ML Aggarwal Solutions for Chapter 13 Rectilinear Figures Class 9 Maths ICSE
Exercise 13.1
1. If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.
Solution
We know that,
Sum of all four angles of a quadrilateral = 360°
Sum of two given angles = 40° + 110° = 150°
So, the sum of remaining two angles = 360° – 150° = 210°
Also given,
Ratio in these angles = 3 : 4
Hence,
Third angle = (210° ×3)/(3 + 4)
= (210° ×3)/7
= 90°
And,
Fourth angle = (210° ×4)/(3 + 4)
= (210° ×4)/7
= 120°
2. If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.
Solution
Given,
In trapezium ABCD in which
∠A : ∠B : ∠C : ∠D = 1 : 2 : 3 : 4
We know,
The sum of angles of the quad. ABCD = 360°
∠A = (360° ×1)/10 = 36°
∠B = (360° ×2)/10 = 72°
∠C = (360° ×3)/10 = 108°
∠D = (360° ×4)/10 = 144°
Now,
∠A + ∠D = 36° + 114° = 180°
Since, the sum of angles ∠A and ∠D is 180° and these are co-interior angles
Thus, AB || DC
Therefore, ABCD is a trapezium.
3. If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.
Solution
Here ABCD is a parallelogram.
Let ∠A = x°
Then, ∠B = (2x/3)° (Given condition)
So,
∠A + ∠B = 180° (As the sum of adjacent angles in a parallelogram is 180°)
Hence, ∠A = 108°
∠B = 2/3 ×108o = 2 ×36° = 72°
⇒ ∠B = ∠D = 72° (opposite angles in a parallelogram is same)
Also,
∠A = ∠C = 108° (opposite angles in a parallelogram is same)
Therefore, angles of parallelogram are 108°, 72°, 108° and 72°.
4. (a) In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.
(b) In figure (2) given below, ABCD is a parallelogram. Find the angles of the AAOD.
(c) In figure (3) given below, ABCD is a rhombus. Find the value of x.
Solution
(a) Since, ABCD is a || gm
We have, AB || CD
∠ADB = ∠DBC (Alternate angles)
⇒ ∠ADB = 80° (Given, ∠DBC = 80°)
Now,
In ∆ADB, we have
∠A + ∠ADB + ∠ABD = 180° (Angle sum property of a triangle)
⇒ 70° + 80° + ∠ABD = 180°
⇒ 150o + ∠ABD = 180o
⇒ ∠ABD = 180o – 150o = 30o
Now,
∠CDB = ∠ABD (Since, AB || CD and alternate angles)
So,
∠CDB = 30o
Hence, ∠ADB = 80o and ∠CDB = 30o.
(b) Given, ∠BOC = 35o and ∠CBO = 77o
In ∆BOC, we have
∠BOC + ∠BCO + ∠CBO = 180o (Angle sum property of a triangle)
⇒ ∠BOC = 180o – 112o = 68o
Now, in || gm ABCD
We have,
∠AOD = ∠BOC (Vertically opposite angles)
Hence, ∠AOD = 68o.
(c) ABCD is a rhombus
So,
∠A + ∠B = 180o (Sum of adjacent angles of a rhombus is 180o)
⇒ 72o + ∠B = 180o (Given, ∠A = 72o)
⇒ ∠B = 180o – 72o = 108o
Hence,
x = ½ B = ½ x 108o = 54o
5. (a) In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.
(b) In figure (2) given below. ABCD is a parallelogram. Find the values of x and y.
(c) In figure (3) given below. ABCD is a rhombus. Find x and y.
Solution
(a) Since, ABCD is a parallelogram
So, AB = CD and BC = AD
⇒ 3x = 2y + 2
⇒ 3x – 2y = 2 …(i)
Also,
AB + BC + CD + DA = 40
⇒ 3x + 2x + 2y + 2 + 2x = 40
⇒ 7x + 2y = 40 – 2
⇒ 7x + 2y = 38 …(ii)
Now, adding (i) and (ii) we get
(3x – 2y) + (7x + 2y) = 2 + 38
⇒ 10x = 40
⇒ x = 40/10 = 4
On substituting the value of x in (i), we get
3(4) – 2y = 2
⇒ 12 – 2y = 2
⇒ 2y = 12 – 2
⇒ y = 10/2 = 5
Hence, x = 4 and y = 5
(b) In parallelogram ABCD, we have
∠A = ∠C (Opposite angles are same in || gm)
⇒ 3x – 20o = x + 40o
⇒ 3x – x = 40o + 20o
⇒ 2x = 60o
⇒ x = 60o/2 = 30o …(i)
Also,
∠A + ∠B = 180o (Sum of adjacent angles in || gm is equal to 180o)
⇒ 3x – 20o + y + 15o = 180o
⇒ 3x + y = 180o + 20o – 15o
⇒ 3x + y = 185o
⇒ 3(30o) + y = 185o [Using (i)]
⇒ 90o + y = 185o
⇒ y = 185o – 90o = 95o
Hence,
x = 30o and 95o
(c) ABCD is a rhombus
So,
AB = CD
⇒ 3x + 2 = 4x – 4
⇒ 3x – 4x = -4 – 2
⇒ -x = -6
⇒ x = 6
Now, in ∆ABD we have
∠BAD = 60o and AB = AD
∠ADB = ∠ABD
So,
∠ADB = (180o – ∠BAD)/2
= (180o – 60o)/2
= 120o/2 = 60o
As ∆ABD is an equilateral triangle, all the angles of the triangle are 60o
Hence, AB = BD
3x + 2 = y – 1
⇒ 3(6) + 2 = y – 1 (Substituting the value of x)
⇒ 18 + 2 = y – 1
⇒ 20 = y – 1
⇒ y = 20 + 1
⇒ y = 21
Thus, x = 6 and y = 21.
6. The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.
Solution
Given, ABCD is a rectangle
We know that the diagonals of rectangle are same and bisect each other
So, we have
AP = BP
∠PAB = ∠PBA (Equal sides have equal opposite angles)
⇒ ∠PAB = 50o (Since, given ∠PBA = 50o)
Now, in ∆APB
∠APB + ∠ABP + ∠BAP = 180o
⇒ ∠APB + 50o + 50o = 180o
⇒ ∠APB = 180o – 100o
⇒ ∠APB = 80o
Then,
∠DPB = ∠APB (Vertically opposite angles)
Hence,
∠DPB = 80o
7. (a) In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.
(b) In figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of the ∆ AOB.
(c) In figure (3) given below, ABCD is rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3:2, find the angles of the ∆ AOD.
Solution
Since, EBC is an equilateral triangle, we have
EB = BC = EC …(i)
Also, ABCD is a square
So, AB = BC = CD = AD …(ii)
From (i) and (ii), we get
EB = EC = AB = BC = CD = AD …(iii)
Now, in ∆ECD
∠ECD = ∠BCD + ∠ECB
= 90o + 60o
= 150o …(iv)
Also, EC = CD [From (iii)]
So, ∠DEC = ∠CDE …(v)
∠ECD + ∠DEC + ∠CDE = 180o [Angles sum property of a triangle]
⇒ 150o + ∠DEC + ∠DEC = 180o [Using (iv) and (v)]
⇒ 2 ∠DEC = 180o – 150o = 30o
⇒ ∠DEC = 30o/2
⇒ ∠DEC = 15o …(vi)
Now, ∠BEC = 60o [BEC is an equilateral triangle]
⇒ ∠BED + ∠DEC = 60o
⇒ xo + 15o = 60o [From (vi)]
⇒ x = 60o – 15o
⇒ x = 45o
Hence, the value of x is 45o.
(b) Given, ABCD is a rectangle
∠ECD = 146o
As ACE is a straight line, we have
146o + ∠ACD = 180o [Linear pair]
⇒ ∠ACD = 180o – 146o = 34o …(i)
And, ∠CAB = ∠ACD …(ii) [Alternate angles]
From (i) and (ii), we have
∠CAB = 34o
⇒ ∠OAB = 34o …(iii)
In ∆AOB
AO = OB [Diagonals of a rectangle are equal and bisect each other]
∠OAB = ∠OBA …(iv) [Equal sides have equal angles opposite to them]
From (iii) and (iv),
∠OBA = 34o …(v)
Now,
∠AOB + ∠OBA + ∠OAB = 180o
⇒ ∠AOB + 34o + 34o = 180o [Using (3) and (5)]
⇒ ∠AOB + 68o = 180o
⇒ ∠AOB = 180o – 68o = 112o
Hence, ∠AOB = 112o, ∠OAB = 34o and ∠OBA = 34o
(c) Here, ABCD is a rhombus and diagonals intersect at O and ∠OAB : ∠OBA = 3 : 2
Let ∠OAB = 2xo
Then,
∠OBA = 2xo
We know that diagonals of rhombus intersect at right angle,
So, ∠OAB = 90o
Now, in ∆AOB
∠OAB + ∠OBA = 180o
⇒ 90o + 3xo + 2xo = 180o
⇒ 90o + 5xo = 180o
⇒ 5xo = 180o – 90o = 90o
⇒ xo = 90o/5 = 18o
Hence,
∠OAB = 3xo = 3 x 18o = 54o
⇒ OBA = 2xo = 2×18o = 36o and ∠AOB = 90o
8. (a) In figure (1) given below, ABCD is a trapezium. Find the values of x and y.
(b) In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and y.
(c) In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.
Solution
(a) Given: ABCD is a trapezium
∠A = x + 20o, ∠B = y, ∠C = 92o, ∠D = 2x + 10o
We have,
∠B + ∠C = 180o [Since AB || DC]
⇒ y + 92o = 180o
⇒ y = 180o – 92o = 88o
Also,
∠A + ∠D = 180o
⇒ x + 20o + 2x + 10o = 180o
⇒ 3x + 30o = 180o
⇒ 3x = 180o – 30o = 150o
⇒ x = 150o/3 = 50o
Hence, the value of x = 50o and y = 88o.
(b) Given: ABCD is an isosceles trapezium BC = AD
∠A = 2x, ∠C = y and ∠D = 3x
Since, ABCD is a trapezium and AB || DC
∠A + ∠D = 180o
⇒ 2x + 3x = 180o
⇒ 5x = 180o
⇒ x = 180o/5 = 36o …(i)
Also, AB = BC and AB || DC
So, ∠A + ∠C = 180o
⇒ 2x + y = 180o
⇒ 2×36o + y = 180o
⇒ 72o + y = 180o
⇒ y = 180o – 72o = 108o
Hence, value of x = 72o and y = 108o.
(c) Given: ABCD is a kite and diagonal intersect at O.
∠DAB = 112o and ∠DCB = 64o
As AC is the diagonal of kite ABCD, we have
∠DCO = 64o/2 = 32o
And, ∠DOC = 90o [Diagonal of kites bisect at right angles]
In ∆OCD, we have
∠ODC = 180o – (∠DCO + ∠DOC)
= 180o – (32o + 90o)
= 180o – 122o
= 58o
In ∆DAB, we have
∠OAB = 112o/2 = 56o
∠AOB = 90o [Diagonal of kites bisect at right angles]
In ∆OAB, we have
∠OBA = 180o – (∠OAB + ∠AOB)
= 180o – (56o + 90o)
= 180o – 146o
= 34o
Hence, ∠ODC = 58o and ∠OBA = 34o.
9. (i) Prove that each angle of a rectangle is 90°.
(ii) If the angle of a quadrilateral are equal, prove that it is a rectangle.
(iii) If the diagonals of a rhombus are equal, prove that it is a square.
(iv) Prove that every diagonal of a rhombus bisects the angles at the vertices.
Solution
(i) Given: ABCD is a rectangle
To prove: Each angle of rectangle = 90o
Proof:
In a rectangle opposite angles of a rectangle are equal
So, ∠A = ∠C and ∠B = ∠C
But, ∠A + ∠B + ∠C + ∠D = 360o [Sum of angles of a quadrilateral]
⇒ ∠A + ∠B + ∠A + ∠B = 360o
⇒ 2(∠A + ∠B) = 360o
⇒ (∠A + ∠B) = 360o/2
⇒ ∠A + ∠B = 180o
But, ∠A = ∠B [Angles of a rectangle]
So, ∠A = ∠B = 90o
Thus,
∠A = ∠B = ∠C = ∠D = 90o
Hence, each angle of a rectangle is 90°.
(ii) Given: In quadrilateral ABCD, we have
∠A = ∠B = ∠C = ∠D
To prove: ABCD is a rectangle
Proof:
∠A = ∠B = ∠C = ∠D
⇒ ∠A = ∠C and ∠B = ∠D
But these are opposite angles of the quadrilateral.
So, ABCD is a parallelogram
And, as ∠A = ∠B = ∠C = ∠D
Therefore, ABCD is a rectangle.
(iii) Given: ABCD is a rhombus in which AC = BD
To prove: ABCD is a square
Proof:
Join AC and BD.
Now, in ∆ABC and ∆DCB we have
∠AB = ∠DC [Sides of a rhombus]
∠BC = ∠BC [Common]
∠AC = ∠BD [Given]
So, ∆ABC ≅ ∆DCB by S.S.S axiom of congruency
Thus,
∠ABC = ∠DBC [By C.P.C.T]
But these are made by transversal BC on the same side of parallel lines AB and CD.
So, ∠ABC + ∠DBC = 180o
∠ABC = 90o
Hence, ABCD is a square.
(iv) Given: ABCD is rhombus.
To prove: Diagonals AC and BD bisects ∠A, ∠C, ∠B and ∠D respectively
Proof:
In ∆AOD and ∆COD, we have
AD = CD [sides of a rhombus are all equal]
OD = OD [Common]
AO = OC [Diagonal of rhombus bisect each other]
So, ∆AOD ≅ ∆COD by S.S.S axiom of congruency
Thus,
∠AOD = ∠COD [By C.P.C.T]
So, ∠AOD + ∠COD = 180o [Linear pair]
∠AOD = 180o
∠AOD = 90o
And, ∠COD = 90o
Thus,
OD ⊥ AC ⇒ BD ⊥ AC
Also, ∠ADO = ∠CDO [By C.P.C.T]
So,
OD bisect ∠D
BD bisect ∠D
Similarly, we can prove that BD bisect ∠B and AC bisect the ∠A and ∠C.
10. ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:
(i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD.
Solution
Given: In parallelogram ABCD in which diagonal AC bisects ∠A
To prove: (i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD
Proof:
(i) As AB || CD, we have [Opposite sides of a || gm]
∠DCA = ∠CAB
Similarly,
∠DAC = ∠DCB
But, ∠CAB = ∠DAC [Since, AC bisects ∠A]
Hence,
∠DCA = ∠ACB and AC bisects ∠C.
(ii) As AC bisects ∠A and ∠C
And, ∠A = ∠C
Hence, ABCD is a rhombus.
(iii) Since, AC and BD are the diagonals of a rhombus and
AC and BD bisect each other at right angles
Hence, AC ⊥ BD
11. (i) Prove that
bisectors of any two adjacent angles of a parallelogram are at right angles.
(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.
(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.
Solution
(i) Given AM bisect angle A and BM bisects angle of || gm ABCD.
To prove: ∠AMB = 90o
Proof:
We have,
∠A + ∠B = 180o [AD || BC and AB is the transversal]
⇒ ½ (∠A + ∠B) = 180o/2
⇒ ½ ∠A + ½ ∠B = 90o
⇒ ∠MAB + ∠MBA = 90o [Since, AM bisects ∠A and BM bisects ∠B]
Now, in ∆AMB
∠AMB + ∠MAB + ∠MBA = 180o [Angles sum property of a triangle]
⇒ ∠AMB + 90o = 180o
⇒ ∠AMB = 180o – 90o = 90o
Hence, bisectors of any two adjacent angles of a parallelogram are at right angles.
(ii) Given: A || gm ABCD in which bisector AR of ∠A meets DC in R and bisector CQ of ∠C meets AB in Q
∠A = ∠C [Opposite angles of || gm are equal]
½ ∠A = ½ ∠C
∠DAR = ∠BCQ [Since, AR is bisector of ½ ∠A and CQ is the bisector of ½ ∠C]
Now, in ∆ADR and ∆CBQ
∠DAR = ∠BCQ [Proved above]
AD = BC [Opposite sides of || gm ABCD are equal]
So, ∆ADR ≅ ∆CBQ, by A.S.A axiom of congruency
Then by C.P.C.T, we have
∠DRA = ∠BCQ
And,
∠DRA = ∠RAQ [Alternate angles since, DC || AB]
Thus, ∠RAQ = ∠BCQ
But these are corresponding angles,
Hence, AR || CQ.
(iii) Given: In quadrilateral ABCD, diagonals AC and BD are equal and bisect each other at right angles
To prove: ABCD is a square
Proof:
In ∆AOB and ∆COD, we have
AO = OC [Given]
BO = OD [Given]
∠AOB = ∠COD [Vertically opposite angles]
So, ∆AOB ≅ ∆COD, by S.A.S axiom of congruency
By C.P.C.T, we have
AB = CD
and ∠OAB = ∠OCD
But these are alternate angles
AB || CD
Thus, ABCD is a parallelogram
In a parallelogram, the diagonal bisect each other and are equal
Hence, ABCD is a square.
12. (i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.
(ii) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution
(i) ABCD is a rectangle and its diagonals AC bisects ∠A and ∠C
To prove: ABCD is a square
Proof:
We know that the opposite sides of a rectangle are equal and each angle is 90o
As AC bisects ∠A and ∠C
So, ∠1 = ∠2 and ∠3 = ∠4
But, ∠A = ∠C = 90o
∠2 = 45o and ∠4 = 45o
And, AB = BC [Opposite sides of equal angles]
But, AB = CD and BC = AD
So, AB = BC = CD = DA
Therefore, ABCD is a square.
(ii) In quadrilateral ABCD diagonals AC and BD are equal and bisect each other at right angle
To prove: ABCD is a square
Proof:
In ∆AOB and ∆BOC, we have
AO = CO [Diagonals bisect each other at right angles]
OB = OB [Common]
∠AOB = ∠COB [Each 90o]
So, ∆AOB ≅ ∆BOC, by S.A.S axiom
By C.P.C.T, we have
AB = BC …(i)
Similarly, in ∆BOC and ∆COD
OB = OD [Diagonals bisect each other at right angles]
OC = OC [Common]
∠BOC = ∠COD [Each 90o]
So, ∆BOC ≅ ∆COD, by S.A.S axiom
By C.P.C.T, we have
BC = CD …(ii)
From (i) and (ii), we have
AB = BC = CD = DA
Hence, ABCD is a square.
13. P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Solution
Given: ABCD is a parallelogram, P and Q are the points on AB and DC. Diagonals AC and BD intersect each other at O.
To prove:
Diagonals of || gm ABCD bisect each other at O
So, AO = OC and BO = OD
Now, in ∆AOP and ∆COQ we have
AO = OC and BO = OD
Now, in ∆AOP and ∆COQ
AO = OC [Proved]
∠OAP = ∠OCQ [Alternate angles]
∠AOP = ∠COQ [Vertically opposite angles]
So, ∆AOP ≅ ∆COQ by S.A.S axiom
Thus, by C.P.C.T
OP = OQ
Hence, O bisects PQ.
14. (a) In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed.
Prove that:
(i) the triangles ABX and QCX are congruent;
(ii)DC = CQ = QP
(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.
Solution
(a) Given: ABCD is parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q and ABPQ is a || gm.
To prove: (i) ∆ABX ≅ ∆QCX
(ii) DC = CQ = QP
Proof:
In ∆ABX and ∆QCX, we have
BX = XC [X is the mid-point of BC]
∠AXB = ∠CXQ [Vertically opposite angles]
∠XCQ = ∠XBA [Alternate angle, since AB || CQ]
So, ABX ≅ ∆QCX by A.S.A axiom of congruence
Now, by C.P.C.T
CQ = AB
But,
AB = DC and AB = QP [As ABCD and ABPQ are ||gms]
Hence,
DC = CQ = QP
(b) In || gm ABCD, P and Q are points on AB and CD respectively, PQ and AC intersect each other at O and AP = CQ
To prove: AC and PQ bisect each other i.e. AO = OC and PO = OQ
Proof:
In ∆AOP and ∆COQ
AP = CQ [Given]
∠AOP = ∠COQ [Vertically opposite angles]
∠OAP = ∠OCP [Alternate angles]
So, ∆AOP ≅ ∆COQ by A.A.S axiom of congruence
Now, by C.P.C.T
OP = OQ and OA = OC
Hence, proved.
15. ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.
Solution
Given: ABCD is a square. P is any point on BC and Q is any point on AB and these points are taken such that AP = DQ
To prove: AP ⊥ DQ
Proof:
In ∆ABP and ∆ADQ, we have
AP = DQ [Given]
AD = AB [Sides of square ABCD]
∠DAQ = ∠ABP [Each 90o]
So, ∆ABP ≅ ∆ADQ by R.H.S axiom of congruency
Now, by C.P.C.T
∠BAP = ∠ADQ
But, ∠BAD = 90o
∠BAD = ∠BAP + ∠PAD ...(i)
⇒ 90o = ∠BAP + ∠PAD
⇒ ∠BAP + ∠PAD = 90o
⇒ ∠BAP + ∠ADQ = 90o
Now, in ∆ADM we have
(∠MAD + ∠ADM) + ∠AMD = 180o [Angles sum property of a triangle]
⇒ 90o + ∠AMD = 180o [From (i)]
⇒ ∠AMD = 180o – 90o = 90o
So, DM ⊥ AP
⇒ DQ ⊥ AP
Hence, AP ⊥ DQ
16. If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.
Solution
Given: ABCD is a || gm is which BP = PQ = QD
To prove: CQ || AP
Proof:
In || gm ABCD, we have
AB = CD [Opposite sides of a || gm are equal]
And BD is the transversal
So, ∠1 = ∠2 …(i) [Alternate interior angles]
Now, in ∆ABP and ∆DCQ
AB = CD [Opposite sides of a || gm are equal]
∠1 = ∠2 [From (i)]
BP = QD [Given]
So, ∆ABP ≅ ∆DCQ by S.A.S axiom of congruency
Then by C.P.C.T, we have
AP = QC
Also, ∠APB = ∠DQC [By C.P.C.T]
-∠APB = -∠DQC [Multiplying both sides by -1]
⇒ 180o – ∠APB = 180o – ∠DQC [Adding 180o both sides]
⇒ ∠APQ = ∠CQP
But, these are alternate angles
Hence, AP || QC
⇒ CQ || AP.
17. A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B ; the four bisectors form a quadrilateral ABCD. Prove that
(i) ABCD is a rectangle.
(ii) CD is parallel to the original parallel lines.
Solution
Given: LM || PQ and AB is the transversal line cutting ∠M at A and PQ at B
AC, AD, BC and BD is the bisector of ∠LAB, ∠BAM, ∠PAB and ∠ABQ respectively.
AC and BC intersect at C and AD and BD intersect at D.
A quadrilateral ABCD is formed.
To prove: (i) ABCD is a rectangle
(ii) CD || LM and PQ
Proof:
(1) ∠LAB + ∠BAM = 180o [LAM is a straight line]
⇒ ½ (∠LAB + ∠BAM) = 90o
⇒ ½ ∠LAB + ½ ∠BAM = 90o
⇒ ∠2 + ∠3 = 90o [Since, AC and AD is bisector of ∠LAB & ∠BAM respectively]
⇒ ∠CAD = 90o
⇒ ∠A = 90o
(2) Similarly, ∠PBA + ∠QBA = 180o [PBQ is a straight line]
⇒ ½ (∠PBA + ∠QBA) = 90o
⇒ ½ ∠PBA + ½ ∠QBA = 90o
⇒ ∠6 + ∠7 = 90o [Since, BC and BD is bisector of ∠PAB & ∠QBA respectively]
⇒ ∠CBD = 90o
⇒ ∠B = 90o
(3) ∠LAB + ∠ABP = 180o [Sum of co-interior angles is 180o and given LM || PQ]
⇒ ½ ∠LAB + ½ ∠ABP = 90o
⇒ ∠2 + ∠6 = 90o [Since, AC and BC is bisector of ∠LAB & ∠PBA respectively]
(4) In ∆ACB,
∠2 + ∠6 + ∠C = 180o [Angles sum property of a triangle]
⇒ (∠2 + ∠6) + ∠C = 180o
⇒ 90o + ∠C = 180o [using (3)]
⇒ ∠C = 180o – 90o
⇒ ∠C = 90o
(5) ∠MAB + ∠ABQ = 180o [Sum of co-interior angles is 180o and given LM || PQ]
⇒ ½ ∠MAB + ½ ∠ABQ = 90o
⇒ ∠3 + ∠7 = 90o [Since, AD and BD is bisector of ∠MAB & ∠ABQ respectively]
(6) In ∆ADB,
∠3 + ∠7 + ∠D = 180o [Angles sum property of a triangle]
⇒ (∠3 + ∠7) + ∠D = 180o
⇒ 90o + ∠D = 180o [using (5)]
⇒ ∠D = 180o – 90o
⇒ ∠D = 90o
(7) ∠LAB + ∠BAM = 180o
∠BAM = ∠ABP [From (1) and (2)]
⇒ ½ ∠BAM = ½ ∠ABP
⇒ ∠3 = ∠6 [Since, AD and BC is bisector of ∠BAM and ∠ABP respectively]
Similarly, ∠2 = ∠7
(8) In ∆ABC and ∆ABD,
∠2 = ∠7 [From (7)]
AB = AB [Common]
∠6 = ∠3 [From (7)]
So, ∆ABC ≅ ∆ABD by A.S.A axiom of congruency
Then, by C.P.C.T we have
AC = DB
Also, CB = AD
(9) ∠A = ∠B = ∠C =∠D = 90o [From (1), (2), (3) and (4)]
AC = DB [Proved in (8)]
CB = AD [Proved in (8)]
Hence, ABCD is a rectangle.
(10) Since, ABCD is a rectangle [From (9)]
OA = OD [Diagonals of rectangle bisect each other]
(11) In ∆AOD, we have
OA = OD [From (10)]
∠9 = ∠3 [Angles opposite to equal sides are equal]
(12) ∠3 = ∠4 [AD bisects ∠MAB]
(13) ∠9 = ∠4 [From (11) and (12)]
But these are alternate angles.
OD || LM
⇒ CD || LM
Similarly, we can prove that
∠10 = ∠8
But these are alternate angles,
So, OD || PQ
⇒ CD || PQ
(14) CD || LM [Proved in (13)]
CD || PQ [Proved in (13)]
18. In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that:
(i) BE bisects ∠B
(ii) ∠AEB is a right angle
Solution
Given: ABCD is a || gm in which bisectors of angle A and B meets in E and AB = 2 AD
To prove: (i) BE bisects ∠B
(ii) ∠AEB = 90o
Proof:
(1) In || gm ABCD
∠1 = ∠2 [AD bisects angles ∠A]
(2) AB || DC and AE is the transversal
∠2 = ∠3 [Alternate angles]
(3) ∠1 = ∠2 [From (1) and (2)]
(4) In ∆ADE, we have
∠1 = ∠3 [Proved in (3)]
DE = AD [Sides opposite to equal angles are equal]
⇒ AD = DE
(5) AB = 2 AD [Given]
AB/2 = AD
AB/2 = DE [using (4)]
DC/2 = DE [AB = DC, opposite sides of a || gm are equal]
So, E is the mid-point of D.
⇒ DE = EC
(6) AD = BC [Opposite sides of a || gm are equal]
(7) DE = BC [From (4) and (6)]
(8) EC = BC [From (5) and (7)]
(9) In ∆BCE, we have
EC = BC [Proved in (8)]
∠6 = ∠5 [Angles opposite to equal sides are equal]
(10) AB || DC and BE is the transerval
∠4 = ∠5 [Alternate angles]
(11) ∠4 = ∠6 [From (9) and (10)]
So, BE is bisector of ∠B
(12) ∠A + ∠B = 180o [Sum of co-interior angles is equal to 180o, AD || BC]
⇒ ½ ∠A + ½ ∠B = 180o/2
⇒ ∠2 + ∠4 = 90o [AE is bisector of ∠A and BE is bisector of ∠B]
(13) In ∆APB,
∠AEB + ∠2 + ∠4 = 180o
⇒ ∠AEB + 90o = 180o
Hence, ∠AEB = 90o
19. ABCD is a parallelogram, bisectors of angles A and B meet at E which lie on DC. Prove that AB.
Solution
Given: ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC in E
To prove: AB = 2 AD
Proof:
In parallelogram ABCD, we have
AB || DC
∠1 = ∠5 [Alternate angles, AE is transversal]
∠1 = ∠2 [AE is bisector of ∠A, given]
Thus,
∠2 = ∠5 …(i)
Now, in ∆AED
DE = AD [Sides opposite to equal angles are equal]
∠3 = ∠6 [Alternate angles]
∠3 = ∠4 [Since, BE is bisector of ∠B (given)]
Thus, ∠4 = ∠6 …(ii)
In ∆BCE, we have
BC = EC [Sides opposite to equal angles are equal]
AD = BC [Opposite sides of || gm are equal]
AD = DE = EC [From (i) and (ii)]
AB = DC [Opposite sides of a || gm are equal]
AB = DE + EC
= AD + AD
Hence,
AB = 2 AD
20. ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO =AP, prove that 3 ∠POB = ∠AOP.
Solution
Given: ABCD is a square and the diagonals intersect at O. P is the point on AB such that AO = AP
To prove: 3 ∠POB = ∠AOP
Proof:
(1) In square ABCD, AC is a diagonal
So, ∠CAB = 45o
∠OAP = 45o
(2) In ∆AOP,
∠OAP = 45o [From (1)]
AO = AP [Sides opposite to equal angles are equal]
Now,
∠AOP + ∠APO + ∠OAP = 180o [Angles sum property of a triangle]
⇒ ∠AOP + ∠AOP + 45o = 180o
⇒ 2∠AOP = 180o – 45o
⇒ ∠AOP = 135o/2
(3) ∠AOB = 90o [Diagonals of a square bisect at right angles]
So, ∠AOP + ∠POB = 90o
⇒ 135o/2 + ∠POB = 90o [From (2)]
⇒ ∠POB = 90o – 135o/2
= (180o – 135o)/2
= 45o/2
3∠POB = 135o/2 [Multiplying both sides by 3]
Hence,
∠AOP = 3 ∠POB [From (2) and (3)]
21. ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution
Given: ABCD is a square in which E, F, G and H are points on AB, BC, CD and DA
Such that AE = BF = CG = DH
EF, FG, GH and HE are joined
To prove: EFGH is a square
Proof:
Since, AE = BF = CG = DH
So, EB = FC = GD = HA
Now, in ∆AEH and ∆BFE
AE = BF [Given]
AH = EB [Proved]
∠A = ∠B [Each 90o]
So, ∆AEH ≅ ∆BFE by S.A.S axiom of congruency
Then, by C.P.C.T we have
EH = EF
And ∠4 = ∠2
But ∠1 + ∠4 = 90o
⇒ ∠1 + ∠2 = 90o
Thus, ∠HEF = 90o
Hence, EFGH is a square.
22. (a) In the Figure (1) given below, ABCD and ABEF are parallelograms. Prove that
(i) CDFE is a parallelogram
(ii) FD = EC
(iii) Δ AFD = ΔBEC.
(b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC.
Solution
To prove: (i) CDFE is a parallelogram
(ii) FD = EC
(iii) Δ AFD = ΔBEC
Proof:
(1) DC || AB and DC = AB [ABCD is a || gm]
(2) FE || AB and FE = AB [ABEF is a || gm]
(3) DC || FE and DC = FE [From (1) and (2)]
Thus, CDFE is a || gm
(4) CDEF is a || gm
So, FD = EC
(5) In ∆AFD and ∆BEC, we have
AD = BC [Opposite sides of || gm ABCD are equal]
AF = BE [Opposite sides of || gm ABEF are equal]
FD = BE [From (4)]
Hence, ∆AFD ≅ ∆BEC by S.S.S axiom of congruency
(b) Given: ABCD is a || gm, ADEF and AGHB are two squares
To prove: FG = AC
Proof:
(1) ∠FAG + 90o + 90o + ∠BAD = 360o [At a point total angle is 360o]
⇒ ∠FAG = 360o – 90o – 90o – ∠BAD
⇒ ∠FAG = 180o – ∠BAD
(2) ∠B + ∠BAD = 180o [Adjacent angle in || gm is equal to 180o]
⇒ ∠B = 180o – ∠BAD
(3) ∠FAG = ∠B [From (1) and (2)]
(4) In ∆AFG and ∆ABC, we have
AF = BC [FADE and ABCD both are squares on the same base]
Similarly, AG = AB
∠FAG = ∠B [From (3)]
So, ∆AFG ≅ ∆ABC by S.A.S axiom of congruency
Hence, by C.P.C.T
FG = AC
23. ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.
Solution
Let each side of the rhombus ABCD be a
∠A = 60o
So, ABD is an equilateral triangle
⇒ BD = AB = a
We know that, the diagonals of a rhombus bisect each other at right angles
So, in right triangle AOB, we have
AO2 + OB2 = AB2 [By Pythagoras Theorem]
AO2 = AB2 – OB2
= a2 – (½ a)2
= a2 – a2/4
= 3a2/4
AO = √(3a2/4) = √3a/2
But, AC = 2 AO = 2 x 3a/2 = 3a
Hence,
AC : BD = √3a : a = √3 : 1
Exercise 13.2
1. Using ruler and compasses only, construct the quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6cm, BC = 3.6cm, CD = 5cm. Measure ∠ BCD.
Solution
Steps of construction:
(i) Draw a line segment AB = 6cm
(ii) At A, draw a ray AX making an angle of 45o and cut off AD = 6cm
(iii) With centre B and radius 3.6 cm and with centre D and radius 5 cm, draw two arcs intersecting each other at C.
(iv) Join BC and DC.
Thus, ABCD is the required quadrilateral.
On measuring ∠BCD, it is 60o.
2. Draw a quadrilateral ABCD with AB = 6cm, BC = 4cm, CD = 4 cm and ∠ BC = ∠ BCD = 90°.
Solution
Steps of construction:
(i) Draw a line segment BC = 4 cm
(ii) At B and C draw rays BX and CY making an angle of 90o each
(iii) From BX, cut off BA = 6 cm and from CY, cut off CD = 4cm
(iv) Join AD.
Thus, ABCD is the required quadrilateral.
3. Using ruler and compasses only, construct the quadrilateral ABCD given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm, ∠BAD = 90° and the diagonal AC = 5.5 cm.
Solution
Steps of construction:
(i) Draw a line segment AB = 5cm
(ii) With centre A and radius 5.5cm and with centre B and radius 2.5cm draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) At A, draw a ray AX making an angle of 90o.
(v) With centre C and radius 6cm, draw an arc intersecting AX at D
(vi) Join CD.
Thus, ABCD is the required quadrilateral.
4. Construct a quadrilateral ABCD in which AB = 3.3 cm, BC = 4.9 cm, CD = 5.8 cm, DA = 4 cm and BD = 5.3 cm.
Solution
Steps of construction:
(i) Draw a line segment AB = 3.3cm
(ii) with centre A and radius 4cm, and with centre B and radius 5.3cm, draw arcs intersecting each other at D.
(iii) Join AD and BD.
(iv) With centre B and radius 4.9 cm and with centre D and radius 5.8cm, draw arcs intersecting each other at C.
(v) Join BC and DC.
Thus, ABCD is the required quadrilateral.
5. Construct a trapezium ABCD in which AD || BC, AB = CD = 3 cm, BC = 5.2cm and AD = 4 cm.
Solution
Steps of construction:
(i) Draw a line segment BC = 5.2cm
(ii) From BC, cut off BE = AD = 4cm
(iii) With centre E and C, and radius 3 cm, draw arcs intersecting each other at D.
(iv) Join ED and CD.
(v) With centre D and radius 4cm and with centre B and radius 3cm, draw arcs intersecting each other at A.
(vi) Join BA and DA
Thus, ABCD is the required trapezium.
6. Construct a trapezium ABCD in which AD || BC, ∠B= 60°, AB = 5 cm. BC = 6.2 cm and CD = 4.8 cm.
Solution
Steps of construction:
(i) Draw a line segment BC = 6.2cm
(ii) At B, draw a ray BX making an angle of 60o and cut off AB = 5cm
(iii) From A, draw a line AY parallel to BC.
(iv) With centre C and radius 4.8cm, draw an arc which intersects AY at D and D’.
(v) Join CD and CD’
Thus, ABCD and ABCD’ are the required two trapeziums.
7. Using ruler and compasses only, construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Solution
Steps of construction:
(i) Draw a line segment BC = 7cm
(ii) A to B, draw a ray Bx making an angle of 75o and cut off AB = 5.1cm
(iii) With centre A and radius 7cm with centre C and radius 5.1cm, draw arcs intersecting each other at D.
(iv) Join AD and CD.
Thus, ABCD is the required parallelogram.
8. Using ruler and compasses only, construct a parallelogram ABCD in which AB = 4.6 cm, BC = 3.2 cm and AC = 6.1 cm.
Solution
Steps of construction:
(i) Draw a line segment AB = 4.6cm
(ii) With centre A and radius 6.1cm and with centre B and radius 3.2cm, draw arcs intersecting each other at C.
(iii) Join AC and BC.
(iv) Again, with centre A and radius 3.2cm and with centre C and radius 4.6cm, draw arcs intersecting each other at C.
(v) Join AD and CD.
Thus, ABCD is the required parallelogram.
9. Using ruler and compasses, construct a parallelogram ABCD give that AB = 4 cm, AC = 10 cm, BD = 6 cm. Measure BC.
Solution
Steps of construction:
(i) Construct triangle OAB such that
OA = ½ x AC = ½ x 10cm = 5cm
OB = ½ x BD = ½ x 6cm = 3cm
As, diagonals of || gm bisect each other and AB = 4cm
(ii) Produce AO to C such that OA = OC = 5cm
(iii) Produce BO to D such that OB = OD = 3cm
(iv) Join AD, BC and CD
Thus, ABCD is the required parallelogram
(v) Measure BC which is equal to 7.2cm
10. Using ruler and compasses only, construct a parallelogram ABCD such that BC = 4 cm, diagonal AC = 8.6 cm and diagonal BD = 4.4 cm. Measure the side AB.
Solution
Steps of construction:
(i) Construct triangle OBC such that
OB = ½ x BD = ½ x 4.4cm = 2.2cm
OC = ½ x AC = ½ x 8.6cm = 4.3cm
Since, diagonals of || gm bisect each other and BC = 4cm
(ii) Produce BO to D such that BO = OD = 2.2cm
(iii) Produce CO to A such that CO = OA = 4.3cm
(iv) Join AB, AD and CD
Thus, ABCD is the required parallelogram.
(v) Measure the side AB, AB = 5.6cm
11. Use ruler and compasses to construct a parallelogram with diagonals 6 cm and 8 cm in length having given the acute angle between them is 60°. Measure one of the longer sides.
Solution
Steps of construction:
(i) Draw AC = 6cm
(ii) Find the mid-point O of AC. [Since, diagonals of || gm bisect each other]
(iii) Draw line POQ such that POC = 60o and OB = OD = ½ BD = ½ x 8cm = 4cm
So, from OP cut OD = 4cm and from OQ cut OB = 4cm
(iv) Join AB, BC, CD and DA.
Thus, ABCD is the required parallelogram.
(v) Measure the length of side AD = 6.1cm
12. Using ruler and compasses only, draw a parallelogram whose diagonals are 4 cm and 6 cm long and contain an angle of 75°. Measure and write down the length of one of the shorter sides of the parallelogram.
Solution
Steps of construction:
(i) Draw a line segment AC = 6cm
(ii) Bisect AC at O.
(iii) At O, draw a ray XY making an angle of 75o at O.
(iv) From OX and OY, cut off OD = OB = 4/2 = 2cm
(v) Join AB, BC, CD and DA.
Thus, ABCD is the required parallelogram.
On measuring one of the shorter sides, we get AB = CD = 3cm
13. Using ruler and compasses only, construct a parallelogram ABCD with AB = 6 cm, altitude = 3.5 cm and side BC = 4 cm. Measure the acute angles of the parallelogram.
Solution
Steps of construction:
(i) Draw AB = 6cm
(ii) At B, draw BP ⊥ AB
(iii) From BP, cut BE = 3.5cm = height of || gm
(iv) Through E draw QR parallel to AB
(v) With B as centre and radius BC = 4cm draw an arc which cuts QR at C.
(vi) Since, opposite sides of || gm are equal
So, AD = BC = 4cm
(vii)With A as centre and radius = 4cm draw an arc which cuts QR at D.
Thus, ABCD is the required parallelogram.
(viii) To measure the acute angle of parallelogram which is equal to 61o.
14. The perpendicular distances between the pairs of opposite sides of a parallelogram ABCD are 3 cm and 4 cm and one of its angles measures 60°. Using ruler and compasses only, construct ABCD.
Solution
Steps of construction:
(i) Draw a straight-line PQ, take a point A on it.
(ii) At A, construct ∠QAF = 60o
(iii) At A, draw AE ⊥ PQ from AE cut off AN = 3cm
(iv) Through N draw a straight line to PQ to meet AF at D.
(v) At D, draw AG ⊥ AD, from AG cut off AM = 4cm
(vi) Through M, draw at straight line parallel to AD to meet AQ in B and ND in C.
Then, ABCD is the required parallelogram
15. Using ruler and compasses, construct a rectangle ABCD with AB = 5cm and AD = 3 cm.
Solution
Steps of construction:
(i) Draw a straight-line AB = 5cm
(ii) At A and B construct ∠XAB and ∠YBA = 90o
(iii) From A and B cut off AC and BD = 3cm each
(iv) Join CD.
Thus, ABCD is the required rectangle.
16. Using ruler and compasses only, construct a rectangle each of whose diagonals measures 6cm and the diagonals intersect at an angle of 45°.
Solution
Steps of construction:
(i) Draw a line segment AC = 6cm
(ii) Bisect AC at O.
(iii) At O, draw a ray XY making an angle of 45o at O.
(iv) From XY, cut off OB = OD = 6/2 = 3cm each
(v) Join AB, BC CD and DA.
Thus, ABCD is the required rectangle.
17. Using ruler and compasses only, construct a square having a diagonal of length 5cm. Measure its sides correct to the nearest millimeter.
Solution
Steps of construction:
(i) Draw a line segment AC = 5cm
(ii) Draw its percentage bisector XY bisecting it at O
(iii) From XY, cut off
OB = OD = 5/2 = 2.5cm
(iv) Join AB, BC, CD and DA.
Thus, ABCD is the required square
On measuring its sides, each = 3.6cm (approximately)
18. Using ruler and compasses only construct A rhombus ABCD given that AB 5cm, AC = 6cm measure ∠BAD.
Solution
Steps of construction:
(i) Draw a line segment AB = 5cm
(ii) With centre A and radius 6cm, with centre B and radius 5cm, draw arcs intersecting each other at C.
(iii) Join AC and BC
(iv) With centre A and C and radius 5cm, draw arc intersecting each other 5cm, draw arcs intersecting each other at D
(v) Join AD and CD.
Thus, ABCD is a rhombus
On measuring, ∠BAD = 106o.
19. Using ruler and compasses only, construct rhombus ABCD with sides of length 4cm and diagonal AC of length 5 cm. Measure ∠ABC.
Solution
Steps of construction:
(i) Draw a line segment AC = 5cm
(ii) With centre A and C and radius 4cm, draw arcs intersecting each other above and below AC at D and B.
(iii) Join AB, BC, CD and DA.
Thus, ABCD is the required rhombus.
20. Construct a rhombus PQRS whose diagonals PR and QS are 8 cm and 6cm respectively.
Solution
Steps of construction:
(i) Draw a line segment PR = 8cm
(ii) Draw its perpendicular bisector XY intersecting it at O.
(iii) From XY, cut off OQ = OS = 6/2 = 3cm each
(iv) Join PQ, QR, RS and SP
Thus, PQRS is the required rhombus.
21. Construct a rhombus ABCD of side 4.6 cm and ∠BCD = 135°, by using ruler and compasses only.
Solution
Steps of construction:
(i) Draw a line segment BC = 4.6cm
(ii) At C, draw a ray CX making an angle of 135o and cut off CD = 4.6cm
(iii) With centres B and D, and radius 4.6cm draw arcs intersecting each other at A.
(iv) Join BA and DA.
Thus, ABCD is the required rhombus.
22. Construct a trapezium in which AB || CD, AB = 4.6 cm, ∠ ABC = 90°, ∠ DAB = 120° and the distance between parallel sides is 2.9 cm.
Solution
Steps of construction:
(i) Draw a line segment AB = 4.6cm
(ii) At B, draw a ray BZ making an angle of 90o and cut off BC = 2.9cm (distance between AB and CD)
(iii) At C, draw a parallel line XY to AB.
(iv) At A, draw a ray making an angle of 120o meeting XY at D.
Thus, ABCD is the required trapezium.
23. Construct a trapezium ABCD when one of parallel sides AB = 4.8 cm, height = 2.6cm, BC = 3.1 cm and AD = 3.6 cm.
Solution
Step construction:
(i) Draw a line segment AB = 4.8cm
(ii) At A, draw a ray AZ making an angle of 90o cut off AL = 2.6cm
(iii) At L, draw a line XY parallel to AB.
(iv) With centre A and radius 3.6cm and with centre B and radius 3.1cm, draw arcs intersecting XY at D and C respectively.
(v) Join AD and BC
Thus, ABCD is the required trapezium.
24. Construct a regular hexagon of side 2.5 cm.
Solution
Steps of construction:
(i) With O as centre and radius = 2.5cm, draw a circle
(ii) take any point A on the circumference of circle.
(iii) With A as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at B.
(iv) With B as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at C.
(v) With C as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at D.
(vi) With D as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at E.
(vii) With E as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at F.
(viii) Join AB, BC, CD, DE, EF and FA.
(ix) ABCDEF is the required Hexagon.
Chapter Test
1. In the given figure, ABCD is a parallelogram. CB is produced to E such that BE=BC. Prove that AEBD is a parallelogram.
Solution
Given ABCD is a || gm in which CB is produced to E such that BE = BC
BD and AE are joined
To prove: AEBD is a parallelogram
Proof:
In ∆AEB and ∆BDC
EB = BC [Given]
∠ABE = ∠DCB [Corresponding angles]
AB = DC [Opposite sides of ||gm]
Thus, ∆AEB ≅ ∆BDC by S.A.S axiom
So, by C.P.C.T
But, AD = CB = BE [Given]
As the opposite sides are equal and ∠AEB = ∠DBC
But these are corresponding angles
Hence, AEBD is a parallelogram.
2. In the given figure, ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || BA. Show that (i) ∠DAC=∠BCA (ii) ABCD is a parallelogram.
Solution
Given: In isosceles triangle ABC, AB = AC. AD is the bisector of ext. ∠PAC and CD || BA
To prove: (i) ∠DAC = ∠BCA
(ii) ABCD is a || gm
Proof:
In ∆ABC
AB = AC [Given]
∠C = ∠B [Angles opposite to equal sides]
Since, ext. ∠PAC = ∠B + ∠C
= ∠C + ∠C
= 2 ∠C
= 2 ∠BCA
So, ∠DAC = 2 ∠BCA
∠DAC = ∠BCA
But these are alternate angles
Thus, AD || BC
But, AB || AC
Hence, ABCD is a || gm.
3. Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.
Solution
Given: ABCD is an isosceles trapezium in which AB || DC and AD = BC
P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
To prove: PQRS is a rhombus
Construction: Join AC and BD
Proof:
Since, ABCD is an isosceles trapezium
Its diagonals are equal
AC = BD
Now, in ∆ABC
P and Q are the mid-points of AB and BC
So, PQ || AC and PQ = ½ AC …(i)
Similarly, in ∆ADC
S and R mid-point of CD and AD
So, SR || AC and SR = ½ AC …(ii)
From (i) and (ii), we have
PQ || SR and PQ = SR
Thus, PQRS is a parallelogram.
Now, in ∆APS and ∆BPQ
AP = BP [P is the mid-point]
AS = BQ [Half of equal sides]
∠A = ∠B [As ABCD is an isosceles trapezium]
So, ∆APS ≅ ∆BPQ by SAS Axiom of congruency
Thus, by C.P.C.T we have
PS = PQ
But there are the adjacent sides of a parallelogram
So, sides of PQRS are equal
Hence, PQRS is a rhombus
Hence, proved
4. Find the size of each lettered angle in the following figures.
Solution
(i) As CDE is a straight line
∠ADE + ∠ADC = 180o
⇒ 122o + ∠ADC = 180o
⇒ ∠ADC = 180o – 122o = 58o …(i)
⇒ ∠ABC = 360o – 140o = 220o …(ii) [At any point the angle is 360o]
Now, in quadrilateral ABCD we have
∠ADC + ∠BCD + ∠BAD + ∠ABC = 360o
⇒ 58o + 53o + x + 220o = 360o [Using (i) and (ii)]
⇒ 331o + x = 360o
⇒ x = 360o – 331o
⇒ x = 29o
(ii) As DE || AB [Given]
∠ECB = ∠CBA [Alternate angles]
⇒ 75o = ∠CBA
⇒ ∠CBA = 75o
Since, AD || BC we have
(x + 66o) + 75o = 180o
⇒ x + 141o = 180o
⇒ x = 180o – 141o
⇒ x = 39o …(i)
Now, in ∆AMB
x + 30o + ∠AMB = 180o [Angles sum property of a triangle]
⇒ 39o + 30o + ∠AMB = 180o [From (i)]
⇒ 69o + ∠AMB + 180o
⇒ ∠AMB = 180o – 69o = 111o …(ii)
Since, ∠AMB = y [Vertically opposite angles]
⇒ y = 111o
Hence, x = 39o and y = 111o
(iii) In ∆ABD
AB = AD [Given]
∠ABD = ∠ADB [Angles opposite to equal sides are equal]
⇒ ∠ABD = 42o [Since, given ∠ADB = 42o]
And,
∠ABD + ∠ADB + ∠BAD = 180o [Angles sum property of a triangle]
⇒ 42o + 42o + y = 180o
⇒ 84o + y = 180o
⇒ y = 180o – 84o
⇒ y = 96o
⇒ ∠BCD = 2 ×26o = 52o
In ∆BCD,
As BC = CD [Given]
∠CBD = ∠CDB = x [Angles opposite to equal sides are equal]
∠CBD + ∠CDB + ∠BCD = 180o
⇒ x + x + 52o = 180o
⇒ 2x + 52o = 180o
⇒ 2x = 180o – 52o
⇒ x = 128o/2
⇒ x = 64o
Hence, x = 64o and y = 90o.
5. Find the size of each lettered angle in the following figures:
Solution
(i) Here, AB || CD and BC || AD
So, ABCD is a || gm
y = 2× ∠ABD
⇒ y = 2× 53o = 106o …(1)
Also, y + ∠DAB = 180o
⇒ ∠DAB = 180o – 106o
= 74o
Thus, x = ½ ∠DAB [As AC bisects ∠DAB]
x = ½ × 74o = 37o
and ∠DAC = x = 37o …(ii)
Also, ∠DAC = z …(iii) [Alternate angles]
From (ii) and (iii),
z = 37o
Hence, x = 37o, y = 106o and z = 37o
(ii) As ED is a straight line, we have
60o + ∠AED = 180o [Linear pair]
⇒ ∠AED = 180o – 60o
⇒ ∠AED = 120o …(i)
Also, as CD is a straight line
50o + ∠BCD = 180o [Linear pair]
⇒ ∠BCD = 180o – 50o
⇒ ∠BCD = 130o …(ii)
In pentagon ABCDE, we have
∠A + ∠B+ ∠AED + ∠BCD + ∠x = 540o [Sum of interior angles in pentagon is 540o]
⇒ 90o + 90o 120o + 130o + x = 540o
⇒ 430o + x = 540o
⇒ x = 540o – 430o
⇒ x = 110o
Hence, value of x = 110o
(iii) In given figure, AD || BC [Given]
60o + y = 180o and x + 110o = 180o
⇒ y = 180o – 60o and x = 180o – 110o
⇒ y = 120o and x = 70o
Since, CD || AF [Given]
∠FAD = 70o …(i)
In quadrilateral ADEF,
∠FAD + 75o + z + 130o = 360o
⇒ 70o + 75o + z + 130o = 360o
⇒ 275o + z = 360o
⇒ z = 360o – 275o = 85o
Hence,
x = 70o, y = 120o and z = 85o
6. In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find (i) ∠DAG (ii) ∠FEG (iii) ∠GAC (iv) ∠AGC.
Solution
Here ABCD and DCFE is a rhombus and square respectively.
So, AB = BC = DC = AD …(i)
Also, DC = EF = FC = EF …(ii)
From (i) and (ii), we have
AB = BC = DC = AD = EF = FC = EF …(iii)
∠ABC = 56o [Given]
∠ADC = 56o [Opposite angle in rhombus are equal]
So, ∠EDA = ∠EDC + ∠ADC = 90o + 56o = 146o
In ∆ADE,
DE = AD [From (iii)]
∠DEA = ∠DAE [Equal sides have equal opposite angles]
∠DEA = ∠DAG = (180o – ∠EDA)/2
= (180o – 146o)/2
= 34o/2
= 17o
⇒ ∠DAG = 17o
Also,
∠DEG = 17o
∠FEG = ∠E – ∠DEG
= 90o – 17o
= 73o
In rhombus ABCD,
∠DAB = 180o – 56o = 124o
⇒ ∠DAC = 124o/2 [Since, AC diagonals bisect the ∠A]
⇒ ∠DAC = 62o
∠GAC = ∠DAC – ∠DAG
= 62o – 17o
= 45o
In ∆EDG,
∠D + ∠DEG + ∠DGE = 180o [Angles sum property of a triangle]
⇒ 90o + 17o + ∠DGE = 180o
⇒ ∠DGE = 180o – 107o = 73o …(iv)
Thus, ∠AGC = ∠DGE …(v) [Vertically opposite angles]
Hence from (iv) and (v), we have
∠AGC = 73o
7. If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.
Solution
Each side of rhombus ABCD is 8cm
So, AB = BC = CD = DA = 8cm
Let ∠A = 60o
So, ∆ABD is an equilateral triangle
Then,
AB = BD = AD = 8cm
As we know, the diagonals of a rhombus bisect each other at right angles
AO = OC, BO = OD = 4 cm and ∠AOB = 90o
Now, in right ∆AOB
By Pythagoras theorem
AB2 = AO2 + OB2
⇒ 82 = AO2 + 42
⇒ 64 = AO2 + 16
⇒ AO2 = 64 – 16 = 48
⇒ AO = √48 = 4√3cm
But, AC = 2 AO
Hence, AC = 2 × 4√3 = 8√3cm.
8. Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.
Solution
Steps of construction:
(i) Draw AB = 5.0cm
(ii) Draw BAP = 45o on side AB
(iii) Take A as centre and radius 2.5cm cut the line AP at D
(iv) Take D as centre and radius 5.0cm draw an arc
(v) Take B as centre and radius equal to 2.5cm cut the arc of step (iv) at C
(vi) Join BC and CD
(vii) ABCD is the required parallelogram
(viii) Draw the bisector of ∠BAD, which cuts the DC at E
(ix) Join EB
(x) Measure ∠AEB which is equal to 90o.