Chapter 5 - Quadrilaterals
Practice Set 5.1
1. Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠XYZ = 135⁰ then what is the measure of ∠XWZ and ∠YZW ? If l(OY) = 5 cm then l(WY) = ?
Given:
□ WXYZ is a parallelogram
Diagonal WY and Diagonal XZ intersect at Point O
∠XYZ = 135⁰
l(OY) = 5 cm
To find:
∠XWZ, ∠YZW and l(WY)
Solution:
□ WXYZ is a parallelogram [Given]
∴ ∠XYZ ≅ ∠XWZ [Opposite angles of a parallelogram are congruent]
∵ ∠XYZ = 135⁰
∴ ∠XWZ = 135⁰ ……(i)
∠YZW + ∠XYZ = 180⁰ [Adjacent angles of a parallelogram are supplementary]
∴ ∠YZW + 135= 180⁰ [From (i)]
∴ ∠YZW = 180⁰ – 135⁰
∴ ∠YZW = 45⁰
Now,
l(OY) ≅ l(WY) [Diagonals of a parallelogram bisect each other]
∵ l(OY) = 5 cm [Given]
∴ l(WY) = 5 cm
l(OY) + l(WY) = l(OY) [From Figure]
5 + 5 = l(OY)
∴ l(OY) = 10 cm
Ans: ∠XWZ = 135⁰, ∠YZW = 45⁰, l(WY) = 10 cm
2. In a parallelogram ABCD, If ∠A = (3x + 12)⁰, ∠B = (2x – 32)⁰ then find the value of x and then find the measures of ∠C and ∠D.
Given:
□ABCD is a parallelogram
∠A = (3x + 12)⁰
∠B = (2x – 32)⁰
To find:
x, ∠C and ∠D
Solution:
□ABCD is a parallelogram [Given]
∴ ∠A + ∠B = 180⁰ [Adjacent angles of a parallelogram are supplementary]
∴ (3x + 12)⁰ + (2x – 32)⁰ = 180⁰
∴ 3x + 12 + 2x – 32 = 180
∴ 5x – 20 = 180
∴ 5x= 180 + 20
∴ 5x = 200
∴ x = \( \frac{200}{5} \)
∴ x = 40
∴ The value of ∠A and ∠B is
∠A = (3x + 12)⁰
∠A = [3(40) + 12]⁰
∴ ∠A = (120 + 12)⁰
∴ ∠A = 132⁰
∠B = (2x – 32)⁰
∴ ∠B = [2(40) – 32]⁰
∴ ∠B = (80 – 32)⁰
∴ ∠B = 48⁰
∴ ∠C ≅ ∠A [Opposite angles of a parallelogram]
∵ ∠C = 135⁰
∴ ∠A = 135⁰
∴ ∠B ≅ ∠D [Opposite angles of a parallelogram]
∵ ∠B = 48⁰
∴ ∠D = 48⁰
Ans: The value of x is 40, and the measures of ∠C and ∠D are 132⁰ and 48⁰ respectively.
3. Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.
Given:
Perimeter of □ABCD = 150 cm
l(AB) = l(AD) + 25 cm
To find:
l(AB), l(BC), l(CD) and l(AD)
Solution:
Let l(AD) = x
∴ l(AB) = l(AD) + 25 cm
∴ l(AB) = x + 25 cm
∴ l(AB) ≅ l(CD) [Opposite sides of a parallelogram are congruent]
∵ l(AB) = x + 25 cm
∴ l(CD) = x + 25 cm
∴ l(AD) ≅ l(BC) [Opposite sides of a parallelogram are congruent]
∵ l(AD) = x
∴ l(BC) = x
Now,
Perimeter of □ABCD = 150 cm [Given]
∴ l(AB + l(BC) + l(DC) + l(AD) = 150
∴ (x + 25) + x + (x + 25) + x = 150
∴ 4x + 50 = 150
∴ 4x = 150 – 50
∴ 4x = 100
∴ x = \( \frac{100}{4} \)
∴ x = 25
Now, the values of
l(AD) = l(BC) = x = 25 cm
l(AB) = l(DC) = x + 25 = 25 + 25 = 50 cm
Ans: The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.
4. If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.
Given:
Adjacent angles of a parallelogram is 1 : 2
To find:
∠A, ∠B, ∠C and ∠D
Solution:
Let the common multiple be x.
∴ ∠A = x⁰ and ∠B = 2x⁰ [Given condition]
□ABCD is a parallelogram [Given]
∴ ∠A + ∠B = 180⁰ [Adjacent angles of a parallelogram are supplementary]
∴ x + 2x = 180
∴ 3x = 180
∴ x = \( \frac{180}{3} \)
∴ x = 60
∴ The value of
∠A = x⁰ = 60⁰
∠B = 2x⁰ = 2 x 60⁰ = 120⁰
∴ ∠A ≅ ∠C [Opposite angles of a parallelogram]
∵ ∠A = 60⁰
∴ ∠C = 60⁰
∴ ∠B ≅ ∠D [Opposite angles of a parallelogram]
∵ ∠B = 120⁰
∴ ∠D = 120⁰
Ans: The measures of the angles of the parallelogram are 60⁰, 120⁰, 60⁰ and 120⁰.
5*. Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that □ ABCD is a rhombus.
Given:
l(AO) = 5
l(BO) = 12
l(AB) = 13
To prove:
□ABCD is a rhombus.
Proof:
l(AO) = 5 [Given]
l(BO) = 12 [Given]
l(AB) = 13 [Given]
Now,
AO² + BO² = 5² + 12²
AO² + BO² = 25 + 144
∴ AO² + BO² = 169 ……(i)
And,
AB² = 13² = 169 ……(ii)
From (i) and (ii)
∴ AB² = AO² + BO²
∴ ∆ AOB is a right-angled triangle [By Converse of Pythagoras theorem]
∴ ∠AOB = 90⁰
∴ seg AC ⊥ seg BD …..(iii) [A – O – C]
∴ □ABCD is a rhombus [The diagonals of a rhombus always bisect each other at 90⁰]
Hence Proved.
6. In the figure 5.12, □PQRS and □ABCR are two parallelograms. If ∠P = 110⁰ then find the measures of all angles of □ABCR.
Given:
∠P = 110⁰
To find:
∠A, ∠B, ∠C and ∠R
Solution:
□PQRS is a parallelogram [Given]
∴ ∠P ≅ ∠R [Opposite angles of a parallelogram]
∵ ∠P = 110⁰
∴ ∠R = 110⁰ …..(i)
□ABCR is a parallelogram [Given]
∴ ∠A + ∠R = 180⁰ [Adjacent angles of a parallelogram are supplementary]
∴ ∠A + 110⁰ = 180⁰ [From (i)]
∴ ∠A= 180⁰ – 110⁰
∴ ∠A = 70⁰
∠C ≅ ∠A [Opposite angles of a parallelogram]
∵ ∠C = 70⁰
∴ ∠A = 70⁰
∠R ≅ ∠B [Opposite angles of a parallelogram]
∵ ∠R = 110⁰
∴ ∠B = 110⁰
Ans: ∠A = 70⁰, ∠B = 110⁰, ∠C = 70⁰ and ∠R = 110⁰
7. In figure 5.13 □ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.
Given:
□ABCD is a parallelogram
BE = AB
To prove:
FC = FB
Proof:
□ABCD is a parallelogram [Given]
∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram]
seg AB ≅ seg BE ……..(ii) [Given]
seg DC ≅ seg BE ……..(iii) [From (i) & (ii)]
side DC || side AB [Opposite sides of a parallelogram]
i.e. side DC || seg AE and seg DE is their transversal [A – B – E]
∴ ∠CDE ≅ ∠AED [Alternate Angles]
∴ ∠CDF ≅ ∠BEF …..(iv) [D – F – E, A – B – E]
In ∆DFC and ∆EFB,
seg DC ≅ seg EB [From (iii)]
∠CDF ≅ ∠BEF [From (iv)]
∠DFC ≅ ∠EFB [Vertically opposite angles]
∴ ∆DFC ≅ ∆EFB [SAA test]
∴ FC ≅ FB [c.s.c.t]
∴ Line ED bisects seg BC at point F
Hence Proved.
Practice Set 5.2
1. In figure 5.22, □ABCD is a parallelogram, P and Q are midpoints of side AB and DC respectively, then prove □APCQ is a parallelogram.
Given:
□ABCD is a parallelogram
P and Q are midpoints of side AB
and DC
To prove:
□APCQ is a parallelogram
Proof:
AB ≅ AP [P is the midpoint of side AB]
∴ AP = \( \frac{1}{2} \) AB …..(i)
QC ≅ DQ [Q is the midpoint of side DC]
∴ QC = \( \frac{1}{2} \) DC …..(ii)
□ABCD is a parallelogram [Given]
∴ AB ≅ DC [Opposite sides of a parallelogram]
∴ \( \frac{1}{2} \) AB = \( \frac{1}{2} \) DC [Multiplying both sides by \( \frac{1}{2} \)]
∴ AP = QC ….(iii) [From (i) and (ii)]
Also,
Side AB || Side DC [Opposite sides of a parallelogram]
i.e. AP || QC ….(iv) [A – P – B, D – Q – C]
We know that,
A quadrilateral is a parallelogram if its opposite sides is parallel and congruent
∴ □APCQ is a parallelogram. [From (iii) & (iv)]
Hence Proved.
2. Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.
Given:
□ABCD is a rectangle
To prove:
Rectangle ABCD is a parallelogram
Proof:
□ABCD is a rectangle
∴ ∠A ≅ ∠B ≅ ∠C ≅ ∠D = 90° [Angles of a rectangle]
We know that,
A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent
∴ Rectangle ABCD is a parallelogram.
Hence Proved.
3. In figure 5.23, G is the point of concurrence of medians of D DEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that □GEHF is a parallelogram.
Given:
Point G (centroid) is the point of concurrence of the medians of △DEF
DG = GH
To prove:
□GEHF is a parallelogram.
Proof:
Let ray DH intersect seg EF at point A such that E – A – F
∴ seg DA is the median of ∆DEF
∴ EA ≅ FA ……(i)
Point G is the centroid of ∆DEF
∴ DG:GI = 2:1 [Centroid divides each median in the ratio 2:1]
∴ DG = 2(GI)
∴ GH = 2(GI) [DG ≅ GH, Given]
∴ GA + HA = 2(GA) [G – A – H]
∴ HA = 2(GA) – GA
∴ HA = GA ….(ii)
We know that,
A quadrilateral is a parallelogram, if its diagonals bisect each other
∴ □GEHF is a parallelogram [From (i) & (ii)]
Hence Proved.
4. Prove that a quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.
Given:
□ABCD is a parallelogram.
Ray AS bisect ∠A
Ray BQ bisect ∠B
Ray CQ bisect ∠C
Ray DS bisect ∠D
To prove:
□PQRS is a rectangle.
Proof:
∠BAS ≅ ∠DAS [Ray AS bisects ∠A]
Let, ∠BAS = x°
∴ ∠DAS = x° ..…(i)
∠ABQ ≅ ∠CBQ [Ray BQ bisects ∠B]
Let, ∠ABQ = y°
∴ ∠CBQ = y° ..…(ii)
∠BCQ ≅ ∠DCQ [Ray CQ bisects ∠C]
Let, ∠BCQ = u°
∴ ∠DCQ = u° ..…(iii)
∠ADS ≅ ∠CDS [Ray DS bisects ∠D]
Let, ∠ADS = v°
∴ ∠CDS = v° ..…(iv)
□ABCD is a parallelogram [Given]
∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BAS + ∠DAS + ∠ABQ + ∠CBQ = 180° [Angle addition property]
∴ x° + x° + y° + y° = 180 [From (i) and (ii)]
∴ 2x° + 2y° = 180
∴ 2(x° + y°) = 180
∴ (x° + y°) = \( \frac{180}{2} \)
∴ x° + y° = 90° ……(v)
Also,
∠C + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠ABQ + ∠CBQ + ∠BCQ + ∠DCQ = 180° [Angle addition property]
∴ y° + y° + u° + u° = 180° [From (ii) and (iii)]
∴ 2y° + 2u° = 180°
∴ 2(y° + u°) = 180
∴ (y° + u°) = \( \frac{180}{2} \)
∴ y° + u° = 90° ……(vi)
Also,
∠C + ∠D = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BCQ + ∠DCQ + ∠ADS + ∠CDS = 180° [Angle addition property]
∴ u° + u° + v° + v° = 180° [From (iii) and (iv)]
∴ 2u° + 2v° = 180°
∴ 2(u° + v°) = 180
∴ (u° + v°) = \( \frac{180}{2} \)
∴ u° + v° = 90° ……(vii)
Also,
∠A + ∠D = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BAS + ∠DAS + ∠ADS + ∠CDS = 180° [Angle addition property]
∴ x° + x° + v° + v° = 180° [From (i) and (iv)]
∴ 2x° + 2v° = 180°
∴ 2(x° + v°) = 180
∴ (x° + v°) = \( \frac{180}{2} \)
∴ x° + v° = 90° ……(viii)
In ∆ARB,
∠RAB + ∠RBA + ∠ARB = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x° + y° + ∠SRQ = 180° [A – S – R, B – Q – R]
∴ 90° + ∠SRQ = 180° [From (v)]
∴ ∠SRQ = 180°- 90°
∴ ∠SRQ = 90° …..(ix)
In ∆DPC,
∠PDC + ∠PCD + ∠DPC = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ v° + u° + ∠SPQ = 180° [P – Q – C, P – S – D]
∴ 90° + ∠SPQ = 180° [From (vii)]
∴ ∠SPQ = 180°- 90°
∴ ∠SPQ = 90° …..(x)
In ∆ASD,
∠ASD + ∠SAD + ∠SDA = 180° [Sum of the measures of angles a triangle is 180°]
∴ ∠ASD + x° + v° = 180°
∴ ∠ASD + 90° = 180° [From (viii)]
∴ ∠ASD = 180°- 90°
∴ ∠ASD = 90°
∴ ∠PSR ≅ ∠ASD [Vertically opposite angles]
∵ ∠ASD = 90°
∴ ∠PSR = 90° …..(xi)
In ∆BQC,
∠BQC + ∠CBQ + ∠BCQ = 180° [Sum of the measures of angles a triangle is 180°]
∴ ∠BQC + y° + u° = 180° [From (vi)]
∴ ∠BQC + 90° = 180°
∴ ∠BQC = 180°- 90°
∴ ∠BQC = 90°
∴ ∠BQC ≅ ∠PQR [Vertically opposite angles]
∵ ∠BQC = 90°
∴ ∠PQR = 90° …..(xii)
∴ In □PQRS,
∠SRQ ≅ ∠SPQ ≅ ∠PSR ≅ ∠PQR = 90° [From (ix), (x), (xi), (xii)]
∴ □PQRS is a rectangle [Each angle is a right angle]
Hence Proved.
5. In figure 5.25, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that □PQRS is a parallelogram.
Given:
□ABCD is a parallelogram
AP = BQ = CR = DS
To prove:
□PQRS is a parallelogram
Proof:
□ABCD is a parallelogram [Given]
∴ ∠B ≅ ∠D ….(i) [Opposite angles of a parallelogram are congruent]
∴ ∠A ≅ ∠C ….(ii) [Opposite angles of a parallelogram are congruent]
Also,
AB ≅ CD [Opposite sides of a parallelogram are congruent]
∴ AP + BP = DR + CR [A – P – B, D – R – C]
∴ AP + BP = DR + AP [Given, AP = CR]
∴ BP = DR ….(iii)
AD ≅ BC [Opposite sides of a parallelogram are congruent]
∴ AS + SD = BQ + QC [A – S – D, B – Q – C]
∴ AS + BQ = BQ + QC [Given, BQ = DS]
∴ AS = QC ….(iv)
In ∆PBQ and ∆RDS
seg BP ≅ seg DR [From (iii)]
∠PBQ ≅ ∠RDS [From (i)]
seg BQ ≅ seg DS [Given]
∴ ∆PBQ ≅ ∆RDS [SAS test]
∴ seg PQ ≅ seg RS …..(v) [c.s.c.t]
In ∆PAS ≅ ∆RCQ,
seg AS ≅ seg QC [From (iv)]
∠PAS ≅ ∠RCQ [From (ii)]
seg AP ≅ seg CR [Given]
∴ ∆PAS ≅ ∆RCQ [SAS test]
∴ seg PS ≅ seg RQ ….(vi) [c.s.c.t]
We know that,
A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent
∴ □PQRS is a parallelogram [From (v) & (vi)]
Hence Proved.
Practice Set 5.3
1. Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find the length of BO and if ∠CAD = 35⁰ then find the measure of ∠ACB.
Given:
□ABCD is a rectangle
AC = 8
∠CAD = 35⁰
To find:
l(BO) and ∠ACB
Solution:
□ABCD is a rectangle [Given]
∴ BD = AC [Diagonals of a rectangle are congruent]
∵ AC = 8 cm [Given]
∴ BD = 8 cm [From (i)]
BO = \( \frac{1}{2} \) BD [Diagonals of a rectangle bisect each other]
∴ BO = \( \frac{1}{2} \) x 8
∴ BO = \( \frac{8}{2} \)
∴ BO = 4 cm
Now,
side AD || side BC and seg AC is their transversal [Opposite sides of a rectangle are parallel]
∴ ∠ACB ≅ ∠CAD [Alternate angles]
∵ ∠CAD = 35°[Given]
∴ ∠ACB = 35°
Ans: BO = 4 cm, ∠ACB = 35°
2. In a rhombus PQRS if PQ = 7.5 then find the length of QR. If ∠QPS = 75⁰ then find the measure of ∠PQR and ∠SRQ.
Given:
□PQRS is a rhombus
PQ = 7.5
∠QPS = 75⁰
To find:
l(QR), ∠PQR and ∠SRQ
Solution:
□PQRS is a rhombus [Given]
∴ QR ≅ PQ [All sides of a rhombus are congruent]
∵ PQ = 7.5 cm
∴ QR = 7.5 cm
Now,
∠QPS + ∠PQR = 180° [Adjacent angles of a rhombus are supplementary]
∴ 75° + ∠PQR = 180°
∴ ∠PQR = 180° – 75°
∴ ∠PQR = 105°
Also,
∠SRQ ≅ ∠QPS [Opposite angles of a rhombus are congruent]
∵ ∠QPS = 75⁰
∴ ∠SRQ = 75°
Ans: QR = 7.5 cm, ∠PQR = 105°& ∠SRQ = 75°
3. Diagonals of a square IJKL intersects at point M, Find the measures of ∠IMJ, ∠JIK and ∠LJK .
Given:
□IJKL is a square
To find:
∠IMJ, ∠JIK and ∠LJK
Solution:
□IJKL is a square [Given]
∴ seg IK ⊥ seg JL [Diagonals of a square are perpendicular to each other]
∴ ∠ IMJ = 90°
and ∠ JIL 90° ……. (i) [Angle of a square]
∠JIK = \( \frac{1}{2} \) ∠JIL [Diagonals of a square bisect the opposite angles]
∠JIK = \( \frac{1}{2} \) (90°) [From (i)]
∴ ∠JIK = 45°
∠IJK = 90° (ii) [Angle of a square]
iii. ∠LJK = \( \frac{1}{2} \) ∠IJK [Diagonals of a square bisect the opposite angles]
∴ ∠LJK = \( \frac{1}{2} \) (90°) [From (ii)]
∴ ∠LJK = 45°
Ans: ∠LJK = 90°, ∠JIK = 45°& ∠LJK = 45°
4. Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of the rhombus and its perimeter
Given:
Let □ABCD be the rhombus
∴ AC = 20 cm, BD = 21 cm
To find:
Side of the rhombus and its Perimeter
Solution:
□ABCD be the rhombus
∴ AO = \( \frac{1}{2} \) AC [Diagonals of a rhombus bisect each other]
∴ AO = \( \frac{1}{2} \) 20
∴ AO = \( \frac{20}{2} \)
∴ AO = 10 cm
And,
∴ BO = \( \frac{1}{2} \) BD [Diagonals of a rhombus bisect each other]
∴ BO = \( \frac{1}{2} \) 21
∴ BO = \( \frac{21}{2} \) cm
In ∆AOB,
∠AOB = 90° [Diagonals of a rhombus are perpendicular to each other]
∴ ∆AOB is a right angled triangle
∴ AB² = AO² + BO² [By Pythagoras theorem]
∴ AB² = (10)² + (\( \frac{21}{2} \))²
∴ AB² = 100 + (\( \frac{441}{4} \))
∴ AB² = (\( \frac{440}{4} \)) + (\(\frac{441}{4} \)) [Equalizing the denominator]
∴ AB² = (\( \frac{440\; +\; 441}{4} \))
∴ AB² = (\( \frac{841}{4} \))
∴ √(AB²) = \( \sqrt{\frac{841}4} \) [Taking square root on both sides]
∴ AB = \( \sqrt{\frac{841}4} \)
∴ AB = (\( \frac{29}{2} \))
∴ AB = 14.5
Now,
Perimeter of □ABCD = 4 x AB
Perimeter of □ABCD = 4 x 14.5
Perimeter of □ABCD = 58 cm
Ans: The side and perimeter of the rhombus are 14.5 cm and 58 cm respectively.
5. State with reasons whether the following statements are ‘true’ or ‘false’.
(i) Every parallelogram is a rhombus.
Ans: False
All the sides of a rhombus are congruent, while only the opposite sides of a parallelogram are congruent, hence every parallelogram cannot be a rhombus.
(ii) Every rhombus is a rectangle.
Ans: False
All the angles of a rectangle are congruent, while only opposite angles of a rhombus are congruent, hence every rhombus cannot be a rectangle.
(iii) Every rectangle is a parallelogram.
Ans: True
The opposite sides of a parallelogram are parallel and congruent. Also, its opposite angles are congruent. The opposite sides of a rectangle are parallel and congruent. Also, all its angles are congruent. Hence, every rectangle is also a parallelogram.
(iv) Every square is a rectangle.
Ans: True
The opposite sides of a rectangle are parallel and congruent. Also, all its angles are congruent. All the sides of a square are parallel and congruent. Also, all its angles are congruent. All the sides of a square are parallel and congruent. Also, all its angles are congruent. Hence, every square is also a rectangle.
(v) Every square is a rhombus.
Ans: True
All the sides of a rhombus are congruent. Also, its diagonals are perpendicular bisectors of each other. All the sides of a square are congruent. Also, its diagonals are perpendicular bisectors of each other. Hence, every square is also a rhombus.
(vi) Every parallelogram is a rectangle.
Ans: False
All the angles of a rectangle are congruent, while only the opposite angles of a parallelogram are congruent. Hence every parallelogram is not a rectangle.
Practice Set 5.4
1. In □IJKL, side IJ || side KL ∠I = 108⁰ ∠K = 53⁰ then find the measures of ∠J and ∠L.
Given:
side IJ || side KL
∠I = 108⁰
∠K = 53⁰
To find:
∠J and ∠L
Solution:
In □IJKL,
side IJ || side KL and side IL is their transversal [Given]
∴ ∠I + ∠L = 180° [Interior angles]
∴ 108° + ∠L = 180°
∴ ∠L = 180° – 108°
∴ ∠L = 72°
Also,
side IJ || side KL and side JK is their transversal [Given]
∴ ∠J + ∠K = 180° [Interior angles]
∴ ∠J + 53° = 180°
∴ ∠J = 180°- 53°
∴ ∠J = 127°
Ans: ∠L = 72°, ∠J = 127°
2. In □ABCD, side BC || side AD, side AB ≅ side DC. If ∠A = 72⁰ then find the measures of ∠B, and ∠D.
Given:
side BC || side AD
side AB ≅ side DC
∠A = 72⁰
To find:
∠B and ∠D
Solution:
In □ABCD,
side BC || side AD and side AB is their transversal [Given]
∴ ∠A + ∠B = 180° [Interior angles]
∴ 72° +∠B = 180°
∴ ∠B = 180° – 72°
∴ ∠B = 108°
In ∆BPA and ∆CQD,
∠BPA ≅ ∠CQD [Both are right angles]
Hypotenuse AB ≅ Hypotenuse DC [Given]
seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]
∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]
∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]
i.e. ∠A = ∠D
∵ ∠A = 72⁰ [Given]
∴ ∠D = 72°
Ans: ∠B = 108°, ∠D = 72°
3. In □ABCD, side BC < side AD, side BC || side AD and if side BA ≅ side CD then prove that ∠ABC ≅ ∠DCB.
Given:
side BC < side AD
side BC || side AD
side BA = side CD
To prove:
∠ABC ≅ ∠DCB
Construction:
Draw seg BP ⊥ side AD, A – P – D
seg CQ ⊥ side AD, A – Q – D
Proof:
In ∆BPA and ∆CQD,
∠BPA ≅ ∠CQB [Both are right angles]
Hypotenuse BA ≅ Hypotenuse CD [Given]
seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]
∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]
∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]
i.e. ∠A = ∠D ….(i)
Now,
side BC || side AD and side AB is their transversal [Given]
∴ ∠A + ∠B = 180°…..(ii) [Interior angles]
Also,
side BC || side AD and side CD is their transversal [Given]
∴ ∠C + ∠D = 180° …..(iii) [Interior angles]
∴ ∠A + ∠B = ∠C + ∠D [From (ii) and (iii)]
∴ ∠A + ∠B = ∠C + ∠A [From (i)]
∴ ∠B = ∠C
i.e. ∠ABC ≅ ∠DCB
Hence proved.
Practice Set 5.5
1. In figure 5.38, points X, Y, Z are the midpoints of side AB, side BC and side AC of ∆ABC respectively. AB = 5 cm, AC = 9 cm and BC = 11 cm. Find the length of XY, YZ, XZ.
Given:
X is the midpoint of side AB
Y is the midpoint of side BC
Z is the midpoint of side AC
AB = 5 cm
AC = 9 cm
BC = 11 cm
To find:
l(XY), l(YZ) and l(XZ)
Solution:
Points X and Y are the midpoints of sides AB and BC respectively [Given]
∴ XY = \( \frac{1}{2} \) AC [Midpoint theorem]
∴ XY = \( \frac{1}{2} \) x 9 [Given]
∴ XY = \( \frac{9}{2} \)
∴ XY = 4.5 cm
Points Y and Z are the midpoints of sides BC and AC respectively [Given]
∴ YZ = \( \frac{1}{2} \) AB [Midpoint theorem]
∴ YZ = \( \frac{1}{2} \) x 5 [Given]
∴ YZ = \( \frac{5}{2} \)
∴ YZ = 2.5 cm
Points X and Z are the midpoints of sides AB and AC respectively [Given]
∴ XZ = \( \frac{1}{2} \) BC [Midpoint theorem]
∴ XZ = \( \frac{1}{2} \) x 11 [Given]
∴ XZ = \( \frac{11}{2} \)
∴ XZ = 5.5 cm
Ans: l(XY) = 4.5 cm, l(YZ) = 2.5 cm & l(XZ) = 5.5 cm
2. In figure 5.39, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR then prove that,
(i) SL = LR,
(ii) LN = (\( \frac{1}{2} \)) SQ
Given:
□PQRS and □MNRL are rectangles
M is the midpoint of side PR
To prove:
- SL = LR
- LN = (\( \frac{1}{2} \)) SQ
Proof:
□PQRS and □MNRL are rectangles [Given]
∴ ∠S ≅ ∠L ≅ ∠Q ≅ ∠N = 90° [Angles of rectangle]
∴ seg ML || seg PS …(i) [Corresponding angles test]
∴ seg MN || seg PQ …(ii) [Corresponding angles test]
In ∆PRS,
Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]
∴ Point L is the midpoint of seg SR ……(ii) [Converse of midpoint theorem]
∴ SL = LR
In ∆PRQ,
Point M is the midpoint of PR and seg MN || seg PQ [Given] [From (ii)]
∴ Point N is the midpoint of seg RQ ……(iii) [Converse of midpoint theorem]
∴ RN = NQ
In ∆RSQ,
Points L and N are the midpoints of seg SR and seg QR respectively [From (ii) and (iii)]
∴ LN = \( \frac{1}{2} \) SQ [Midpoint theorem]
Hence proved.
3. In figure 5.40, ∆ABC is an equilateral triangle. Points F,D and E are midpoints of side AB, side BC, side AC respectively. Show that ∆FED is an equilateral triangle.
Given:
∆ABC is an equilateral triangle
Point F is the midpoint of side AB
Point D is the midpoint of side BC
Point E is the midpoint of side AC
To prove:
∆FED is an equilateral triangle.
Proof:
∆ABC is an equilateral triangle [Given]
∴ AB ≅ BC ≅ AC ….(i) [Sides of an equilateral triangle]
Point F is the midpoint of side AB and Point D is the midpoint of side BC [Given]
∴ FD = \( \frac{1}{2} \) AC [Midpoint theorem] …..(ii)
Point D is the midpoint of side BC and Point E is the midpoint of side AC [Given]
∴ DE = \( \frac{1}{2} \) AB [Midpoint theorem]
i.e. DE = \( \frac{1}{2} \) AC [From (i)] …..(iii)
Point F is the midpoint of side AB and Point E is the midpoint of side AC [Given]
∴ FE = \( \frac{1}{2} \) BC
i.e. FE = \( \frac{1}{2} \) AC [From (i)] …..(iv)
∴ FD = DE = FE [From (ii), (iii) and (iv)]
∴ ∆FED is an equilateral triangle
Hence proved.
4. In figure 5.41, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \( \frac{PM}{PR} \) = \( \frac{1}{3} \)
[Hint : draw DN || QM]
Given:
seg PD is a median of ∆PQR
Point T is the midpoint of seg PD.
To Prove:
\( \frac{PM}{PR} \) = \( \frac{1}{3} \)
Construction:
Draw seg DN || seg QM such that P – M – N and M – N – R
Proof:
In ∆PDN,
Point T is the midpoint of seg PD and seg TM || seg DN [Given]
∴ Point M is the midpoint of seg PN [Construction and Q – T – M]
∴ PM ≅ MN [Converse of midpoint theorem]
In ∆QMR,
Point D is the midpoint of seg QR and seg DN || seg QM [Construction]
∴ Point N is the midpoint of seg MR [Converse of midpoint theorem]
∴ RN ≅ MN …..(ii)
∴ PM ≅ MN ≅ RN …..(iii) [From (i) and (ii)]
Now,
PR = PM + MN + RN
∴ PR = PM + PM + PM [From (iii) ]
∴ PR = 3PM
∴ \( \frac{PM}{PR} \) = \( \frac{1}{3} \)
Hence Proved.
Problem Set 5
1. Choose the correct alternative answer and fill in the blanks.
(i) If all pairs of adjacent sides of a quadrilateral are congruent then it is called ….
(A) rectangle
(B) parallelogram
(C) trapezium
(D) rhombus
Answer:
OPTION (D) – rhombus
(ii) If the diagonal of a square is 12 √2 cm then the perimeter of square is ……
(A) 24 cm
(B) 24 √2 cm
(C) 48 cm
(D) 48 √2 cm
Solution:
In ∆ABC,
AC² = AB² + BC²
∴ (12 √2)² = AB² + AB² [□ABCD is a square]
∴ 2AB² = 12² × (√2)²
∴ 2AB² = 144 × 2
∴ 2AB² = 288
∴ AB² = \( \frac{288}{2} \)
∴ AB² = 144
∴ √(AB²) = √144 [Taking square root on both sides]
∴ AB = 12
∴ Perimeter of □ABCD = 4 x 12
∴ Perimeter of □ABCD = 48 cm
OPTION (C) – 48 cm
(iii) If opposite angles of a rhombus are (2x)⁰ and (3x – 40)⁰ then value of x is …
(A) 100 ⁰
(B) 80 ⁰
(C) 160 ⁰
(D) 40 ⁰
Solution:
2x = 3x – 40 [Opposite angles of a rhombus are congruent]
∴ 2x – 3x = –40
∴ -x = –40
∴ x = 40°
OPTION (D) – 40°
2. Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
Given:
Let □ABCD be the rectangle.
∴ AB = 7 cm
and BC = 24 cm
To find:
Length of its diagonal
Solution:
In ∆ABC,
∠B = 90° [Angle of a rectangle]
∴ ∆ABC is a right angled triangle
∴ AC² = AB² + BC² [By Pythagoras theorem]
∴ AC² = 7² + 24²
∴ AC² = 49 + 576
∴ AC² = 625
∴ √(AC²) = √625 [Taking square root on both sides]
∴ AC = 25 cm
Ans: The length of the diagonal of the rectangle is 25 cm.
3. If the diagonal of a square is 13 cm then find its side.
Given:
□PQRS and □MNRL are rectangles
M is the midpoint of side PR
To prove:
- SL = LR
- LN = (\( \frac{1}{2} \)) SQ
Proof:
□PQRS and □MNRL are rectangles [Given]
∴ ∠S ≅ ∠L ≅ ∠Q ≅ ∠N = 90° [Angles of rectangle]
∴ seg ML || seg PS …(i) [Corresponding angles test]
∴ seg MN || seg PQ …(ii) [Corresponding angles test]
In ∆PRS,
Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]
∴ Point L is the midpoint of seg SR ……(ii) [Converse of midpoint theorem]
∴ SL = LR
In ∆PRQ,
Point M is the midpoint of PR and seg MN || seg PQ [Given] [From (ii)]
∴ Point N is the midpoint of seg RQ ……(iii) [Converse of midpoint theorem]
∴ RN = NQ
In ∆RSQ,
Points L and N are the midpoints of seg SR and seg QR respectively [From (ii) and (iii)]
∴ LN = \( \frac{1}{2} \) SQ [Midpoint theorem]
Hence proved.
4. Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of each side.
Given:
Let □STUV be the parallelogram
Ratio of two adjacent sides of a parallelogram is 3 : 4
Perimeter of □STUV = 112 cm
To find:
Length of each side
Solution:
Let the common multiple be x.
∴ ST = 3x cm and TU ≅ 4x cm
∴ ST ≅ UV = 3x cm and TU = SV = 4x cm …..(i) [Opposite sides of a parallelogram]
Perimeter of □STUV = 112 [Given]
∴ ST + TU + UV + SV = 112
∴ 3x + 4x + 3x + 4x = 112 [From (i)]
∴ 14x = 112
∴ x = \( \frac{112}{14} \)
∴ x = 8
The measure of each sides are,
∴ ST ≅ UV = 3x = 3 x 8 = 24 cm [From (i)]
∴ TU ≅ SV = 4x = 4 x 8 = 32 cm [From (i)]
Ans: The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm
5. Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
Given:
□PQRS is a rhombus
PR = 20 cm
QS = 48 cm
To find:
Length of side PQ
Solution:
In □PQRS,
∴ PT = \( \frac{1}{2} \) PR [Diagonals of a rhombus bisect each other]
∴ PT = \( \frac{1}{2} \) x 20
∴ PT = \( \frac{20}{2} \)
∴ PT = 10 cm
Also,
QT = \( \frac{1}{2} \) QS [Diagonals of a rhombus bisect each other]
∴ QT = \( \frac{1}{2} \) x 48
∴ QT = \( \frac{48}{2} \)
∴ QT = 24 cm
In ∆PTQ,
∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]
∴ ∆PQT is a right angled triangle
∴ PQ² = PT² + QT² [By Pythagoras theorem]
∴ PQ² = 10² + 24²
∴ PQ² = 100 + 576
∴ PQ² = 676
∴ √(PQ²) = √676 [Taking square root of both sides]
∴ PQ = 26 cm
Ans: The length of side PQ is 26 cm.
6. Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50⁰ then find the measure of ∠MPS.
Given:
□PQRS is a rectangle
∠QMR = 50⁰
To find:
∠MPS
Solution:
□PQRS is a rectangle [Given]
∴ PM = \( \frac{1}{2} \) PR …(i) [Diagonals of a rectangle bisect each other]
and MS = \( \frac{1}{2} \) QS …(ii) [Diagonals of a rectangle bisect each other]
Also,
PR ≅ QS …..(iii) [Diagonals of a rectangle are congruent]
∴ PM ≅ MS ….(iv) [From (i), (ii) and (iii)]
In ∆PMS,
PM ≅ MS [From (iv)]
∴ ∠MSP = ∠MPS = x° …..(v) [Isosceles triangle theorem]
∠QMR ≅ ∠PMS = 50° [Vertically opposite angles]
∵ ∠QMR = 50⁰
∴ ∠PMS = 50° ……(vi)
In ∆MPS,
∠PMS + ∠MPS + ∠MSP = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 50° + x⁰ + x⁰ = 180° [From (v) and (vi)]
∴ 50° + 2x⁰ = 180
∴ 2x = 180 – 50
∴ 2x = 130
∴ x = \( \frac{130}{2} \)
∴ x = 65°
∴ ∠MPS = x = 65° [From (v)]
Ans: ∠MPS = 65°
7. In the adjacent Figure 5.42, if seg AB || seg PQ , seg AB ≅ seg PQ, seg AC || seg PR, seg AC ≅ seg PR then prove that, seg BC || seg QR and seg BC ≅ seg QR.
Given:
seg AB || seg PQ
seg AB ≅ seg PQ
seg AC || seg PR
seg AC ≅ seg PR
To prove:
seg BC || seg QR and seg BC ≅ seg QR
Proof:
In □ABQP,
seg AB || seg PQ [Given]
seg AB ≅ seg PQ [Given]
∴ □ABQP is a parallelogram [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ segAP || segBQ …..(i)
∴ seg AP ≅ seg BQ …..(ii) [Opposite sides of a parallelogram are congruent]
In □ACRP,
seg AC || seg PR [Given]
seg AC ≅ seg PR [Given]
∴ □ACRP is a parallelogram [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg AP || seg CR …(iii)
∴ seg AP ≅ seg CR …(iv) [Opposite sides of a parallelogram are congruent]
In □BCRQ,
seg BQ || seg CR
seg BQ ≅ seg CR
∴ □BCRQ is a parallelogram [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg BC || seg QR
∴ seg BC ≅ seg QR [Opposite sides of a parallelogram are congruent]
Hence proved.
8*. In Figure 5.43, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = \( \frac{1}{2} \) (AB + DC)
Given:
□ABCD is a trapezium
AB || DC
Point P is the midpoint of seg AD
Point Q is the midpoint of seg BC
To prove:
PQ || AB and PQ = \( \frac{1}{2} \) (AB + DC)
Construction:
Join points A and Q. Extend seg AQ and let it meet produced DC at R.
Proof:
seg AB || seg DC and seg BC is their transversal [Given]
∴ ∠ABC ≅ ∠RCB [Alternate angles]
∴ ∠ABQ ≅ ∠RCQ ….(i) [B – Q – C]
In ∆ABQ and ∆RCQ,
∠ABQ ≅∠RCQ [From (i)]
seg BQ ≅ seg CQ [Q is the midpoint of seg BC]
∠BQA ≅ ∠CQR [Vertically opposite angles]
∴ ∆ABQ ≅ ∆RCQ [ASA test]
seg AB ≅ seg CR …(ii) [c. s. c. t.]
seg AQ ≅ seg RQ [c. s. c. t.]
∴ Q is the midpoint of seg AR ….(iii)
In ∆ADR,
Point P is the midpoint of seg AD and Point Q is the midpoint of seg AR [Given and from (iii)
∴ seg PQ || seg DR [Midpoint theorem]
i.e. seg PQ || seg DC ……..(iv) [D-C-R]
But, seg AB || seg DC …….(v) [Given]
∴ seg PQ || seg AB [From (iv) and (v)]
In ∆ADR,
PQ = \( \frac{1}{2} \) DR [Midpoint theorem]
PQ = \( \frac{1}{2} \) (DC + CR) [D – C – R]
PQ = \( \frac{1}{2} \) (DC + AB)
i.e. PQ = \( \frac{1}{2} \) (AB + DC)
Hence Proved.
9. In the adjacent figure 5.44, □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
Given:
□ABCD is a trapezium
AB || DC
Point M is the midpoint of diagonal AC Point N is the midpoint of diagonal DB
To prove:
MN || AB
Construction:
Join D and M. Extend seg DM to meet seg AB at point E such that A – E – B
Proof:
seg AB || seg DC and seg AC is their transversal [Given]
∴ ∠CAB ≅ ∠ACD [Alternate angles]
∴ ∠MAE ≅ ∠MCD ….(i) [C – M – A, A – E – B]
In ∆AME and ∆CMD,
∠AME ≅ ∠CMD [Vertically opposite angles]
seg AM ≅ seg CM [M is the midpoint of seg AC]
∠MAE ≅∠MCD [From (i)]
∴ ∆AME ≅ ∆CMD [ASA test]
∴ seg ME ≅ seg MD [c.s.c.t]
∴ Point M is the midpoint of seg DE. …(ii)
In ∆DEB,
Point M is the midpoint of diagonal DE and Point N is the midpoint of diagonal DB [Given and from (ii)]
∴ seg MN || seg EB [Midpoint theorem]
∴ seg MN || seg AB [A – E – B]
Hence proved.