PSEB Solutions for Class 9 Maths Chapter 8 Quadrilaterals
PSEB 9th Class Maths Solutions Chapter 8 Quadrilaterals
Ex 8.1
Question 1.
The angles of a quadrilateral are in the ratio 3: 5 : 9: 13. Find all the angles of the quadrilateral.
Answer:
Let, ABCD be a given quadrilateral.
∴ ∠A : ∠B : ∠C : ∠D = 3 : 5 : 9 : 13
Sum of ratios = 3 + 5 + 9 + 13 = 30
In quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360°
∴ ∠A = 3/30 × 360° = 3 × 12 = 36°
∴ ∠B = 5/30 × 360° = 5 × 12 = 60°
∴ ∠C = 9/30 × 360° = 9 × 12 = 108°
∴ ∠D = 13/30 × 360° = 13 × 12 = 156°
Thus the angles of the given quadrilateral are 36°, 60°, 108° and 156°.
Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer:
In parallelogram ABCD, diagonals are equal.
∴ AC = BD.
In ∆ DAB and ∆ CBA.
DA = CB (Theorem 8.2)
AB = BA (Common)
DB = CA (Given)
∴ ∆ DAB ≅ ∆ CBA sss rule)
∴ ∠ DAB = ∠CBA (CPCT)
In parallelogram ABCD, AD || BC and AB is their transversal.
∴ ∠ DAB + ∠ CBA = 180°
(Interior angles on the same side of transversal)
Thus, in parallelogram ABCD, two angles ∠A and∠B are right angles. Hence, all the angles are right angle.
Hence, the parallelogram ABCD having equal diagonals is a rectangle.
Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer:
In quadrilater ABCD. diagonals AC and BD bisect each other at M at right angles.
∴ AM = CM, BM = DM and
∠AMB = ∠CMB = ∠CMD = ∠AMD = 90°.
In ∆ AMB and ∆ CMB,
AM = CM
∠ AMB = ∠CMB
BM = BM (Common)
∴ ∆ AMB ≅ ∆ CMB (SAS rule)
∴ AB = CB (CPCT)
Similarly, proving ∆ BMC ≅ ∆ DMC and ∆ DMA ≅ ∆ BMA, we get BC = DC and DA = BA.
Thus, in quadrilateral. ABCD.
AB = BC CD = DA.
Therefore, quadrilateral ABCD is a rhombus.
Thus, if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Answer:
ABCD is a square in which diagonals AC and BD intersect at M.
Every square is a parallelogram.
∴ AC and BD bisect each other. …………… (1)
In ∆ DAB and ∆ CBA,
DA = CB (Sides of a square)
∠ DAB = ∠ CBA (Right angles in a square)
AB = BA (Common)
∴ ∆ DAB ≅ ∆ CBA (SAS rule)
∴ BD = AC (CPCT) ……………….. (2)
Now, in ∆ AMB and ∆ CMB,
AM = CM (BD bisects AC at M).
BM = BM (Common)
AB = CB (Sides of a square)
∴ ∆ AMB ≅ ∆ CMB (SSS rule)
∴ ∠ AMB = ∠CMB (CPCT)
But, ∠ AMB and ∠ CMB form a linear pair.
∴ ∠ AMB + ∠ CMB = 180°
Hence, ∠AMB = ∠ CMB = 90° (3)
(1), (2) and (3) taken together proves that the diagonals of a square are equal and bisect each other at right angles.
Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer:
In quadrilateral ABCD, diagonals AC and BD are equal and bisect each other at right angles.
∴ AC = BD,
MA = MC = MB = MD = 1/2 AC = 1/2 BD and
∠AMB = ∠CMB= ∠DMC = ∠DMA= 90°.
In ∆ AMB and ∆ CMB,
AM = CM
∠ AMB = ∠ CMB (Right angles)
BM = BM (Common)
∴ ∆ AMB ≅ ∆ CMB (SAS rule)
∴ AB = CB (CPCT)
Similarly, we can prove that BC = DC and
DA = BA.
Thus, in quadrilateral ABCD,
AB = BC = CD = DA …………… (1)
Now, in ∆ DAB and ∆ CBA,
DA = C B
BD = AC (Given)
AB = BA (Common)
∴ ∆ DAB ≅ ∆ CBA (SSS rule)
∴ ∠DAB = ∠CBA (CPCT)
Thus, in quadrilateral ABCD, ∠A = ∠B.
Similarly, we can prove that ∠B = ∠C and ∠C = ∠D.
Thus, in quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D.
Moreover. In quadrilateral ABCD,
∠A + ∠B + ∠C + ∠D = 360°
∴ ∠A = ∠B = ∠C = ∠D = 360°/4 = 90° ……………… (2)
Thus, (1) and (2) taken together proves that in quadrilateral ABCD, all the sides are equal and all the angles are equal.
Therefore, quadrilateral ABCD is a square.
Thus, if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.
Answer:
Diagonal AC of parallelogram ABCD bisects ∠A.
∴ ∠DAC = ∠BAC …………… (1)
Now, ∠BAC and ∠DCA are alternate angles formed by transversal AC of AB || CD.
∴ ∠BAC = ∠DCA …………… (2)
Similarly, ∠DAC and ∠BCA are alternate angles formed by transversal AC of AD || BC.
∴ ∠DAC = ∠BCA ……………… (3)
From (1), (2) and (3),
∠DCA = ∠BCA.
But, ∠DCA + ∠BCA = ∠BCD (Adjacent angles)
∴ AC bisects ∠C also.
In parallelogram ABCD,
∠A = ∠C (Theorem 8.4)
∴ 1/2∠A = 1/2∠C
∴ ∠ DAC = ∠ DCA
∴ In ∆ DAC, DA = DC (Sides opposite to equal angles)
Moreover, in parallelogram ABCD,
AB = CD and BC = DA (Theorem 8.2)
∴ AB = BC = CD = DA
Thus. In parallelogram ABCD, all the sides are equal.
Hence, ABCD is a rhombus.
Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Answer:
ABCD is a rhombus
∴ AB || DC, BC || AD and AB = BC = CD = DA.
AB || DC and AC is their transversal.
∴ ∠CAB = ∠ACD (Alternate angles)
In, ∆ DAC, CD = DA
∴ ∠ACD = ∠CAD
Then, ∠CAB = ∠CAD
But, ∠CAB + ∠CAD = ∠ DAB (Adjacent angles)
∴ ∠ CAB = ∠CAD = 1/2 ∠DAB
This shows that AC bisects ∠A.
Again, BC || AD and AC is their transversal.
∴ ∠ BCA = ∠ DAC (Alternate angles)
In, ∆ DAC, DA = DC
∴ ∠ DAC = ∠ DCA
Then, ∠BCA = ∠DCA
But, ∠ BCA + ∠ DCA = ∠ DCB (Adjacent angles)
∴ ∠ BCA = ∠ DCA = 1/2∠ DCB
This shows that AC bisects ∠C.
Thus, AC bisects ∠A as well as ∠C.
Similarly, taking BD as transversal of AB || DC, and BC || AD, it can be proved that BD bisects ∠B as well as ∠D.
Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square. (ii) Diagonal BD bisects ∠B as well as ∠D.
Answer:
In rectangle ABCD, AB = CD, BC = AD, AB || CD and BC || AD.
AC bisects ∠A as well as ∠C.
∴ ∠DAC = ∠BAC = 1/2∠A and
∠ DCA = ∠ BCA = 1/2∠C
Now, AB || CD and AC is their transversal.
∴ ∠ BAC = ∠ DCA (Alternate angles)
∴ ∠ DAC = ∠ DCA
Thus, in ∆ DAC, ∠DAC = ∠DCA
∴ AD = CD (Sides opposite to equal angles)
From this, we get AB = BC = CD = DA.
Also, in rectangle ABCD,
∠A = ∠B = ∠C = ∠D = 90°
Hence, ABCD is a square. …..Result (i)
In ∆ BCD, BC = CD
∴ ∠ CBD = ∠ CDB
Moreover, AB || CD and BD is their transversal.
∴ ∠ CDB = ∠ ABD (Alternate angles)
∴ ∠ CBD = ∠ ABD
Now, ∠ CBD + ∠ ABD = ∠ ABC
∴ ∠ CBD = ∠ ABD = 1/2 ∠ ABC
Thus, BD bisects ∠B.
Similarly, diagonal BD bisects ∠ D.
Hence, diagonal BD bisects ∠B as well as ∠D …….. Result (ii)
Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:
(i) ∆ APD ≅ ∆ CQB
(ii) AP = CQ
(iii) ∆ AQB ≅ ∆ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.
Answer:
ABCD is a parallelogram.
∴ AD || BC and BD is their transversal.
∴ ∠ADB = ∠CBD (Alternate angles)
∴ ∠ADP = ∠CBQ …………… (1)
Similarly, CD || BA and BD is their transversal.
∴ ∠ ABD = ∠ CDB (Alternate angles)
∴ ∠ABQ = ∠CDP ……………… (2)
In ∆ APD and ∆ CQB,
AD = CB (Opposite sides of a parallelogram)
∠ ADP = ∠ CBQ [by (1)]
DP = BQ (Given)
∴ ∆ APD ≅ ∆ CQB (SAS rule) ……. Result (i)
∴ AP = CQ (CPCT) …… Result (ii)
In ∆ AQB and ∆ CPD,
AB = CD (Opposite sides of a parallelogram)
∠ ABQ = ∠ CDP [by (2)]
BQ = DP (Given)
∴ ∆ AQB ≅ ∆ CPD (SAS rule) …….. Result (iii)
∴ AQ = CP (CPCT) ………….. Result (iv)
Now, in quadrilateral APCQ, AP = CQ and AQ = CP
Hence, by theorem 8.3, APCQ is a parallelogram. ………. Result (v)
Question 10.
ABCD is a parallelogram and AP and Cg are perpendiculars from vertices A and C on diagonal BD (see the given figure). Show that
(i) ∆ APB ≅ ∆ CQD
(ii) AP = CQ
Answer:
In parallelogram ABCD, AB || CD and BD is their transversal.
∴ ∠ ABD = ∠ CDB (Alternate angles)
∴ ∠ABP = ∠CDQ ……………. (1)
Now, in ∆ APB and ∆ CQD,
AB = CD (Opposite sides of a parallelogram)
∠ ABP = ∠ CDQ [by (1)]
∠ APB = ∠ CQD (Right angles)
∆ APB ≅ ∆ CQD (AAS rule) ………… Result (i)
∴ AP = CQ (CPCT) ……….. Result (ii)
Question 11.
In ∆ ABC and ∆ DBF, AB = DE, AB || DE, j BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that:
(i) Quadrilateral ABED is a parallelogram
(ii) Quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) Quadrilateral ACFD is a parallelogram
(v ) AC = DF
(vi) ∆ ABC ≅ ∆ DEF.
Answer:
In quadrilateral ∆ BED, AB = DE and AB || DE. Thus, in quadrilateral ABED, sides in one s pair of opposite sides are equal and parallel. Hence, by theorem 8.8, quadrilateral ABED is a parallelogram. …… Result (i)
Similarly, in quadrilateral BEFC, BC = EF and BC || EF.
Hence, by theorem 8.8, quadrilateral BEFC is a parallelogram. …………. Result (ii)
In parallelogram ABED, AD || BE and in parallelogram BEFC, BE || CE Thus, AD and CF both are parallel to BE.
∴ AD || CF ……….(1)
In parallelogram ABED, AD = BE and in parallelogram BEFC, BE = CF.
∴ AD = CF ……… (2)
Taking (1) and (2) together, we get
AD || CF and AD = CF ………. Result (iii)
In quadrilateral ACFD, AD || CF and AD = CF. Hence, by theorem 8.8, quadrilateral ACFD is a parallelogram. ………. Result (iv)
AC and DF are opposite sides of parallelogram ACFD.
∴ AC = DF ………….. Result (v)
Now, in ∆ ABC and ∆ DEF,
AB = DE (Given)
BC = EF (Given)
AC = DF [by result (v)l
∴ ∆ ABC ≅ ∆ DEF (SSS rule) …… Result (vi)
Question 12.
ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that:
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ ABC ≅ ∆ BAD
(iv) diagonal AC = diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.)
Answer:
AB is extended to E, and AB || CD.
∴ AE || CD
In quadrilateral ADCE, AE || CD and by consturction CE || DA.
∴ Quadrilateral ADCE is a parallelogram.
∴ AD = CE
Moreover, AD = BC (Given)
∴ BC = CE
In ∆ BCE, BC = CE
∴ ∠CBE = ∠CEB
∴ ∠CBE = ∠CEA ………….. (1)
In parallelogram ADCE, AD || CE and AE is their transversal.
∴ ∠ DAE + ∠ CEA = 180° (Interior angles on the same side of transversal)
∴ ∠ DAE + ∠ CBE = 180° [by (1)]
∴ ∠ DAE = 180° – ∠ CBE …………… (2)
Moreover, ∠ ABC + ∠ CBE = 180° (Linear pair)
∴ ∠ ABC = 180° – ∠ CBE …………. (3)
From (2) and (3),
∠ DAE = ∠ ABC
∴ ∠A = ∠B ……… Result (i)
AB || CD and AD is their transversal.
∴ ∠A + ∠D = 180°
∴ ∠D = 180°- ∠A ………….. (4)
AB || CD and BC is their transversal.
∴ ∠B + ∠C = 180°
∴ ∠C = 180°- ∠B
∴ ∠C = 180° – ∠ A [by result (i)] ……… (5)
From (4) and (5),
∠C = ∠D …….. Result (ii)
Draw diagonals AC and BD.
In ∆ ABC and ∆ BAD,
BC = AD (Given)
∠ ABC = ∠ BAD [by result (i)]
AB = BA (Common)
∴ ∆ ABC ≅ ∆ BAD (SAS rule) ………. Result (iii)
∴ AC = BD (CPCT)
Thus, diagonal AC = diagonal BD … Result (iv)
Note: A trapezium in which non-parallel sides are equal is called an isosceles trapezium. As proved above, in an isosceles trapezium, the diagonals are equal and the angles on each parallel side are equal.
Ex 8.2
Question 1.
ABCD is a quadrilateral in which F Q, R and S are midpoints of the sides AB, BC, CD and DA respectively (see the given figure 1). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Answer:
In ∆ DAC, S and R are the midpoints of DA and DC respectively.
Through C draw a line parallel to AD which intersects line SR at T.
In ∆ DRS and ∆ CRT
∠ DRS = ∠ CRT (Vertically opposite angles)
∠ RSD = ∠ RTC (Alternate angles formed by transversal ST of DS || TC)
DR = CR (R is the midpoint of DC.)
∴ ∆ DRS ≅ ∆ CRT (AAS rule)
∴ DS = CT and SR = RT (CPCT)
As S is the midpoint of DA, we have DS = SA.
∴ SA = CT
And, by construction, SA || CT.
∴ Quadrilateral SACT is a parallelogram.
∴ ST || AC
∴ SR || AC ………… (1)
Now, SR = RT gives SR = 1/2 ST
In parallelogram SACT, ST = AC.
∴ SR = 1/2 AC ……………. (2)
Taking (1) and (2) together,
SR || AC and SR = 1/2 AC ….. Result (1)
Similarly, in ∆ ABC, P and Q are the midpoints of AB and BC respectively. ,
∴ PQ || AC and PQ = 1/2 AC
Now, SR = 1/2 AC and PQ = 1/2 AC
∴ PQ = SR …… Result (ii)
Similarly, SR || AC and PQ || AC.
∴ PQ || SR
Thus, in quadrilateral PQRS, PQ = SR and PQ || SR.
Hence, by theorem 8.8, PQRS is a parallelogram. … Result (iii)
Question 2.
ABCD is a rhombus and F Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Answer:
ABCD is a rhombus and F Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.
∴ In ∆ ABC, PQ || AC and PQ = 1/2 AC.
∴ In ∆ ADC, SR || AC and SR = 1/2 AC.
Hence, in quadrilateral PQRS, PQ || SR and PQ = SR.
∴ Quadrilateral PQRS is a parallelogram.
Now, since ABCD is a rhombus, AC and BD bisect each other at right angles at M.
∴ ∠ AMB = 90°
Now, AC || PQ and MN is their transversal.
∴ ∠ AMN + ∠ MNP = 180° (Interior angles on the same side of transversal)
∴ ∠ AMB + ∠MNP = 180°
∴ 90° + ∠ MNP = 180°
∴ ∠ MNP = 90°
In ∆ ABD, P and S are the midpoints of AB and AD respectively.
∴ PS || BD and NP is their transversal.
∴ ∠ DNP + ∠ NPS = 180°
∴ ∠ MNP + ∠ NPS =180°
∴ 90° + ∠ NPS = 180°
∴ ∠ NPS = 90°
∴ ∠ SPQ = 90°
Thus, in parallelogram PQRS, one angle ∠P is a right angle.
Hence, quadrilateral PQRS is a rectangle.
Question 3.
ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer:
Question 4.
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the midpoint of BC.
Answer:
Suppose line EF drawn through E and parallel to AB intersects BD at M.
EF || AB and AB || DC
∴ EF || DC
Trapezium ABCD is divided into two triangles, ∆ ABD and ∆ BCD, by diagonal BD.
In ∆ ABD, E is the midpoint of AD and a line through E and parallel to AB intersects BD at M.
Hence, by theorem 8.10, M is the midpoint of BD.
Now, in ∆ BCD, M is the midpoint of BD and a line through M and parallel to CD intersects BC at F.
Hence, by theorem 8.10, F is the midpoint of BC.
Note: The following result about the length of EF can also be derived:
EF = 1/2 (AB + CD)
Moreover, if X and Y are the midpoints of the diagonals of above trapezium ABCD, then XY = 1/2|AB – CD|.
Question 5.
In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.
Answer:
E and F are the midpoints of AB and CD respectively.
∴ AE = 1/2 AB and CF = 1/2 CD
In parallelogram ABCD, AB = CD and AB || CD.
∴ AE = CF and AE || CF
Hence, quadrilateral AECF is a parallelogram.
∴ AF || EC
∴ AP || EQ
In ∆ ABP E is the midpoint of AB and EQ || AR
∴ Q is the midpoint of PB. (Theorem 8.10)
∴PQ = QB …………… (1)
Similarly, in ∆ DQC, F is the midpoint of DC and FP || CQ.
∴ P is the midpoint of DQ. (Theorem 8.10)
∴ DP = PQ …………….. (2)
From (1) and (2), DP = PQ = QB.
Moreover, DP + PQ + QB = BD.
Thus, AF and EC trisect the diagonal BD.
Question 6.
Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Answer:
In quadrilateral ABCD, P Q, R and S are the midpoints of sides AB, BC, CD and DA respectively.
In ∆ ABC, P and Q are the midpoints of AB and BC respectively.
∴ PQ || AC and PQ = 1/2 AC …………….. (1)
In ∆ ADC, S and R are the midpoints of DA and DC respectively.
∴ SR || AC and SR = 1/2 AC ……………… (2)
From (1) and (2),
PQ = SR and PQ || SR.
Thus, in quadrilateral PQRS, sides in one pair of opposite sides are equal and parallel. Hence, quadrilateral PQRS is a parallelogram. The diagonals of a parallelogram bisect each other. [Theorem 8.6]
∴ PR and SQ bisect each other.
Thus, the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Question 7.
ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the midpoint of AC.
(ii) MD ⊥ AC
(iii) CM = MA = 1/2 AB
Answer:
In ∆ ABC, ∠ C is a right angle and M is the midpoint of hypotenuse AB. A line through M and parallel to BC intersects AC at D.
Hence, by theorem 8.10, DM bisects AC.
∴ D is the midpoint of AC. ….. Result (i)
In ∆ ABC, ∠ C is a right angle.
∴ ∠ C = 90°
Now, BC || DM and DC is their transversal.
∴ ∠ MDC + ∠ DCB = 180° (Interior angles on the same side of transversal)
∴ ∠ MDC + 90° = 180°
∴ ∠ MDC = 90°
Thus, MD is perpendicular to AC.
∴ MD ⊥ AC …… Result (ii)
Now, in ∆ ADM and ∆ CDM,
AD = CD (D is the midpoint of AC)
∠ ADM = ∠ CDM (Right angles)
DM = DM (Common)
∴ ∆ ADM ≅ ∆ CDM (SAS rule)
∴ AM = CM (CPCT) ……………. (1)
Now, M is the midpoint of AB.
∴ AM = 1/2 AB …… (2)
< Prom (1) and (2),
CM = MA = 1/2 AB …… Result (iii)
MCQ
Multiple Choice Questions and Answer
Answer each question by selecting the proper alternative from those given below each question to make the statement true:
Question 1.
The ratio of four angles in order of a quadrilateral is 2 : 4 : 5 : 4. Then, the measure of the smallest angle of the quadrilateral is
A. 120°
B. 96°
C. 48°
D. 60°
Answer:
C. 48°
Question 2.
In quadrilateral PQRS, ∠P = 5x, ∠Q = 3x, ∠R = 4x and ∠S = 6x. Then, the measure of the greatest angle of quadrilateral PQRS is …………… .
A. 100°
B. 60°
C. 80°
D. 120°
Answer:
D. 120°
Question 3.
In quadrilateral ABCD, ∠A + ∠B = 150°.
Then ∠C + ∠D =
A. 105°
B. 210°
C. 150°
D. 300°
Answer:
B. 210°
Question 4.
In trapezium PQRS, PQ || RS. If ∠P = 150°, then ∠S = …………. .
A. 75°
B. 150°
C. 60°
D. 30°
Answer:
D. 30°
Question 5.
The perimeter of parallelogram ABCD is 22 cm.
If AB = 4 cm, then BC = ……………. cm.
A. 7
B. 6
C. 5.5
D. 4
Answer:
A. 7
Question 6.
In parallelogram ABCD, ∠A – ∠B = 30°. Then, ∠C = ……………… .
A. 105°
B. 75°
C. 150°
D. 60°
Answer:
A. 105°
Question 7.
In parallelogram ABCD, the bisectors of ∠A and ∠B intersect at M. If ∠A = 80°, then ∠AMB = ……………. .
A. 40°
B. 50°
C. 80°
D. 90°
Answer:
D. 90°
Question 8.
In parallelogram ABCD, the ratio ∠A : ∠B : ∠C : ∠D can be
A. 3 : 4 : 5 : 6
B. 2 : 3 : 3 : 2
C. 2 : 3 : 2 : 3
D. 2 : 3 : 5 : 8
Answer:
C. 2 : 3 : 2 : 3
Question 9.
In parallelogram ABCD, 3 ∠ A = 2 ∠ B. Then, ∠ D = ………………. .
A. 120°
B. 108°
C. 72°
D. 60°
Answer:
B. 108°
Question 10.
In ∆ ABC, E and F are the midpoints of AB and AC respectively. If EF = 4 cm, then BC = …………… cm.
A. 8
B. 2
C. 4
D. 12
Answer:
A. 8
Question 11.
In ∆ ABC, P is the midpoint of AB and Q is the midpoint of AC. Then, PQCB is a ………….. .
A. trapezium
B. parallelogram
C. rectangle
D. rhombus
Answer:
A. trapezium
Question 12.
In ∆ ABC, D, E and F are the midpoints of AB, BC and CA respectively. If the perimeter of ∆ DEF is 30 cm, then the perimeter of ∆ ABC is ……………. cm.
A. 15
B. 30
C. 45
D. 60
Answer:
D. 60
Question 13.
∆ ABC is an equilateral triangle. D, E and F are the midpoints of AB, BC and CA respectively. If AB = 8 cm, the perimeter of ∆ DEF is …………… cm.
A. 24
B. 12
C. 6
D. 48
Answer:
B. 12
Question 14.
ABCD is a rectangle. If AB = 5 cm and BC = 12
cm, then BD = ………………. cm
A. 17
B. 13
C. 8.5
D. 1
Answer:
B. 13
Question 15.
ABCD is a rhombus. If AC = 10 cm and BD = 24 cm, the perimeter of ABCD is …………………. cm.
A. 13
B. 26
C. 52
D. 48
Answer:
C. 52