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OP Malhotra Class 9 ICSE Maths Solutions Chapter 11 Rectilinear Figures
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 11, Rectilinear Figures. Here students can easily find step by step solutions of all the problems for Rectilinear Figures, Exercise 11a, 11b 11c and 11d Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 11 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Rectilinear Figures Exercise 11(a) Solution :
Question no – (1)
Solution :
(i) In rectangle, diagonals are equal and bisect each other at ‘o’
So, AO = BO
∠OAB = ∠OBA = 18°
∠ABC = 90°
x = (90° – 18°) = 72°
(ii) In rectangle, PQRS, diagonals bisects each others at T
∠PTQ = 120° (Given)
∴ ∠TPQ + ∠TQP = 180° – 120°
= 60°
But, PT = QT
∴ ∠TPQ = ∠TQP = 60/2 = 30°
∴ ∠SRT = ∠TPQ
Her, So, b = 30°
and, a + b = 90°
∴ a = 90° – 30° = 60°
(iii) From figure, Rhombus EF GH,
∠EFG = 140°
But ∠EFG + ∠FGH = 180°
= 140 + ∠FGH = 180°
= ∠FGH = 180° – 140°
= 40°
Diagonals bisects each other at right angles,
x = 1/2 × 40°
= 20°
(iv) Here, if the diagonals of rhombus bisect each other at right angles.
∴ ∠PLQ = 90°
In PLQ
∴ ∠PRQ + ∠LAP = 90°
= 24 6+ X = 90°
= X = 90° – 34 °
= 56°
AND, diagonals QS bisect ∠Q and ∠S
So, x = y
y = 56°
(v) From figure, square ABCD,
Diagonal AC intersect x DY
∠DXC = 112°
If the diagonals AC bisects the angles A and C,
Then ∠ACB = ∠XCY = 45°
∠CXY + ∠CXD = 120°
= ∠C × Y = 180° – 112°
= 68°
Now, in △CXY
= ∠C XY + ∠XYC + ∠XCY = 180°
= 68° + d + 45° = 180°
= d = 67°
(vi) In figure CDEF
DF is Diagonal, EN is line segment which intersects. DF at M such that,
∠EMF = 38, and Let, ∠MNE = x
In ∠EMF = ∠DMN
∴ ∠MHN = 38
Diagonals DF bisects the ∠Q and ∠F
∴ ∠MDN = 45
In, △DMN
∠MNC = ∠DMN + ∠MDN
= x = 38 + 45
= 83
(vii) From Figure,
AB || DC and AD = BC
and, ∠A = a, ∠C = 75°
If ABCD is an isolates trapezium , then
∠A + C = 180°
= ∠B + ∠D = 180°
∴ ∠A + ∠C = 180°
= a + 75° = 180°
= a = (180 – 75°)
= 105°
(ix) ABCD is an isosceles
trapezium where AB || DC
AC = BD are diagonals
these intersects at ‘0’
= ∠0DC = ∠0CD = 180°
Now, in ∠0CD,
a + ∠0DC + ∠0CD
= a + 34 + 34 = 180°
= a + 180° – 68°
= 112°
and, ∠ADB = ∠ACB = 6
∴ △0BC, a = 72° + B
= 112° – 72° = B
= ∠b = 40°
and ∠a = 112°
Question no – (2)
Solution :
Here, In a parallelogram ABCD,
∠A = ∠B = 1 : 5
Let, ∠A = x and ∠B = 5x
∴ ∠A + ∠B = 180°
= x + 5x = 180°
= 6x = 180°
= x = 180/6 = 30°
∴ ∠A = 30°
= ∠3 = 5x
= 30° = 150°
∴ ∠C = ∠A and ∠3 = ∠D
Question no – (3)
Solution :
Let, smallest angle is = x
and other greatest = (2x – 20°)
∴ x + 2x – 20 = 180°
= 3x = 180° + 20°
= x = 20/3°
∴ smallest angle 200/3
Now, other greatest angle
= 2 × 200/3° – 20
= 400/3 – 20
= 400 – 60/3
= 340/3
= 113 1/3°
Therefore, the angles are, 200/3°, 200/3°, 113 1/3°, 113 1/3°
Question no – (4)
Solution :
Here, ABCD is a rhombus
∠A = 50° Diagonal AC and BD
bisect at point ‘0’
∠AOB = 90°
If the diagonal bisect the opposite angles.
∴ ∠0AB = 50°/2 = 25°
∴ ∠0BA + (90 + 25) = 180°
= ∠0BA = 180° – 115°
= 65°
Therefore, the angles are, 25°, 65°, 90°
Question no – (5)
Solution :
In ABCD rectangle, AC and BD diagonals are bisect at point ‘P’
∠ABD = 50°
In △APB, AP = BP
= ∠PAB = ∠PBA
∴ ∠APB = 180° – (50° + 50°)
= 180 – 100°
= 80°
∴ ∠CPP = ∠APD …(opposite angles)
∴ ∠CPD = 80°
Question no – (6)
Solution :
Here, In parallelogram, ABCD,
Let, ∠B = x
and, ∠A = 2/3 x
= ∠A + ∠B = 180°
∴ 2/3x + x = 180°
= 2x + 3x/3 = 180°
= x = 3 × 180/5
= 108°
∴ ∠B = 108°
Question no – (8)
Solution :
The diagonal of rhombus if ABCD bisect each other,
= AO = OC = 24/2 = 12 cm
= BO = OD = 18/2 = 9 cm
and, ∠AOB = 90
∴ In △AOB
AB2 = AO2 + BO2
= (12)2 + (9)2
= 144 + 81
= 225
∴ AB = 15
Therefore, Side AB = BC = CD = DA = 15 cm.
Question no – (9)
Solution :
In rhombus ABCD, AC and BD are diagonals
Both interest at right angles
AO = OC = 8/2 = 4 cm and BO = OD
∠AOB = 90
Now, in △AOB,
AB2 = AO2 + BO2
= (5)2 = 42 + B02
= 302 = 25 – 16
= 9
= BO = 3
∴ BD = 2 × BO
= 2 × 3
= 6 cm
∴ Area rhombus,
= 8 × 6/2
= 24 CM2
Question no – (10)
Solution :
PORS is a rhombus
= In rhombus PQRS
= PQ = 3 cm
SL = 2.5 cm
Diagonals bisect at point ‘0’
(i) If it is given that PQ = 3c, calculate the perimeter of PORS.
= Perimeter,
= 4 × 3
= 12 cm
(ii) If the height of the rhombus is 2.5 cm, calculate the area.
= Area of PQRS
= (3 × 2.5) cm2
= 7.5 cm2
(iii) If the diagonals cut at O, state the size of the angle POQ in degrees. (ICSE)
= If the diagonals bisect each other at right angle,
∴ ∠POQ = 90°
Question no – (12)
Solution :
Here, In figure, EFGH is isosceles trapezium
∴ EH = FG
and, ∠HEF = ∠GFE
= 2y2 – 25 = y2 + 24
= y2 = 24 + 45
= 49
= y = – 7
∴ y = 7 and, y = – 7
Question no – (13)
Solution :
In rhombus PQRS, PR and QS are diagonals which bisect at point ‘0’
∠SRT = 152°
∴ ∠PRS + ∠SRT = 180°
= ∠PRS = 180° – ∠SRT
= 180° – 152°
= 28°
∴ ∠SPQ = ∠SRQ = 2∠PRS
= 2 × 28°
= 56°
But here,
∠PQR + ∠SRQ = 180°
= ∠PQR = 180° – ∠SRQ
= 180° – 56°
= 124°
∴ x = 1/2 ∠PQR
= 1/2 × 124°
= 62°
∴ x = 62°
and, y = 92°
= z = 28°
Question no – (15)
Solution :
Here, ABCD is square and EBC is an equilateral triangle. ED is Join.
In ECD, CD = CE
∴ ∠CED = ∠CDE
But ∠DCE = ∠DCB + ∠BCE
= 90° + 60°
= 150°
So, ∠CED + ∠CDE = 180° – 150°
= 30°
∴ 2 ∠CED = 30°
= ∠CED = 30/2 = 15°
and, ∠BED = 60° – 15°
= 45°
Question no – (16)
Solution :
ABCD is square and ABO is an equilateral triangle.
In OAD,
∠OAD = ∠BAD – ∠OAB
= 90 – 60
= 30
∴ ∠AOD = ∠ADO
∴ ∠AOD = 150/2
= 75
Similarly, In △OBC,
∠BOC = 75°
and, ∠DOC + ∠AOD + ∠BOC + ∠AOB = 360°
= ∠DOC + 75° + 75° + 60° = 360°
= ∠DOC = 360° – 210°
= 150°
Next Chapter Solution :
👉 Chapter 13 👈
