Rs Aggarwal 2018 for Class 7 Math Chapter 17 - Constructions

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Rs Aggarwal 2018 Solutions for Class 7 Math Chapter 17 Constructions are provided here with simple step-by-step explanations. These solutions for Constructions are extremely popular among Class 7 students for Math Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 7 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 204:

Question 1:

Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.

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:

Question 2:

Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.

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:

Question 3:

Draw a line l and draw another line m parllel to l at a distance of 4.3 cm from it.

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:

Question 1:

Construct a âˆ†ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.

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:

Question 2:

Construct a âˆ†PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of ∠P.

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:

Question 3:

Construct an equilateral triangle each of whose sides measures 6.2 cm. Measure each of its angles.

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:

Question 4:

Construct a âˆ†ABC in which AB = AC = 4.8 cm and BC = 5.3 cm. Measure ∠B and ∠C. Draw AD ⊥ BC.

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:

Question 5:

Construct a âˆ†ABC in which AB = 3.8 cm, ∠A = 60° and AC = 5 cm.

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:

Question 6:

Construct a âˆ†ABC in which BC = 4.3 cm, ∠C = 45° and AC = 6 cm.

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:

Question 7:

Construct a âˆ†ABC in which AB = AC = 5.2 cm and ∠A = 120°. Draw AD ⊥ BC.

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:

Question 8:

Construct a âˆ†ABC in which BC = 6.2 cm, ∠B = 60° and ∠C = 45°.

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:

Question 9:

Construct a âˆ†ABC in which BC = 5.8 cm, ∠B = ∠C = 30°. Measure AB and AC. What do you observe?

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:

Question 10:

Construct a âˆ†ABC in which AB = 7 cm, ∠A = 45° and ∠C = 75°.

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:

Question 11:

Construct a âˆ†ABC in which BC = 4.8 cm, ∠C = 90° and AB = 6.3 cm.

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:

Question 12:

Construct a right-angled triangle one side of which measures 3.5 cm and the length of whose hypotenuse is 6 cm.

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:

Question 13:

Construct a right triangle having hypotenuse of length 5.6 cm and one of whose acute angles measures 30°.

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:

Question 1:

Mark (âœ“) against the correct answer
The supplement of 45° is

(a) 45°
(b) 75°
(c) 135°
(d) 155°

Answer:

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

Page No 208:

Question 2:

Mark (âœ“) against the correct answer
The complement of 80° is

(a) 100°
(b) 10°
(c) 20°
(d) 280°

Answer:

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

Page No 208:

Question 3:

Mark (âœ“) against the correct answer
An angle is its own complement. The measure of the angle is

(a) 30°
(b) 45°
(c) 90°
(d) 60°

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:

Question 4:

Mark (âœ“) against the correct answer
An angle is one-fifth of its supplement. The measure of the angle is

(a) 30°
(b) 15°
(c) 75°
(d) 150°

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:

Question 5:

Mark (âœ“) against the correct answer
An angle is 24° more than its complement. The measure of the angle is

(a) 47°
(b) 57°
(c) 53°
(d) 66°

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:

Question 6:

Mark (âœ“) against the correct answer
An angle is 32° less than its supplement. The measure of the angle is

(a) 37°
(b) 74°
(c) 148°
(d) none of these

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:

Question 7:

Mark (âœ“) against the correct answer
Two supplementary angles are in the ratio 3 : 2. The smaller angle measures

(a) 108°
(b) 81°
(c) 72°
(d) none of these

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:

Question 8:

Mark (âœ“) against the correct answer
In the given figure, AOB is a straight line and the ray OC stands on it.
If ∠BOC = 132°, then ∠AOC = ?

(a) 68°
(b) 48°
(c) 42°
(d) none of these

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:

Question 9:

Mark (âœ“) against the correct answer
In the given figure, AOB is a straight line, ∠AOC = 68° and ∠BOC = x°.
The value of x is

(a) 32
(b) 22
(c) 112
(d) 132

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:

Question 10:

Mark (âœ“) against the correct answer
In the adjoining figure, what value of x will make AOB a straight line?

(a) x = 30
(b) x = 35
(c) x = 25
(d) x = 40

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:

Question 11:

Mark (âœ“) against the correct answer
In the given figure, what value of x will make AOB a straight line?

(a) x = 50
(b) x = 100
(c) x = 60
(d) x = 80

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:

Question 12:

Mark (âœ“) against the correct answer
In the given figure, it is given that AOB is a straight line and 4x = 5y.
What is the value of x?

(a) 100
(b) 105
(c) 110
(d) 115

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:

Question 13:

Mark (âœ“) against the correct answer
In the given figure, two straight lines AB and CD intersect at a point O and ∠AOC = 50°. Then, ∠BOD = ?

(a) 40°
(b) 50°
(c) 130°
(d) 60°

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:

Question 14:

Mark (âœ“) against the correct answer
In the given figure, AOB is a straignt line, ∠AOC = (13x − 8)°, ∠COD = 50° and ∠BOD = (x + 10)°. The value of x is

(a) 32
(b) 42
(c) 36
(d) 52

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:

Question 15:

Mark (âœ“) against the correct answer
In âˆ†ABC, side BC has been produced to D. If ∠ACD = 132° and ∠A = 54°, then ∠B = ?

(a) 48°
(b) 78°
(c) 68°
(d) 58°

Answer:

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

Page No 209:

Question 16:

Mark (âœ“) against the correct answer
In âˆ†ABC, side BC has been produced to D. If ∠BAC = 45° and ∠ABC = 55°, then ∠ACD = ?

(a) 80°
(b) 90°
(c) 100°
(d) 110°

Answer:

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

Page No 209:

Question 17:

Mark (âœ“) against the correct answer
In the given figure, side BC of âˆ†ABC is produced to D such that ∠ABC = 70° and ∠ACD = 120°. Then, ∠BAC = ?

(a) 60°
(b) 50°
(c) 70°
(d) 35°

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:

Question 18:

Mark (âœ“) against the correct answer
In the given figure, rays OA, OB, OC and OD are such that ∠AOB = 50°, ∠BOC = 90°, ∠COD = 70° and ∠AOD = x°.
Then, the value of x is

(a) 50°
(b) 70°
(c) 150°
(d) 90°

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:

Question 19:

Mark (âœ“) against the correct answer
In the given figure, ∠A = 50°, CE || BA and ∠ECD = 60°
Then, ∠ACB = ?

(a) 50°
(b) 60°
(c) 70°
(d) 80°

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:

Question 20:

Mark (âœ“) against the correct answer
In âˆ†ABC, if ∠A = 65° and ∠C = 85°, then ∠B = ?

(a) 25°
(b) 30°
(c) 35°
(d) 40°

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:

Question 21:

Mark (âœ“) against the correct answer
The sum of all angles of a triangle is

(a) 90°
(b) 100°
(c) 150°
(d) 180°

Answer:

$(d) 180^{0}$

Page No 210:

Question 22:

Mark (âœ“) against the correct answer
The sum of all angles of a quadrilateral is

(a) 180°
(b) 270°
(c) 360°
(d) 480°

Answer:

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

Page No 210:

Question 23:

Mark (âœ“) against the correct answer
In the given figure, AB || CD. ∠OAB = 150° and ∠OCD = 120°.
Then ∠AOC = ?

(a) 80°
(b) 90°
(c) 70°
(d) 100°

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:

Question 24:

Mark (âœ“) against the correct answer
In the given figure, PQ || RS. ∠PAB = 60° and ∠ACS = 100°.
Then ∠BAC = ?

(a) 40°
(b) 60°
(c) 80°
(d) 50°

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:

Question 25:

Mark (âœ“) against the correct answer
In the given figure, AB || CD || EF, ∠ABG = 110°, ∠GCD = 100° and ∠BGC = x°.
Then x = ?

(a) 35
(b) 50
(c) 30
(d) 40

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:

Question 26:

The sum of any two sides of a triangle is always

(a) equal to the third side
(b) less than the third side
(c) greater than or equal to the 3rd side
(d) greater than the 3rd side

Answer:

$(d) greater than the 3 rd side$

Page No 210:

Question 27:

The diagonals of a rhombus

(a) are always equal
(b) never bisect each other
(c) always bisect each other at an acute angle
(d) always bisect each other at right angles

Answer:

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

Page No 210:

Question 28:

Mark (âœ“) against the correct answer
In âˆ†ABC, ∠B = 90°, AB = 5 cm and AC = 13 cm. Then, BC = ?

(a) 8 cm
(b) 18 cm
(c) 12 cm
(d) none of these

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:

Question 29:

Mark (âœ“) against the correct answer
In a âˆ†ABC, it is given that ∠B = 37°, and ∠C = 29°. Then, ∠A = ?

(a) 86°
(b) 66°
(c) 114°
(d) 57°

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:

Question 30:

Mark (âœ“) against the correct answer
The angles of a triangle are in the ratio 2 : 3 : 7. The measure of the largest angle is

(a) 84°
(b) 98°
(c) 105°
(d) 91°

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:

Question 31:

Mark (âœ“) against the correct answer
In a âˆ†ABC, if 2∠A = 3∠B = 6∠C, then ∠B = ?

(a) 30°
(b) 90°
(c) 60°
(d) 45°

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:

Question 32:

Mark (âœ“) against the correct answer
In a âˆ†ABC, if ∠A + ∠B = 65° and ∠B +∠C = 140°. Then, = ∠B?

(a) 25°
(b) 35°
(c) 40°
(d) 45°

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:

Question 33:

Mark (âœ“) against the correct answer
In a âˆ†ABC, ∠A − ∠B = 33° and ∠B −∠C = 18°. Then, = ∠B?

(a) 35°
(b) 55°
(c) 45°
(d) 57°

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:

Question 34:

Mark (âœ“) against the correct answer
The angles of a triangle are (3x)° ,(2x − 7)° and (4x − 11)°. Then, x = ?

(a) 18
(b) 20
(c) 22
(d) 30

Answer:

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

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Question 35:

Mark (âœ“) against the correct answer
âˆ†ABC is right-angled at A. If AB = 24 cm and AC = 7 cm then BC = ?

(a) 31 cm
(b) 17 cm
(c) 25 cm
(d) 28 cm

Answer:

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

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Question 36:

Mark (âœ“) against the correct answer
A ladder is placed in such a way that its foot is 15 m away from the wall and its top reaches a window 20 m above the ground. The length of the ladder is

(a) 35 m
(b) 25 m
(c) 18 m
(d) 17.5 m

Answer:

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

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Question 37:

Mark (âœ“) against the correct answer
Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?

(a) 13 m
(b) 14 m
(c) 15 m
(d) 12.8 m

Answer:

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

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Question 38:

Mark (âœ“) against the correct answer
âˆ†ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm. Then, AB = ?

(a) 2.5 cm
(b) 5 cm
(c) 10 cm
(d) $5 \sqrt{2} cm$

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:

Question 1:

In the given figure, AB || CD, ∠ABO = 60° and ∠CDO = 40°. Then, find ∠BOD.

Answer:

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

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Question 2:

In the given figure, CE || BA. If ∠BAC = 70° and ∠ECD = 50°, find ∠ACB.

Answer:

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

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Question 3:

In the given figure, two straight lines AB and CD intersect at a point O such that ∠AOC = 50°.
Find: (i) ∠BOD (ii) ∠BOC.

Answer:

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

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Question 4:

In the given figure, AOB is a straight line and OC is ray such that∠AOC = (3x + 20)° and ∠BOC = (2x − 10)°. Find the value of x and hence find (i) ∠AOC and ∠BOC.

Answer:

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

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Question 5:

In a âˆ†ABC, If ∠A = 65°, ∠B = 45°, find ∠C.

Figure

Answer:

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

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Question 6:

In the given figure, x : y = 2 : 3 and ∠ACD = 120°. Find the values of x,y and z.

Answer:

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

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Question 7:

Two legs of a right triangle are 8 cm and 15 cm long. Find the length of the hypotenuse of the triangle.

Answer:

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

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Question 8:

In the adjoining figure, ABC is a triangle in which AD is the bisector of ∠A. If AD ⊥ BC, show that âˆ†ABC is isosceles.

Answer:

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

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Question 9:

Construct a âˆ†ABC in which BC = 5.3 cm, ∠B = 60° and AB = 4.2 cm. Also, draw the perpendicular bisector of AC.

Answer:

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

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Question 10:

Mark (âœ“) against the correct answer
The supplement of 35° is

(a) 55°
(b) 65°
(c) 145°
(d) 165°

Answer:

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

Page No 212:

Question 11:

Mark (âœ“) against the correct answer
In the given figure, AOB is a straignt line, ∠AOC = 56° and ∠BOC = x°. The value of x is

(a) 34
(b) 44
(c) 144
(d) 124

Answer:

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

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Question 12:

Mark (âœ“) against the correct answer
In âˆ†ABC, side BC has been produced to D such that ∠ACD = 125° and ∠BAC = 60°. Then ∠ABC = ?

(a) 55°
(b) 60°
(c) 65°
(d) 70°

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:

Question 13:

Mark (âœ“) against the correct answer
In a âˆ†ABC, If ∠B = 40° and ∠C = 35°, then ∠A = ?

(a) 50°
(b) 55°
(c) 105°
(d) 150°

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:

Question 14:

Mark (âœ“) against the correct answer
In a âˆ†ABC, If 2∠A = 3∠B = 6∠C, then ∠B = ?

(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

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

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Question 15:

Mark (âœ“) against the correct answer
In a âˆ†ABC, If A − B = 33° and B − C = 18°, then ∠B = ?

(a) 35°
(b) 55°
(c) 45°
(d) 57°

Answer:

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

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Question 16:

Mark (âœ“) against the correct answer
âˆ†ABC is an isosceles right triangle in which ∠A = 90° and BC = 6 cm. Then AB = ?

(a) $2 \sqrt{2} cm$
(b) $3 \sqrt{2} cm$
(c) $4 \sqrt{2} cm$
(d) $2 \sqrt{3} cm$

Answer:

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

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Question 17:

Fill in the blanks.

(i) The sum of the angles of a triangle is ...... .
(ii) The sum of any two sides of a triangle is always ...... than the third side.
(iii) In âˆ†ABC, if ∠A = 90°, then BC² = (......) + (......).
(iv) In âˆ†ABC, AB = AC and AD ⊥ BC, then BD = ...... .
(v) In the given figure, side BC of âˆ†ABC is produced to D and CE || BA. If ∠BAC = 50°
then ∠ACE = ...... .

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 âˆ†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 âˆ†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 âˆ†ABC is produced to D and CE || BA.
If ∠BAC = 50°, then ∠ACE = 50°.

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

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Question 18:

Write 'T' for true and 'F' for false

(i) If two parallel lines are cut by a transversal, then the alternate interior angles are equal.
(ii) If two lines intersect each other, then the vertically opposite angles are equal.
(iii) Each acute angle of an isosceles right triangle measures 60°.
(iv) A right triangle cannot have an obtuse angle.

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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