Class 7: Maths Chapter 17 solutions. Complete Class 7 Maths Chapter 17 Notes.
Contents
RS Aggarwal Solutions for Class 7 Maths Chapter 17–Constructions
RS Aggarwal 7th Maths Chapter 17, Class 7 Maths Chapter 17 solutions
Ex 17A
Question 1.
Solution:
Steps of construction :
(i) Draw a line segment AB
(ii) From a point P outside AB, draw a line PQ meeting AB at Q.
(iii) At P, draw a line PQ making an angle
∠QPC equal to ∠PQB with the help of compass and ruler and produce it to D.
Then the line CD is parallel to AB. Which is the required line.
Question 2.
Solution:
Steps of construction :
(i) Draw a line AB and take a point P on it.
(ii) From P, draw a perpendicular PX and cut off PQ = 3.5 cm.
(iii) From Q, draw a perpendicular line CD and produce it to both sides.
Then, CD is the required line which is parallel to AB.
Question 3.
Solution:
Steps of construction :
(i) Draw a line / and take a point P on it.
(ii) At P, draw a perpendicular line PX and cut off PQ = 4.3 cm.
(iii) From Q, draw a line m which is perpendicular on PX, and produce it to both sides.
Then m is the required line which is parallel to l.
Ex 17B
Question 1.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6cm
(ii) At B, draw an arc of the radius 5cm.
(iii) At C, draw another arc of the radius 5.4cm which intersects the first arc at A
(iv) Join AB and AC
(v) With centre B and C and radius more than half of BC, draw arcs intersecting each other at L and M.
(vi) Join LM which intersects BC at Q and produce it to P.
Then PQ is perpendicular bisector of side BC.
Question 2.
Solution:
Steps of construction :
(i) Draw a line segment QR = 6cm
(ii) With centre Q and radius 4.4 cm draw an arc.
(iii) With centre R and radius 5.3 cm, draw another arc intersecting the first arc at P.
(iv) Join PQ and PR.
(v) With centre P and a suitable radius, draw an arc meeting PR at E and PQ at F.
(vi) With centres E and F, with same radius, draw two arcs intersecting each other at G.
(vii) Join PG and produce it to meet QR at S. Then PS is the bisector of ∠P.
Question 3.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.2 cm.
(ii) With centres B and C radius 6.2 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC.
∆ABC is the required equilateral triangle.
On measuring, each angle is equal to 60°.
Question 4.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5.3 cm.
(ii) With centre B and C, and radius 4.8 cm, draw arcs intersecting each other at A
(iii) Join AB and AC, Then ∆ABC is the required triangle.
(iv) Now, with centre A and a suitable radius draw an arc intersecting BC at L and M.
(v) Then with centre L and M, draw two arcs intersecting eachother at E.
(vi) Join AE intersecting BC at D. Then AD is perpendicular to BC.
On measuring ∠B and ∠C, each is equal to 55°.
Question 5.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8cm.
(ii) At A draw a ray AX making an angle of 60°.
(iii) Cut off AC = 5 cm from AX
(iv) Join CB.
Then ∆ABC is the required triangle.
Question 6.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.3 cm.
(ii) At C, draw a ray CY making an angle equal to 45°.
(iii) With centre C, and radius 6cm, draw an arc intersecting CY at A.
(iv) Join AB.
Then ∆ABC is the required triangle.
Question 7.
Solution:
(i) Draw a line segment AB = 5.2cm.
(ii) At A, draw a ray AX making an angle equal to 120°.
(iii) From AX cut off AC = 5.2cm.
(iv) Join BC.
Then ∆ABC is the required triangle.
(v) With centre A and some suitable radius draw an arc intersecting BC at L and M.
(vi) With centres L and M, draw two arcs intersecting each other at E.
(vii) Join AE intersecting BC at D Then AD is the perpendicular to BC.
Question 8.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.2 cm.
(ii) At B draw a ray BX making an angle of 60°.
(iii) At C draw another ray CY making an angle of 45° which intersect the ray BX at A. .
Then ∆ABC is the required triangle.
Question 9.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5.8 cm.
(ii) At B draw a ray BX making an angle of 30°
(iii) At C draw another ray CY making an angle of 30° intersecting the BX at A
Then ∆ABC is the required triangle.
On measuring AB and AC, AB = 3.5cm and AC = 3.5cm.
AB = AC
∆ABC is an isosceles triangle.
Question 10.
Solution:
Steps of construction :
In ∆ABC, ∠A = 45° and ∠C = 75°
But ∠A + ∠B + ∠C = 180°
⇒ 45° + ZB + 75° = 180°
⇒ ∠B = 180° – 45° – 75°
⇒ ∠B = 180° – 120° = 60°
(i) Draw a line segment AB = 7cm.
(ii) At A, draw a ray AX making an angle of 45°.
(iii) At B, draw another ray BY making an angle of 60° which intersects AX at C
Then ∆ABC is the required triangle.
Question 11.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.8 cm
(ii) At C, draw a ray CX making an angle of 90°.
(iii) With centre B, an radius 6.3cm draw an arc intersecting CX at A.
(iv) Join AB.
Then ∆ABC is the required triangle.
Question 12.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.5cm
(ii) At B, draw a ray BX making an angle of 90°.
(iii) With centre C and radius 6cm draw an arc intersecting BX at A
(iv) Join AC.
Then ∆ABC is the required triangle.
Question 13.
Solution:
One acute angle = 30°, then
second acute angle will be = 90° – 30° = 60° (Sum of acute angles = 90°)
Steps of construction :
(i) Draw a line segment BC = 6cm.
(ii) At B draw a ray BX making an angle of 30°.
(iii) At C draw another ray CY making an angle of 60° which intersects BX at A
Then ∆ABC is the required triangle.
Ex 17C Solutions
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(c) Supplement of 45° is 135°
135°+ 45° = 180°
Question 2.
Solution:
(b) Complement of 80° is 10°
10° + 80° = 90°
Question 3.
Solution:
(b) The angle is its own complement.
The measure of the angles will be 45° (45° + 45° = 90°)
Question 4.
Solution:
(a) The angle is one-fifth of its supplement
Let angle be x, then
x + 5x = 180°
⇒ 6x = 180°
⇒ x = 30°
Angle is 30°
Question 5.
Solution:
(b) Let angle is x
Then its complement angle=x-24° But x + x- 24° = 90°
⇒ 2x = 90° + 24° = 114°
⇒ x = 57°
The required angle is 57°
Question 6.
Solution:
(b) Let required angle = x
Then its supplement angle = x + 32
But x + x + 32° = 180°
⇒ 2x = 180° – 32 = 148°
⇒ x = 74°
Required angle = 74°
Question 7.
Solution:
(c) Two supplementary angle are in the ratio = 3 : 2
Let first angle = 3x
Second angle = 2x
But 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Smaller angle = 2x = 2 x 36° = 72°
Question 8.
Solution:
(b) In the figure ∠BOC = 132°
But ∠AOC + ∠BOC =180° (Linear pair)
⇒ ∠AOC + 132° = 180°
⇒ ∠AOC = 180° – 132° = 48°
Question 9.
Solution:
(c) In the figure, ∠AOC = 68°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 68° + x = 180°
⇒ x = 180° – 68° = 112°
Question 10.
Solution:
(b) In the figure,
AOB is a straight line
∠AOC + ∠BOC = 180° (Linear pair)
⇒ 2x – 10° + 3x + 15° = 180°
⇒ 5x = 180° + 10° – 15° = 175°
⇒ x = 35°
x = 35
Question 11.
Solution:
(d) In the figure,
AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 55° + x + 45° = 180°
⇒ x + 100° = 180°
⇒ x = 180° – 100° = 80°
Question 12.
Solution:
(a) AOB is a straight line
x + y = 180°
But 4x = 5y
Question 13.
Solution:
(b) AB and CD intersect each other at O and ∠AOC = 50°
∠BOD = ∠AOC = 50° (Vertically opposite angles)
Question 14.
Solution:
(a) AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 3x – 8° + 50° + x + 10° = 180°
⇒ 4x = 180° + 8° – 50° – 10°
⇒ 4x = 128°
⇒ x = 32°
Question 15.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ACD = 132° and ∠A = 54°
Ext. ∠ACD = ∠A + ∠B
⇒ 132° = 54° + ∠B
⇒ ∠B = 132° – 54° = 78°
Question 16.
Solution:
(c) In ∆ABC,
Side BC is produced to D
∠A = 45°, ∠B = 55°
Ext. ∠ACD = ∠A + ∠B = 45° + 55° = 100°
Question 17.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ABC = 70° and ∠ACD = 120°
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 120° = ∠BAC + 70°
⇒ ∠BAC = 120° – 70° = 50°
Question 18.
Solution:
(c) In the figure,
∠AOB = 50°, ∠BOC = 90°
∠COD = 70°, ∠AOD = x.
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)
⇒ 50° + 90° + 70° + x = 360°
⇒ 210 + x = 360°
⇒ x = 360° – 210°
⇒ x = 150°
Question 19.
Solution:
(c) In the figure,
Side BC of ∆ABC is produced to D
CE || BA is drawn
∠A = 50° and ∠ECD = 60°
AB || CE
∠ABC = ∠ECD (corresponding angle) = 60°
But in ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angles of a triangles)
⇒ 50° + 60° + ∠ACB = 180°
⇒ ∠ACB = 180° – 50° – 60° = 70°
Question 20.
Solution:
(b) In ∆ABC,
∠A = 65°, ∠C = 85°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° +∠B+ 85° = 180°
⇒ 150° + ∠B = 180°
⇒ ∠B = 180° – 150° = 30°
Question 21.
Solution:
(d) Sum of angles of a triangle = 180°
Question 22.
Solution:
(c) Sum of angles of a quadrilateral = 360°
Question 23.
Solution:
(b) In the figure, AB || CD
∠OAB = 150°, ∠OCD = 120°
From O, draw OE || AB or CD
AB || DE
∠OAB + ∠AOE =180°
⇒ 150° + ∠AOE = 180°
⇒ ∠AOE = 180° – 150° = 30°
Similarly DE || CD
∠EOC + ∠OCD = 180°
⇒ ∠EOC + 120° = 180°
⇒ ∠EOC = 180° – 120° = 60°
Now ∠AOC = ∠AOE + ∠EOC = 30° + 60° = 90°
Question 24.
Solution:
(a) In the given figure,
PQ || RS,
∠PAB = 60° and ∠ACS = 100°
PQ || RS
∠ABC = ∠PAB (alternate angles) = 60°
But Ext. ∠ACS = ∠BAC + ∠ABC
⇒ 100° = ∠BAC + 60°
⇒ ∠BAC = 100° – 60° = 40°
Question 25.
Solution:
(c) In the figure, AB || CD || EF
∠ABG =110° and ∠GCD = 100°
∠BGC = x°
AB || EF
∠ABG + ∠BGE = 180°
⇒ 110° + ∠BGE = 180°
⇒ ∠BGE = 180° – 110° = 70°
Similarly CD || EF
∠GCD + ∠CGF = 180°
⇒ 100° + ∠CGF = 180°
⇒ ∠CGF = 180° – 100° = 80°
But ∠BGE + ∠BGC + ∠CGF = 180°
⇒ 70° + x + 80° = 180°
⇒ 150° + x = 180°
⇒ x = 180° – 150° = 30°
Question 26.
Solution:
(d) Sum of any two sides of a triangle is always greater than the third side
Question 27.
Solution:
(d) The diagonals of a rhombus always bisect each other at right angles.
Question 28.
Solution:
(c) In ∆ABC, ∠B = 90°
AB = 5 cm and AC = 13 cm
But AC² = AB² + BC² (By Pythagoras Theorem)
⇒ (13)² = (5)² + BC²
⇒ 169 = 25 + BC2
⇒ BC² = 169 – 25 = 144 = (12)²
BC = 12 cm
Question 29.
Solution:
(c) In ∆ABC, ∠B = 37°, ∠G = 29°
But ∠A + ∠B + ∠C = 180° (angles of a triangle)
⇒ ∠A + 37° + 29° = 180°
⇒ ∠A + 66° = 180°
⇒ ∠A = 180° – 66° = 114°
Question 30.
Solution:
(c) The ratio of angles of a triangle is 2 : 3 : 7
But sum of angles of a triangle = 180°
Question 31.
Solution:
In ∆ABC,
Let 2∠A = 3∠B = 6∠C = x
Question 32.
Solution:
(a) In ∆ABC,
∠A + ∠B = 65°, ∠B + ∠C = 140°
∠A = 65°
∠C = 140° – ∠B
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° – ∠B + 140° – ∠B + ∠B = 180°
⇒ 205° – ∠B = 180°
⇒ ∠B = 205° – 180° = 25°
Question 33.
Solution:
(b) In ∆ABC, ∠A – ∠B = 33°
and ∠B – ∠C = 18°
∠A = 33° + ∠B and ∠C = ∠B – 18°
But ∠A + ∠B + ∠C = 180°
⇒ 33° + ∠B + ∠B + ∠B – 18° = 180°
⇒ 3∠B = 180° – 33° + 18° = 165°
⇒ ∠B = 55°
Question 34.
Solution:
(c) In ∆ABC
∠A + ∠B + ∠C= 180° (Sum of angles of a triangle)
But angles are (3x)°, (2x – 7)° and (4x – 11)°
3x + (2x – 7) + (4x – 11)° = 180°
⇒ 3x + 2x – 7 + 4x – 11° = 180°
⇒ 9x – 18° = 180°
⇒ 9x = 180° + 18° = 198°
⇒ x = 22°
Question 35.
Solution:
(c) ∆ABC is a right angled, ∠A = 90°
AB = 24 cm, AC = 7 cm
but BC² = AB² + AC²
⇒ BC² = (24)² + (7)² = 576 + 49 = 625 = (25)²
BC = 25 cm
Question 36.
Solution:
(b) Let AB is a ladder and A is the window
BC = 15 m, AC = 20 m
Now in right ∆ABC
AB² = BC² + AC² = (15)² + (20)² = 225 + 400 = 625 = (25)²
AB = 25 m
Length of ladder = 25 m
Question 37.
Solution:
(a) Let AB and CD are two poles such that
AB = 6 m, CD = 11 m
and distance between two poles BD = 12m
From A, draw AE || BD
AE = BD = 12m
CE = CD – ED = 11 – 6 = 5 m
Now in right ∆AEC
AC² = AE² + CE² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13 m
Distance between tops of poles = 13 m
Question 38.
Solution:
(d) ∆ABC is an isosceles triangle
∠C = 90°,
AC = 5 cm
BC = AC = 5 cm
In right ∆ABC
AB² = AC² + BC² = (5)² + (5)² = 25 + 25 = 50 = 2 x 25
AB = √(2 x 25) = 5√2 cm
RS Aggarwal Solutions for Class 7 Maths Chapter 17: Download PDF
RS Aggarwal Solutions for Class 7 Maths Chapter 17–Constructions
Download PDF: RS Aggarwal Solutions for Class 7 Maths Chapter 17–Constructions PDF
Chapterwise RS Aggarwal Solutions for Class 7 Maths :
- Chapter 1–Integers
- Chapter 2–Fractions
- Chapter 3–Decimals
- Chapter 4–Rational Numbers
- Chapter 5–Exponents
- Chapter 6–Algebraic Expressions
- Chapter 7–Linear Equations in One Variable
- Chapter 8–Ratio and Proportion
- Chapter 9–Unitary Method
- Chapter 10–Percentage
- Chapter 11–Profit and Loss
- Chapter 12–Simple Interest
- Chapter 13–Lines and Angles
- Chapter 14–Properties of Parallel Lines
- Chapter 15–Properties of Triangles
- Chapter 16–Congruence
- Chapter 17–Constructions
- Chapter 18–Reflection and Rotational Symmetry
- Chapter 19–Three-Dimensional Shapes
- Chapter 20–Mensuration
- Chapter 21–Collection and Organisation of Data (Mean, Median and Mode)
- Chapter 22–Bar Graphs
- Chapter 23–Probability
About RS Aggarwal Class 7 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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