 1/2 BC
∴ FE || BD and FE = BD
∴ FEDB is a parallelogram.
Similarly, CDFE and AFDE are also parallelograms.
∴ ∠B = ∠DEF, ∠C = ∠DFE and ∠FDE = ∠A
⇒ ∠DEF = 60°, ∠DFE = 70° and ∠FDE = 50°

Question: In the given parallelogram ABCD, the sum of any two consecutive angles is 180° and opposite angles are equal. Find the value of ∠A.

Answer: In ΔBCD, we have
∠BDC + ∠DCB + ∠CBD = 180° [Angle sum property of a triangle]
⇒ 5a + 9a + 4a = 180°
⇒ 18a = 180° ⇒ a = 10°
∴ ∠C = 9 × 10° = 90°
Since, opposite angles of a parallelogram are equal
Therefore, ∠A = ∠C ⇒ ∠A = 90°

Question: In given figure, ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = 1/4 AC. If PQ produced meet BC at R, then prove that R is a midpoint of BC.

Answer: Suppose AC and BD intersect at O.

Then, OC = 1/2 AC 
Now, CQ = 1/4 AC [Given]
⇒ CQ = 1/2 OC
In ΔCOD, P and Q are the midpoints of DC and OC respectively.
∴ PQ || DO [By mid-point theorem]
Also, in ΔCOB, Q is the midpoint of OC and QR || OB
∴ R is the midpoint of BC. [By converse of mid-point theorem]

Question: In the following figure, AL and CM are medians of ΔABC and LN || CM. Prove that BN = 1/4 AB. 

Answer: We have, AL and CM are medians of ΔABC, i.e., L and M are the mid-points of BC and AB respectively.
∴ LC = BL = 1/2 BC and BM = AM = 1/2 AB …(i)
In ΔBMC, L is the mid-point of BC and LN || CM. So, by converse of mid-point theorem, N is mid-point of BM.
i.e., BN = NM = 1/2 BM …(ii)
From (i) and (ii), we get
BN = 1/2 (1/2AB) ⇒ BN = 1/4 AB 

Question: PQRS is a parallelogram and ∠SPQ = 60°. If the bisectors of ∠P and ∠Q meet at point A on RS, prove that A is the mid-point of RS.

Answer: ∠P + ∠Q = 180° (Adjacent angles of parallelogram)
⇒ 60° + ∠Q = 180° ⇒ ∠Q = 120°
Since, PA and QA are bisectors of angles P and Q
∴ ∠SPA = ∠APQ = 1/2 ∠P = 1/2 × 60° = 30°
And ∠RQA = ∠AQP = 1/2 ∠Q = 1/2 × 120° = 60°
Now, SR || PQ and AP is transversal.
∴ ∠SAP = ∠APQ = 30° [Alternate interior angles]
In ΔASP, we have
∠SAP = ∠APS = 30°
⇒ SP = AS …(i)
(Sides opposite to equal angles are equal)
Similarly, QR = AR …(ii)
But, QR = SP [Opposite sides of parallelogram] …(iii)
From (i), (ii) and (iii), we have AS = AR
⇒ A is the mid-point of SR.

Question: In the given quadrilateral ABCD, X and Y are points on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram.

Answer: Since BXDY is a parallelogram.
∴ XO = YO …(i)
and DO = BO …(ii) [Diagonals of a parallelogram bisect each other]
Also, AX = CY (Given)…(iii)
Adding (i) and (iii), we have XO + AX = YO + CY
⇒ AO = CO …(iv)
From (ii) and (iv), we have
AO = CO and DO = BO
Thus, ABCD is a parallelogram, because diagonals AC and BD bisect each other at O.

Question: Rima has a photo-frame without a photo in the shape of a triangle with sides a, b, c in length. She wants to find the perimeter of a triangle formed by joining the mid-points of the sides of the photo-frame. Find the perimeter of the triangle formed by joining the mid-points of the frame.
Answer:  Let the photo-frame be ABC such that BC = a,
CA = b and AB = c and the mid-points of AB, BC and CA are D, E and F respectively.

We have to determine the perimeter of ΔDEF.
In ΔABC, DF is the line-segment joining the mid-points of sides AB and AC.
By mid-point theorem, DF || BC and DF = BC/2 = a/2
Similarly, DE = AC/2 = b/2 and EF = AB/2 = c/2
∴ Perimeter of ΔDEF = DF + DE + EF
= a/2 + b/2 + c/2 = a + b + c/2

Question: In the given figure, ABCD is a square, side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP.

Answer: Since, ABCD is a square.
∴ AB = BC = CD = DA
Also, PA = AB = BQ
∴ AB = BC = CD = DA = PA = BQ
In ΔPDA and ΔQCB, PA = BQ (Given)
AD = BC (Sides of a square)
∠A = ∠B (Each 90°)
ΔPDA ≅ ΔQCB (By SAS congruency rule)
⇒ PD = QC (By C.P.C.T.) …(i)
∠PDA = ∠QCB (By C.P.C.T.) …(ii)
Now, ∠PDC = ∠PDA + ∠ADC
= ∠PDA + 90° …(iii)
∠QCD = ∠QCB + ∠BCD
= ∠QCB + 90° …(iv)
From (ii), (iii) and (iv), we have ∠PDC = ∠QCD
Now, in ΔPDC and DQCD, PD = QC (Given)
DC = DC (Common)
∠PDC = ∠QCD (Proved above)
ΔPDC ≅ DQCD (By SAS congruency rule)
∴ DQ = CP [By C.P.C.T.)

Question: In the given figure, M and P are the midpoints of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, then find (BC + AC) – AB.

Answer: We have,
MN = 1/2 BC, MP = 1/2 AC and NP = 1/2 AB [By midpoint theorem]
⇒ BC = 6 cm, AC = 5 cm and AB = 7 cm.
The value of (BC + AC) – AB
= (6 + 5) – 7 = 4 cm.

Question: In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that
∠COD = 1/2 (∠A + ∠B).
Answer: In ΔCOD, we have

∠COD + ∠1 + ∠2 = 180°
⇒ ∠COD = 180° – (∠1 + ∠2)
⇒ ∠COD = 180° – (1/2∠C + 1/2∠D)
⇒ ∠COD = 180° – 1/2(∠C + ∠D)
⇒ ∠COD = 180° – 1/2 {360° – (∠A + ∠B)} [∵ ∠A + ∠B + ∠C + ∠D = 360°]
⇒ ∠COD = 1/2 (∠A + ∠B)

Question: In a parallelogram PQRS, if ∠QRS = 2x, ∠PQS = 4x and ∠PSQ = 4x, then find the angles of the parallelogram.
Answer: ∠SPQ = ∠QRS = 2x (∵ Opposite angles of a parallelogram are equal)
In ΔPSQ, ∠PSQ + ∠PQS + ∠SPQ = 180°

⇒ 4x + 4x + 2x = 180°
⇒ 10x = 180°
⇒ x = 18°
Now, ∠PSR = ∠PQR (∵ Opposite angles of a parallelogram are equal)
⇒ 4x + ∠QSR = 4x + ∠SQR
⇒ ∠QSR = ∠SQR …(i)
In DSRQ, ∠SRQ + ∠RSQ + ∠SQR = 180°
⇒ 2 × 18° + 2 ∠RSQ = 180° [From (i)]
⇒ 2 ∠RSQ = 180° – 36° = 144° ⇒ ∠RSQ = 72°
Hence, ∠P = ∠R = 2 × 18° = 36°,
∠Q = ∠S = 4x + 72° = 4 × 18° + 72° = 144°

Long Answer Type Questions:

Question: P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ^ BD. Prove that PQRS is a square.
Answer: In quadrilateral ABCD, AC ⊥ BD and AC = BD.
In ΔADC, S and R are the midpoints of the sides AD and DC respectively.

∴ SR || AC and SR = 1/2 AC …(i) [By mid-point theorem]
In ΔABC, P and Q are the midpoints of AB and BC respectively.
∴ PQ|| AC and PQ = 1/2 AC …(ii) [By mid-point theorem]
From (i) and (ii), PQ || SR
and PQ = SR = 1/2 AC …(iii)
Similarly, in ΔABD,
SP || BD and SP = 1/2 BD [By mid-point theorem]
∴ SP = 1/2 AC [∴ AC = BD] …(iv)
Now in ΔBCD, RQ || BD and RQ = 1/2 BD [By mid-point theorem]
∴ RQ = 1/2 AC [∴ BD = AC] …(v)
From (iv) and (v), SP = RQ = 1/2 AC …(vi)
From (iii) and (vi), PQ = SR = SP = RQ …(vii)
∴ All four sides are equal.
Now, in quadrilateral OERF,
OE || FR and OF || ER
∴ ∠EOF = ∠ERF = 90° [AC ⊥ DB]
∴ ∠QRS = 90° …(viii)
From (vii) and (viii), we get PQRS is a square.

Question: Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides.
Answer:  Let, trapezium ABCD in which, AB || DC and P and Q are the mid-points of its diagonals AC and BD respectively.
We have to prove (i) PQ || AB and PQ || DC

(ii) PQ = 1/2( AB − DC) 
Join D and P and produce DP to meet AB at R.
(i) Since AB || DC and transversal 
AC cuts them at A and C respectively.
∴ ∠1 = ∠2 (Alternate angles) ….(1)
In ΔAPR and ΔCPD,
∠1 = ∠2 (From (1))
AP = CP (Q P is the mid-point of AC)
∠3 = ∠4 (Vertically opposite angles)
∴ ΔAPR ≅ ΔCPD (By ASA congruence rule)
⇒ AR = DC and PR = DP (By C.P.C.T.)
In ΔDRB, P and Q are the mid-points of side DR and DB respectively.
∴ PQ || RB (By mid-point theorem)
⇒ PQ || AB (Q RB is a part of AB)
⇒ PQ || AB and PQ || DC (Q AB || CD)
(ii) In ΔDRB, P and Q are the mid-points of side DR and DB respectively.
∴ PQ = 1/2 RB (By mid-point theorem)
⇒ PQ = 1/2 (AB − AR) ⇒ PQ = 1/2(AB −DC) [From part (i), AR = DC]