Rs Aggarwal 2020 2021 for Class 7 Maths Chapter 17 - Constructions

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Rs Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 17 Constructions are provided here with simple step-by-step explanations. These solutions for Constructions are extremely popular among Class 7 students for Maths Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 204:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw a line AB . \end{array} \begin{array}{l} 2 . Take a point Q on AB and a point P outside AB, and join PQ. \\ 3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z. \\ 4. With P as the centre and the same radius, draw an arc cutting QP at Y . \\ 5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E. \\ 6. Join PE and produce it on both the sides to get the required line . \end{array}$

Page No 204:

Answer:

$\begin{array}{l} Steps for construction: \\ 1. Let AB be the given line . \\ 2. Take any two points P and Q on AB . \\ 3 . Construct ∠ B P E = 90 ° and ∠ B Q F = 90 ° \\ 4. With P as the centre and the radius equal to 3 .5 cm, cut PE at R. \\ 5 . With Q as the centre and the radius equal to 3 .5cm, cut QF at S. \\ 6 . Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3 .5 cm from it . \end{array}$

Page No 204:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Let l be the given line . \\ 2 . Take any two points A and B on line l. \\ 3. Construct ∠ B AE = 90 ° and ∠ ABF=90 ° \\ 4. With A as the centre and the radius equal to 4 .3 cm, cut AE at C . \\ 5 . With B as the centre and the radius equal to 4 .3 cm, cut BF at D . \\ 6 . Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4 .3 cm from it . \end{array}$

Page No 207:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw a line segment (AB) of length 5 cm. \\ 2 . Draw an arc of radius 5 .4 cm from the centre (A). \\ 3 . With B as the centre, draw another arc of radius 3 .6 cm, cutting the previous arc at C. \\ 4 . Join AC and BC . \\ 5 . Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC . \\ 6 . Similarly, taking C as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q. \\ 7. Join PQ. \end{array} Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.$

Page No 207:

Answer:

$\begin{array}{l} Steps of construction: \\ 1. Draw a line segment QR of length 6 cm . \\ 2 . Draw arcs of 4 .4 cm and 5 .3 cm from Q and R, respectively. They intersect at P. \\ 3 . Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively. \\ 4 . With S as the centre and the radius more than half of ST, draw an arc . \\ 5. With T as the centre and the same radius, draw another arc cutting the previously drawn arc at X . \\ 6. Join P and X . \end{array} Then, PX is the bisector of ∠ P .$

Page No 207:

Answer:

$\begin{array}{l} Steps of construction: \\ 1. Draw AB of length 6 .2 cm . \\ 2 . By taking the centres as A and B, draw equal arcs of 6 .2 cm on the same side of AB, cutting each other at C . \\ 3 . Join AC and BC . \end{array}$

When we will measure angles of triangle using protractor then we find that all angles are equal to 60 $°$

Page No 207:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw BC=5 .3 cm \\ 2 . Draw an arc of radius 4 .8 cm from the centre, B . \\ 3. Draw another arc of radius 4 .8 cm from the centre, C . \\ 4 . Both of these arcs intersect at A . \\ 5 . Join AB and AC . \\ 6 . With A as the centre and any radius, draw an arc cutting BC at M and N. \\ 7 . With M as the centre and the radius more than half of MN, draw an arc. \\ 8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at P. \\ 9. Join AP, cutting BC at D. \\ Then, AD ⊥ B C \end{array}$

Page No 207:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw AB of length 3 .8 cm . \\ 2 . Draw ∠BAZ=60° \\ 3 . With the centre as A, cut ray AZ at 5 cm at C. \\ 4 Join BC. \end{array} Then, ABC is the required triangle.$

Page No 208:

Answer:

$\begin{array}{l} \begin{array}{l} \begin{array}{l} Steps of construction: \\ 1. Draw AC= 6 cm \end{array} \\ 2. Draw ∠ ACZ = 45 ° \end{array} \\ 3 . With C as the centre, cut ray BZ at 4.3 cm at point B . \\ \begin{array}{l} 4. Join AB. \\ Then, ABC is the required triangle . \end{array} \end{array}$

Page No 208:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw AB=5.2 cm \\ 2 . Draw ∠BAX=120° \\ 3 . With A as the centre, cut the ray AX at 5.3 cm at point C. \\ 4 . Join BC. \\ 5 . With A as the centre and any radius, draw an arc cutting BC at M and N . \\ 6. With M as the centre and the radius more than half of MN, draw an arc. \\ 7 . With N as the centre and the same radius as before, draw another arc cutting the previously drawn arc at P. \\ 8 . Join AP meeting BC at D . \end{array} ∴ AD ⊥ BC$

Page No 208:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw BC=6.2 cm \\ 2 . Draw ∠BCX=45° \\ 3 . Draw ∠CBY=60° \end{array} 4 . The ray CX and BY intersect at A . Then, ABC is the required triangle .$

Page No 208:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw BC=5 .8 cm \end{array} 2 . Draw ∠ BCY = 30 ° 3 . Draw ∠ CBX = 30 ° 4 . The ray BX and CY intersect at A . Then, ABC is the required triangle . On measuring AB and AC : AB = AC = 3.4 cm$

Page No 208:

Answer:

$By angle sum property : ∠ B = 180 ° - ∠ A - ∠ C = 180 ° - 45 ° - 75 ° = 60 ° \begin{array}{l} Steps of construction: \\ 1 . Draw AB=7cm \\ 2 Draw ∠BAX= 45° \\ 3. Draw ∠ABY= 60° \\ 4.The ray AX and BY intersect at C. \end{array} Then, ABC is the required triangle.$

Page No 208:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 .Draw BC=4 .8 cm \\ 2 .Draw a perpendicular on C such that ∠ {C is equal to 90°.}^{} \\ 3. Draw an arc of radius 6 .3 cm from the centre B. \\ 4 . Join AB. \end{array}$

Page No 208:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw AB=3 .5 cm \\ 2 . Construct ∠ ABX = 90 ° \\ 3. With centre A, d raw an arc of radius 6 cm cutting BX at C. \\ 4 . Join AC. \end{array} Then, ABC is the required triangle.$

Page No 208:

Answer:

$\begin{array}{l} Here, ∠ A=30° and ∠ C=90 ° \\ By angle sum property: \\ ∠ B=60 ° \end{array} {1. Draw the hypotenuse AB of length 5.6 cm.}^{} 2. Draw ∠BAX=30° and ∠ABY=60° 3 . The ray AX and BY intersect at C . Then, ABC is the required triangle .$

Page No 208:

Answer:

$(c) 135 ° \begin{array}{l} {Supplement of 45}^{°} {=180}^{°} - 45^{°} \\ {=135}^{°} \end{array}$

Page No 208:

Answer:

$(b) 10 ° \begin{array}{l} {Complement of 80}^{°} = 90^{°} - 80^{°} \\ {=10}^{°} \end{array}$

Page No 208:

Answer:

$\begin{array}{l} (b) 45 ° \end{array} Suppose the angle is x ° . Then, the complement is also x ° . Complement of x ° = 90 ° - x ° \Rightarrow x ° = 90 ° - x ° \Rightarrow x ° + x ° = 90 ° \Rightarrow 2 x ° = 90 ° \Rightarrow x = \frac{90}{2} \Rightarrow x = 45$

Page No 208:

Answer:

$(a) 30 ° Suppose the angle is x . x = \frac{(180 - x)}{5} \Rightarrow 5 x = 180 - x \Rightarrow 5 x + x = 180 \Rightarrow x = \frac{180}{6} \Rightarrow x = 30 °^{}$

Page No 208:

Answer:

$(b) 57 ° S uppose the angle is x . x = 90 - x + 24 \Rightarrow x + x = 114 \Rightarrow 2 x = 114 \Rightarrow x = \frac{114}{2} \Rightarrow x = 57 °^{}$

Page No 208:

Answer:

$(b) 74 ° Suppose the angle is x . x = 180 - x - 32 \Rightarrow x + x = 148 \Rightarrow 2 x = 148 \Rightarrow x = \frac{148}{2} \Rightarrow x = 74 °$

Page No 208:

Answer:

$(c) 72 ° \begin{array}{l} Supplementary angles: \\ 3 x + 2 x = 180 \\ => x = \frac{180}{5} \end{array} \Rightarrow x = 36 ° \begin{array}{l} Smaller angle = (2 \times 36 °) \\ =72 °^{} \end{array}$

Page No 208:

Answer:

$(b) 48 ° ∠ A O C + ∠ B O C = 180 ° (linear pair) ∠ A O C = 180 ° - ∠ B O C = 180 ° - 132 ° = 48 °$

Page No 208:

Answer:

$(x) 112 ∠ A O C + ∠ A O B = 180 ° (linear pair) 68 ° + x ° = 180 ° \Rightarrow x ° = 180 ° - 68 ° \Rightarrow x ° = 112 °$

Page No 208:

Answer:

$\begin{array}{l} (c) x = 35 \\ (2 x - 10) + (3 x + 15) = 180 \\ = > 2 x - 10 + 3 x + 15 = 180 \\ = > 5 x + 5 = 180 \\ = > 5 x = 180 - 5 \\ = > 5 x = 175 \\ = > x = \frac{17 5^{35}}{5^{1}} \\ = > x = 35 \end{array}$

Page No 209:

Answer:

$(d) x = 80 x + 55 + 45 = 180 (linear pair) \Rightarrow x = 180 - 55 - 45 \Rightarrow x = 180 - 100 \Rightarrow x = 80$

Page No 209:

Answer:

$(a) 100 \begin{array}{l} x + y = 180 (linear pair) \\ => x + \frac{4}{5} x = 180 ° \\ =>9 x = 5 \times 180 \\ => x = 100 \end{array}$

Page No 209:

Answer:

$(b) 50 ° \begin{array}{l} Here, ∠ AOC and ∠ BOD are vertically opposite angles . \\ ∴ ∠ AOC= ∠ BOD \\ Given, ∠ {AOC=50}^{0} \\ ∴ ∠ {BOD=50}^{0} \end{array}$

Page No 209:

Answer:

$(a) 32 \begin{array}{l} (3 x - 8) ° + (x + 10) ° + 50 ° = 180 ° (linear pair) \\ =>4 x ° + 52 ° = 180 ° \\ =>4 x ° = 128 ° \\ => x ° = 32 ° \end{array} ∴ x = 32$

Page No 209:

Answer:

$(b) 78 ° \begin{array}{l} ∠ ACD= ∠ ABC+ ∠ BAC (exterior angle property) \\ => ∠ ABC=132° - 54 ° = 78 ° \end{array}$

Page No 209:

Answer:

$(c) 100 ° \begin{array}{l} ∠ ACB = ∠ ABC+ ∠ BAC (exterior angle property) \\ = (45 ° + 55 °) \\ =100 ° \end{array}$

Page No 209:

Answer:

$(b) 50 ° \begin{array}{l} ∠ BCA = 180^{0} - 120^{0} (linear pair) \\ = 60^{0} \\ ∠ BAC = 180^{0} - (60^{0} + 70^{0}) (angle sum property of triangles) \\ = 50^{0} \end{array}$

Page No 209:

Answer:

$(c) 150 ° \begin{array}{l} x^{0} + 70^{0} + 50^{0} + 90^{0} = 360^{0} (complete angle) \\ = > x^{0} = 360^{0} - 210^{0} \\ = 150^{0} \end{array}$

Page No 209:

Answer:

$(c) 70 ° \begin{array}{l} Here, ∠ ACE = ∠ BAC = 50^{0} [alternate angles] \\ ∠ACB+∠ACE+∠DCE=180° (linear pair) \\ ∠ ACB = 180^{0} - (50 ° + 60 °) \\ = 180 ° - 110 ° \end{array} = 70 °$

Page No 209:

Answer:

$(b) 30 ° \begin{array}{r} ∠ A+ ∠ B+ ∠ C = 180^{0} \\ = > ∠ B = 180^{0} - (65^{0} + 85^{0}) \\ = > ∠ B = 180^{0} - 150^{0} \\ = > ∠ B = 30^{0} \end{array}$

Page No 209:

Answer:

$(d) 180^{0}$

Page No 210:

Answer:

$\begin{array}{l} {(c) 360}^{0} \end{array}$

Page No 210:

Answer:

$(b) 90 ° \begin{array}{l} Draw a parallel line through O and produce AB and CD on R and P, respectively . \\ ∴ ∠ OCD = ∠ {COQ=120}^{0} (alternate angles) \end{array} ∠ {COS=180}^{0} - 120^{0} (linear pair) = 60^{0} Similarly, ∠ AOQ= ∠ {BAO=150}^{0} (alternate angles) ∠ {AOS=180}^{o} - 150^{0} (linear pair) = 30^{0} ∠ A O C = ∠ A O S + ∠ C O S ∴ ∠ {AOC=60}^{0} + 30^{0} = 90^{0}$

Page No 210:

Answer:

$(a) 40 ° \begin{array}{l} ∠ PAC = ∠ ACS = 100^{0} [alternate angles] \\ ∠ PAB + ∠ BAC = 100^{0} \\ = > ∠ BAC = 100 ° - 60 ° = 40 ° \end{array}$

Page No 210:

Answer:

$(c) 30 \begin{array}{l} Here, ∠ DCG + ∠ CGF = 180^{0} (angles on the same side of a transversal line are supplementary) \\ = > ∠ CGF = 180^{0} - 100 ° = 80 ° \\ ∠ ABG = ∠ BGF = 110^{0} [alternate angles] \\ x^{0} + ∠ CGF = 110^{0} \\ = > x^{0} = 110^{0} - 80^{0} \\ = > x^{0} = 30^{0} \end{array} ∴ x = 30$

Page No 210:

Answer:

$(d) greater than the 3 rd side$

Page No 210:

Answer:

$(d) The diagonals of a rhombus always bisect each other at right angles .$

Page No 210:

Answer:

$(c) 12 cm \begin{array}{l} In a right angle triangle: \\ {AC}^{2} = {AB}^{2} + {BC}^{2} (Pythagoras theorem) \\ = > {BC}^{2} = 13^{2} - 5^{2} \\ {=> BC}^{2} = 169 - 25 \\ = > {BC}^{2} = 144 \\ = > BC =±12 \end{array} The length cannot be negative. ∴ BC= 12 cm$

Page No 210:

Answer:

$(c) 114 ° \begin{array}{l} In triangle ABC: \\ ∠ A+ ∠ B+ ∠ {C=180}^{0} \\ = > ∠ {A=180}^{0} - (37^{0} + 29^{0}) \\ = > ∠ {A=180}^{0} - (66^{0}) \\ = 114^{0} \end{array}$

Page No 210:

Answer:

$(c) 105 ° Suppose the angles of a triangle are 2 x, 3 x and 7 x . \begin{array}{l} Sum of the angles of a triangle is 180°. \\ 2 x + 3 x + 7 x = 180 \\ = > 12 x = 180 \\ = > x = 15^{0} \\ Measure of the largest angle = 15^{0} \times 7 = 105^{0} \end{array}$

Page No 210:

Answer:

$(c) 60 ° G i v e n : 2 ∠ A = 3 ∠ B or ∠ A= \frac{3}{2} ∠ B 3 ∠ B = 6 ∠ C, or ∠C= \frac{1}{2} ∠B \begin{array}{l} In a △ ABC: \\ ∠ A+ ∠ B+ ∠ C = 180^{0} \\ => \frac{3}{2} ∠ B+ ∠ B + \frac{1}{2} ∠ B = 180^{0} \\ => \frac{3 ∠ B+2 ∠ B+ ∠ B}{2} = 180^{^{0}} \\ => \frac{6 ∠ B}{2} = 180^{0} \\ => ∠ B= \frac{360^{0}}{6} \\ => ∠ {B=60}^{0} \end{array}$

Page No 210:

Answer:

$(a) 25 ° Given : ∠ A + ∠ B = 65 ° ∠ A = 65 ° - ∠ B . . . (i) ∠ B + ∠ C = 140 ° ∠ C = 140 ° - ∠ B . . . (i i) In ∆ ABC : ∠ A + ∠ B + ∠ C = 180 ° Putting the value of ∠ B and ∠ C : \Rightarrow 65 ° - ∠ B + ∠ B + 140 ° - ∠ B = 180 ° \Rightarrow - ∠ B = 180 ° - 205 ° \Rightarrow ∠ B = 25 °$

Page No 210:

Answer:

$(b) 55 ° \begin{array}{l} In △ ABC: \\ ∠ A + ∠ B + ∠ C = 180^{0} . . . (i) \\ Given: \\ ∠ A - ∠ B = 33^{0} = > ∠ A = ∠ B + 33^{0} . . . (i i) \\ ∠ B - ∠ C = 18^{0} = > ∠ C = ∠ B - 18^{0} . . . (i i i) \\ Using (i i) and (i i i) in equation (i) : \\ = > ∠ {B + 33}^{0} + ∠ B + ∠ B - 18^{0} = 180^{0} \\ => 3 ∠ {B + 15}^{0} = 180^{0} \\ => 3 ∠ {B = 165}^{0} \\ => ∠ B = \frac{165^{0}}{3} = 55^{0} \end{array}$

Page No 211:

Answer:

$(c) 22 \begin{array}{l} Sum of the angles of a triangle is 180°. \\ (3 x) ° + (2 x - 7) ° + (4 x - 11) ° = 180 ° \\ = > 9 x ° - 18 ° = 180 ° \\ = > 9 x ° = 198 ° \\ = > x ° = 22 ° \end{array} \Rightarrow x = 22$

Page No 211:

Answer:

$(c) 25 cm \begin{array}{l} In a right angle triangle ABC: \\ {AC}^{2} = {BC}^{2} + {AB}^{2} \\ {=>BC}^{2} = 24^{2} + 7^{2} \\ {=>BC}^{2} = 576 + 49 \\ {=>BC}^{2} = 625 \\ =>BC = \pm 25 cm \end{array} Since the length cannot be negative, we will negelect - 25 . ∴ BC = 25 cm$

Page No 211:

Answer:

$(b) 25 m \begin{array}{l} In right triangle ABC: \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ {=15}^{2} + 20^{2} \\ {=> AC}^{2} = 625 \\ =>AC = \pm 25 \\ Since the length cannot be negative, we will negelect - 25 . \end{array} ∴ Length of the ladder = 25 m$

Page No 211:

Answer:

$(a) 13 m Suppose there are two poles AE and BD . EC = AB = 12 m (ABCE is a rectangle) AE = BC = 6 m (ABCE is a rectangle) DC = BD - AE = 11 - 6 = 5 m In the right angled triangle ECD : {ED}^{2} = {EC}^{2} + {DC}^{2} (Pythagoras theorem) {ED}^{2} = 5^{2} + 12^{2} {ED}^{2} = 25 + 144 {ED}^{2} = 169 ED = \pm 13 The length cannot be negative . ∴ ED = 13 m$

Page No 211:

Answer:

$(d) 5 \sqrt{2} cm \begin{array}{l} In right angled isoceles triangle, right angled at C, A C is equal to B C and A B is the hypotenuse. \\ \begin{array}{l} \begin{array}{l} {AB}^{2} = {AC}^{2} + {BC}^{2} \\ {=5}^{2} + 5^{2} \\ =50 \\ ∴ AB= \sqrt{2 \times 25} = 5 \sqrt{2} cm \end{array} \end{array} \end{array}$

Page No 212:

Answer:

$\begin{array}{l} ∠ ABO = 60^{^{0}} \\ ∠ {CDO =40}^{0} \\ ∠ ABO = ∠ BOC [alternate angles] \\ {=60}^{0} \\ ∠ CDO = ∠ DOC = 40^{0} [alternate angles] \\ ∠ BOD = ∠ BOC + ∠ DOC = 60^{0} + 40^{0} = 100^{0} \end{array}$

Page No 212:

Answer:

$\begin{array}{l} Here, AB II EC \\ ∴ ∠ BAC = ∠ ACE = 70^{0} (alternate angles) \\ ∠ BCA+ ∠ ACE+ ∠ ECD=18 0^{0} \\ ∠ BCA=18 0^{0} - 120^{0} \\ ∠ BCA = 60^{0} \end{array}$

Page No 212:

Answer:

$\begin{array}{l} (i) ∠ AOC = ∠ {BOD =50}^{0} [vertically opposite angles] \\ (ii) ∠ BOC = 180^{0} - 50^{0} (linear pair) \\ {=130}^{0} \end{array}$

Page No 212:

Answer:

$\begin{array}{l} Here, 3 x + 20 + 2 x - 10 = 180 \\ = > 5 x + 10 = 180 \\ =>5 x = 170 \\ = > x = 34 \\ ∠ AOC = (3 \times 34 + 20) ° \\ = (102 + 20) ° \\ =122 ° \\ ∠ BOC =(2 \times 34 - 10) ° \\ =(68 - 10) ° \\ = 58 ° \end{array}$

Page No 212:

Answer:

$\begin{array}{l} In Δ ABC, ∠ A+ ∠ B+ ∠ {C=180}^{0} \\ {=>65}^{0} + 45^{0} + ∠ C = 180^{0} \\ => ∠ {C=180}^{0} - 110^{0} = 90^{0} \end{array}$

Page No 212:

Answer:

$\begin{array}{l} Let x = 2 k and y=3 k \\ ∴ 2 k + 3 k = 120^{0} (e xterior angle property) \\ = > 5 k = 120^{0} \\ = > k = 24^{0} \\ ∴ x = 2 \times 24^{0} = 48^{0} and y=3 \times 24^{0} = 72^{0} \\ In Δ ABC: \\ ∠ A + ∠ B + ∠ C = 180^{0} \\ = > 48^{0} + 72^{0} + ∠ {C=180}^{0} \\ => ∠ C = 180^{0} - 120^{0} \\ => ∠ C = 60^{0} \\ ∴ z = 60^{0} \end{array}$

Page No 212:

Answer:

$Since it is a right triangle, by using the Pythagoras theorem: \begin{array}{l} Length of the hypotenuse = \sqrt{8^{2} + 15^{2}} \\ = \sqrt{64 + 225} \\ = \sqrt{289} \\ = \pm 17 cm \end{array} The length of the side can not be negative. ∴ Length of the hypotenuse = 17 cm$

Page No 212:

Answer:

$\begin{array}{l} Given: \\ ∠ BAD = ∠ DAC ...(i) \\ To show that △ ABC is isoceles, we should show that ∠ B = ∠ C. \\ ∴ AD ⊥ BC, ∠ ADB = ∠ ADC = 90 °^{} \end{array} ∠ A D C = ∠ A D B ∠ B A D + ∠ A B D = ∠ D A C + ∠ A C D (e x t e r i o r a n g l e p r o p e r t y) ∠ D A C + ∠ A B D = ∠ D A C + ∠ A C D (from equation (i)) ∠ A B D = ∠ A C D This is because opposite angles of a triangle ∆ ABC are equal . Hence, ABS is an isosceles triangle .$

Page No 212:

Answer:

$\begin{array}{l} Steps of construction: \\ 1 . Draw BC = 5.3 cm \\ 2 . Construct ∠ CBX = 60 °^{} \\ 3 . With B as the centre and radius 4.2 cm, cut the ray BX at point A. \\ 4 . Join A and C . \\ 5 . With A as the centre and radius more than half of AC, draw an arc on either side of AC . \end{array} 6 . With C as the centre and the same radius, draw another arc cutting the previously drawn arc at M and N . 7 . Join M and N .$

Page No 212:

Answer:

$(c) 145 ° Supplement of 35°= 180 ° - 35 ° = 145 °$

Page No 212:

Answer:

$(d) 124 \begin{array}{l} x^{0} + 56^{0} = 180^{0} (linear pair) \\ => x^{0} = 180^{0} - 56^{0} \\ {=124}^{0} \end{array} ∴ x=124$

Page No 212:

Answer:

$(c) 65 ° \begin{array}{l} ∠ ACD = 125^{0} \\ ∠ A C D = ∠ C A B + ∠ A B C (∵ the exterior angles are equal to the sum of its interior opposite angles) \\ ∴ ∠ ABC = 125^{0} - 60^{0} = 65^{0} \end{array}$

Page No 213:

Answer:

$(c) 105 ° \begin{array}{l} ∠ A+ ∠ B+ ∠ {C=180}^{0} \\ => ∠ A = 180^{0} - (40^{0} + 35^{0}) \\ => ∠ A = 105^{0} \end{array}$

Page No 213:

Answer:

$\begin{array}{l} (c) 60 ° G i v e n : 2 ∠ A = 3 ∠ B ∠ A = \frac{3}{2} ∠ B . . . (i) 3 ∠ B = 3 ∠ C ∠ C = \frac{1}{2} ∠ B . . . (ii) \end{array} ∠ A + ∠ B + ∠ C = 180 ° \begin{array}{l} => \frac{3}{2} ∠ B + ∠ B + \frac{1}{2} ∠ B = {180°}^{} \\ => (\frac{3}{2} + 1 + \frac{1}{2}) ∠ B = 180° \\ => \frac{6}{2} ∠ B = 180 ° \\ => ∠ B = \frac{180 °}{3} = 60 ° \end{array}$

Page No 213:

Answer:

$(b) 55 ° \begin{array}{l} In △ ABC: \\ A+B+C = 180 °^{} . . . (i) \\ G i v e n, A - B = 33 ° \end{array} A = 33 ° + B . . . (ii) B - C = 18 ° C = B + 18 ° . . . (iii) Putting the value s of A and B in equation (i) : \Rightarrow B + 33 ° + B + B - 18 ° = 180 ° \Rightarrow 3 B = 180 ° - 15 \Rightarrow B = \frac{165 °}{3} = 55 °$

Page No 213:

Answer:

$(b) 3 \sqrt{2} cm \begin{array}{l} Here, AB = AC \\ In right angled isoceles triangle: \\ {BC}^{2} = {AB}^{2} + {AC}^{2} \\ {=>BC}^{2} = 2 {AB}^{2} \\ => 36 = 2 {AB}^{2} \\ {=>AB}^{2} = \frac{3 6^{18}}{2} \\ =>AB= \sqrt{18} \\ =>AB=3 \sqrt{2} cm \end{array}$

Page No 213:

Answer:

(i) The sum of the angles of a triangle is 180°.
(ii) The sum of any two sides of a triangle is always greater than the third side.
(iii) In ∆ABC, if ∠A = 90°, then:
BC² = (AB²) + (BC²)

$\begin{array}{l} In △ ABC, if ∠ A = 90 °, then by phythagoras theorem : \end{array} {BC}^{2} = {AB}^{2} + {AC}^{2}$

(iv) In ∆ABC:
AB = AC
AD ⊥ BC
Then, BD = DC

$\begin{array}{l} BD = DC \end{array} This is because in an isosceles triangle, the perpendicular dropped from the vertex joining the equal sides, bisects the base.$

(v) In the given figure, side BC of ∆ABC is produced to D and CE || BA.
If ∠BAC = 50°, then ∠ACE = 50°.

$\begin{array}{l} AB II CE \\ ∴ ∠ BAC = ∠ ACE = 50 °^{} (alternate angles) \end{array}$

Page No 213:

Answer:

$\begin{array}{l} (i) True \\ (ii) True \\ (iii) {False. Each acute angle of an isoceles right triangle measures 45°.}^{} \\ (iv) True. This is because the sum of three angles of a triangle must be 180°. \end{array} So, if one of {the angles is 90°, the sum of the other two angles must also be 90°}^{} . So, none of the angles can be greater than 90°.$

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