myhelper: RS Aggarwal 2019,2020 solution class 7 chapter 17 Construction Exercise 17C

RS Aggarwal 2019,2020 solution class 7 chapter 17 Construction Exercise 17C

Exercise 17C

Page-208

Question 1:

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The supplement of 45° is

(a) 45°
(b) 75°
(c) 135°
(d) 155°

Answer 1:

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

Question 2:

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The complement of 80° is

(a) 100°
(b) 10°
(c) 20°
(d) 280°

Answer 2:

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

Question 3:

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An angle is its own complement. The measure of the angle is

(a) 30°
(b) 45°
(c) 90°
(d) 60°

Answer 3:

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

Question 4:

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An angle is one-fifth of its supplement. The measure of the angle is

(a) 30°
(b) 15°
(c) 75°
(d) 150°

Answer 4:

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

Question 5:

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An angle is 24° more than its complement. The measure of the angle is

(a) 47°
(b) 57°
(c) 53°
(d) 66°

Answer 5:

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

Question 6:

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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 6:

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

Question 7:

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Two supplementary angles are in the ratio 3 : 2. The smaller angle measures

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

Answer 7:

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

Question 8:

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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 8:

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

Question 9:

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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 9:

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

Question 10:

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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 10:

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

Page-209

Question 11:

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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 11:

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

Question 12:

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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 12:

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

Question 13:

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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 13:

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

Question 14:

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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 14:

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

Question 15:

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In ∆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 15:

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

Question 16:

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In ∆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 16:

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

Question 17:

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In the given figure, side BC of ∆ABC is produced to D such that ∠ABC = 70° and ∠ACD = 120°. Then, ∠BAC = ?

(a) 60°
(b) 50°
(c) 70°
(d) 35°

Answer 17:

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

Question 18:

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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 18:

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

Question 19:

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In the given figure, ∠A = 50°, CE || BA and ∠ECD = 60°
Then, ∠ACB = ?

(a) 50°
(b) 60°
(c) 70°
(d) 80°

Answer 19:

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

Question 20:

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In ∆ABC, if ∠A = 65° and ∠C = 85°, then ∠B = ?

(a) 25°
(b) 30°
(c) 35°
(d) 40°

Answer 20:

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

Question 21:

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The sum of all angles of a triangle is

(a) 90°
(b) 100°
(c) 150°
(d) 180°

Answer 21:

$(d) 180^{0}$

Page-210

Question 22:

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The sum of all angles of a quadrilateral is

(a) 180°
(b) 270°
(c) 360°
(d) 480°

Answer 22:

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

Question 23:

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In the given figure, AB || CD. ∠OAB = 150° and ∠OCD = 120°.
Then ∠AOC = ?

(a) 80°
(b) 90°
(c) 70°
(d) 100°

Answer 23:

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

Question 24:

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In the given figure, PQ || RS. ∠PAB = 60° and ∠ACS = 100°.
Then ∠BAC = ?

(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer 24:

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

Question 25:

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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 25:

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

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 26:

$(d) greater than the 3 rd side$

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 27:

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

Question 28:

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In ∆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 28:

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

Question 29:

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In a ∆ABC, it is given that ∠B = 37°, and ∠C = 29°. Then, ∠A = ?

(a) 86°
(b) 66°
(c) 114°
(d) 57°

Answer 29:

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

Question 30:

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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 30:

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

Question 31:

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In a ∆ABC, if 2∠A = 3∠B = 6∠C, then ∠B = ?

(a) 30°
(b) 90°
(c) 60°
(d) 45°

Answer 31:

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

Question 32:

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In a ∆ABC, if ∠A + ∠B = 65° and ∠B +∠C = 140°. Then, = ∠B?

(a) 25°
(b) 35°
(c) 40°
(d) 45°

Answer 32:

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

Question 33:

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In a ∆ABC, ∠A − ∠B = 33° and ∠B −∠C = 18°. Then, = ∠B?

(a) 35°
(b) 55°
(c) 45°
(d) 57°

Answer 33:

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

Page-211

Question 34:

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The angles of a triangle are (3x)° ,(2x − 7)° and (4x − 11)°. Then, x = ?

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

Answer 34:

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

Question 35:

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∆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 35:

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

Question 36:

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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 36:

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

Question 37:

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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 37:

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

Question 38:

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∆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 38:

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