myhelper: RD Sharma solution class 7 chapter 15 Properties of triangles Objective Type Questions

RD Sharma solution class 7 chapter 15 Properties of triangles Objective Type Questions

Objective Type Questions

Page-15.31

Question 1:

If the measures of the angles of a triangle are (2x)° , (3x − 5)° and (4x − 13)°. Then the value of x is
(a) 22
(b) 18
(c) 20
(d) 30

Answer 1:

(2x)° + (3x − 5)^∘ + (4x − 13)^∘ = 180° [Angle sum property of triangle]
⇒ 2x + 3x − 5 + 4x − 13 = 180°
⇒ 9x − 18 = 180
⇒ 9x = 198
⇒ x = 22
Hence, the correct answer is option (a).

Question 2:

The angles of a triangle are in the ratio 2 : 3 : 7. The measure of the largest angle is
(a) 84°
(b) 91°
(c) 105°
(d) 98°

Answer 2:

Let the angles of the triangle be 2x, 3x and 7x.
Now, 2x + 3x + 7x = 180° [Angle sum property of triangle]
⇒ 12x = 180°
⇒ x = 15°
∴ Largest angle = 7x = 7 × 15° = 105°
Hence, the correct answer is option (c).

Question 3:

In a △ABC, if 2∠A = 3∠B = 6∠C, then th s measure of the smallest angle is
(a) 90°
(b) 60°
(c) 40°
(d) 30°

Answer 3:

We have
2∠A = 3∠B = 6∠C

∴ 3 ∠ B = 2 ∠ A and 6 ∠ C = 2 ∠ A \Rightarrow ∠ B = \frac{2}{3} ∠ A and ∠ C = \frac{2}{6} ∠ A = \frac{1}{3} ∠ A Now, ∠ A + ∠ B + ∠ C = 180 ° [Angle sum property of triangle] \Rightarrow ∠ A + \frac{2}{3} ∠ A + \frac{1}{3} ∠ A = 180 ° \Rightarrow 3 ∠ A + 2 ∠ A + ∠ A = 180 ° \times 3 \Rightarrow 6 ∠ A = 540 ° \Rightarrow ∠ A = 90 °

∴ Smallest angle =

∠ C = \frac{1}{3} ∠ A = \frac{1}{3} \times 90 ° = 30 °

Hence, the correct answer is option (d).

Question 4:

In a △ABC,if ∠A + ∠B = 150° and ∠B + ∠C = 75°, then ∠B =
(a) 35°
(b) 45°
(c) 55°
(d) 25°

Answer 4:

∠A + ∠B + ∠C = 180° [Angle sum property of triangle]
⇒ 150° + ∠C = 180°
⇒ ∠C = 30°
Now, ∠B + ∠C = 75°
⇒ ∠B + 30° = 75°
⇒ ∠B = 45°
Hence, the correct answer is option (b).

Question 5:

In a △ABC, if ∠A − ∠B = 33° and ∠B − ∠C = 18°, then ∠B =
(a) 35°
(b) 45°
(c) 55°
(d) 25°

Answer 5:

∠A − ∠B = 33° and ∠B − ∠C = 18°
⇒ ∠A = ∠B + 33° and ∠C = ∠B − 18°
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of triangle]
⇒ ∠B + 33° + ∠B + ∠B − 18° = 180°
⇒ 3∠B + 15° = 180°
⇒ 3∠B = 165°
⇒ ∠B = 55°
Hence, the correct answer is option (d).

Question 6:

If the measures of the angles of a triangle are

(2 x - 5) °, {(3 x - \frac{1}{2})}^{°} and {(30 - \frac{x}{2})}^{°}

, then x =
(a)

\frac{311}{9}

(b)

\frac{309}{9}

(c)

\frac{310}{9}

(d)

\frac{301}{9}

Answer 6:

(2 x - 5) ° + {(3 x - \frac{1}{2})}^{°} + {(30 - \frac{x}{2})}^{°} = 180 ° [Angle sum property of triangle] \Rightarrow 2 x - 5 + 3 x - \frac{1}{2} + 30 - \frac{x}{2} = 180 \Rightarrow 2 x + 3 x - \frac{x}{2} - 5 - \frac{1}{2} + 30 = 180 \Rightarrow 4 x + 6 x - x - 10 - 1 + 60 = 180 \times 2 \Rightarrow 9 x + 49 = 360 \Rightarrow 9 x = 311 \Rightarrow x = \frac{311}{9}

Hence, the correct answer is option (a).

Page-15.32

Question 7:

In Fig. 59, the value of x is
(a) 84
(b) 74
(c) 94
(d) 57

Answer 7:

∠CAD = ∠ABC + ∠ACB [Exterior angle property]
⇒ 123° = 39° + x°
⇒ 84° = x°
⇒ x = 84
Hence, the correct answer is option (a).

Question 8:

In Fig. 60, the values of x and y are
(a) x = 20, y = 130
(c) x = 20, y = 140
(b) x = 40, y = 140
(d) x = 15, y = 140

Answer 8:

∠ACB + ∠ACD = 180°
⇒ 40° + y° = 180°
⇒ y° = 140°
⇒ y = 140
Now, ∠ACD = ∠ABC + ∠BAC [Exterior angle property]
⇒ 3x° + 4x° = y°
⇒ 7x° = 140°
⇒ x = 20
Hence, the correct answer is option (c).

Question 9:

In Fig. 61, the value of xis
(a) 72
(b) 50
(c) 58
(d) 48

Answer 9:

∠A + ∠B + ∠C = 180° [Angle sum property of triangle]
⇒ 50° + 72° + ∠C = 180°
⇒ ∠C + 122° = 180°
⇒ ∠C = 58°
Now, x° = ∠C [Vertically opposite angles]
⇒ x° = 58°
⇒ x = 58
Hence, the correct answer is option (c).

Question 10:

In Fig. 62, if AB || DE, then the value of x is

(a) 25
(b) 35
(c) 40
(d) 45

Answer 10:

∠ACD + ∠ACB = 180°         [Linear angles]
⇒ 91° + ∠ACB = 180°
⇒ ∠ACB = 89°
Since, AB || DE
∠DEC = ∠CAB = 46°           [Alternate angles]
Now,
∠ACB + ∠CAB + ∠ABC = 180°         [Angle sum property of triangle]
⇒ 89° + 46° + x° = 180°
⇒ x° = 45°
⇒ x = 45
Hence, the correct answer is option (d).

Page-15.33

Question 11:

In Fig. 63, if AB || CD, the value of x is

(a) 25
(b) 35
(c) 15
(d) 20

Answer 11:

Since, AB || DE
∠DCB = ∠CBA = 3x° [Alternate angles]
Now,
∠ACB + ∠CAB + ∠CBA = 180° [Angle sum property of triangle]
⇒ 55° + 2x° + 3x° = 180°
⇒ 5x° = 125°
⇒ x = 25
Hence, the correct answer is option (a).

Question 12:

In Fig. 64, if AB || CD, the values of x and y are
(a) x = 21, y = 28
(b) x = 21, y = 38
(c) x = 38, y = 21
(d) x = 22, y = 38

Answer 12:

∠AEC + ∠AEB = 180°         [Linear angles]
⇒ 79° + ∠AEB = 180°
⇒ ∠AEB = 101°
Since, AB || CD
∠ABE = ∠ECD = 58°           [Alternate angles]
Now, In △AEB
∠AEB + ∠EAB + ∠ABE = 180°         [Angle sum property of triangle]
⇒ 101° + 58° + x° = 180°
⇒ x° = 21°
⇒ x = 21
Now, In △AEB
∠AEC + ∠CAE + ∠CEA = 180°         [Angle sum property of triangle]
⇒ 79° + y° + 3x° = 180°
⇒ 79° + y° + 3(21)° = 180°
⇒ 79° + y° + 63° = 180°
⇒ y° = 38°
⇒ y = 38
Hence, the correct answer is option (b).

Question 13:

In Fig. 65, if AB || CE, then the values of x and y are
(a) x = 26, y = 144
(b) x = 36, y = 154
(c) x = 154, y = 36
(d) x = 144, y = 26

Answer 13:

∠DCA + ∠ACB = 180°         [Linear angles]
⇒ 134° + ∠ACB = 180°
⇒ ∠ACB = 46°
Now, In △ABC
∠BAC + ∠ACB+ ∠ABC = 180°         [Angle sum property of triangle]
⇒ 62° + 46° + 2x° = 180°
⇒ 2x° = 72°
⇒ x = 36
Since, AB || CE
∴ ∠ECB = ∠CBA= (2x)° = (2 × 36)° = 72° [Alternate angles]
Now, ∠DCE + ∠ECB = 180°         [Linear angles]
⇒ y° + 72° = 180°
⇒ y = 108
Disclaimer: No Option is correct

Page-15.34

Question 14:

In Fig. 66, if AF || DE, then x =
(a) 37
(b) 57
(c) 47
(d) 67

Answer 14:

Since, AF || DE
∠EDC = ∠ACB = 109° [Corresponding angles]
Now, In △ABC
∠ACB + ∠CAB + ∠CBA = 180° [Angle sum property of triangle]
⇒ 109° + 24° + x° = 180°
⇒ x° = 47°
⇒ x = 47
Hence, the correct answer is option (c).

Question 15:

In Fig. 67, the values of x and y are
(a) x = 130, y = 120
(b) x = 120, y = 130
(c) x = 120, y = 120
(d) x = 130, y = 130

Answer 15:

In △ABD
∠ADB + ∠BAD + ∠ABD = 180°         [Angle sum property of triangle]
⇒ 61° + 59° + ∠ABD = 180°
⇒ ∠ABD = 60°
∠ABD + ∠DBC = 180°         [Linear pair angles]
⇒ 60° + y° = 180°
⇒ y = 120
Now, ∠ADB = ∠GDE = 61°           [Vertically opposite angles]
Now, In △GDE
∠GDE + ∠DGE + ∠GED = 180°         [Angle sum property of triangle]
⇒ 61° + 69° + ∠GED = 180°
⇒ ∠GED = 50°
Now, ∠GED + ∠GEF = 180°         [Linear pair angles]
⇒ 50° + x° = 180°
⇒ x = 130
Hence, the correct answer is option (a).

Question 16:

In Fig. 68, the values of x and y are
(a) x = 120, y = 150
(b) x = 110, y = 160
(c) x = 150, y = 120
(d) x = 110, y = 160

Answer 16:

In △DEF
∠DEF + ∠DFE + ∠EDF = 180°         [Angle sum property of triangle]
⇒ 110° + 40° + ∠EDF = 180°
⇒ ∠EDF = 30°
Now, ∠EDF + ∠FDA = 180°         [Linear pair angles]
⇒ 30° + x° = 180°
⇒ x = 150
Now, ∠EDF = ∠ADB = 30°           [Vertically opposite angles]
Now, In △ABD
∠ADB + ∠DAB + ∠ABD = 180°         [Angle sum property of triangle]
⇒ 30° + 90° + ∠ABD = 180°
⇒ ∠ABD = 60°
Now, ∠ABD + ∠DBC = 180°         [Linear pair angles]
⇒ 60° + y° = 180°
⇒ y = 120
Hence, the correct answer is option (c).

Question 17:

In Fig. 69, if AB || CD, then the values of x and y are
(a) x = 106, y = 307
(b) x = 307, y = 106
(C) x =107, y = 306
(d) x = 105, y = 308

Answer 17:

In △CDE
∠CDE + ∠CED + ∠ECD = 180°         [Angle sum property of triangle]
⇒ 53° + 53° + ∠ECD = 180°
⇒ ∠ECD = 74°
Since, AB || CD
∴ ∠ECD = ∠CGB = 74°           [Corresponding angles]
Now, ∠CGB + ∠BGF = 180°         [Linear pair angles]
⇒ 74° + x° = 180°
⇒ x = 106
Now, In △EGB
∠EGB + ∠BEG + ∠EBG = 180°         [Angle sum property of triangle]
⇒ 74° + 53° + ∠EBG = 180°
⇒ ∠EBG = 53°
Now, ∠EBG + Reflex∠EBG = 360°         [Complete angle]
⇒ 53° + y° = 360°
⇒ y = 307
Hence, the correct answer is option (a).

Page-15.35

Question 18:

In Fig. 70, if AB || CD, then the values of x and y are
(a) x = 24, y = 48
(b) x = 34, y = 68
(c) x = 24, y = 68
(d) x = 34, y = 48

Answer 18:

∠AGE + ∠BGE = 180°         [Linear pair angles]
⇒ 121° + ∠BGE = 180°
⇒ ∠BGE = 59°
Since, AB || CD
∴ ∠BGE = ∠GHD = 59°           [Corresponding angles]
⇒ x° + 25° = 59°
⇒ x = 34
In △GHI
∠GHI + ∠GIH + ∠HGI = 180°         [Angle sum property of triangle]
⇒ 34° + 78° + y° = 180°
⇒ y = 68
Hence, the correct answer is option (b).

Question 19:

In Fig. 71, if AB || CD, then the values of x, y and z are
(a) x = 56, y = 47, z = 77
(b) x = 47, y = 56, z = 77
(c) x = 77, y = 56, z = 47
(d) x = 56, y = 77, z = 47

Answer 19:

∠AFE + ∠EFG = 180°         [Linear pair angles]
⇒ 124° + ∠EFG = 180°
⇒ ∠EFG = 56°
Since, AB || CD
∴ ∠EFG = ∠FHK = 56°           [Corresponding angles]
⇒ x = 56
Now, ∠QKH + ∠GKH = 180°         [Linear pair angles]
⇒ 103° + ∠GKH = 180°
⇒ ∠GKH = 77°
Since, AB || CD
∴ ∠EGF = ∠GKH = 77°           [Corresponding angles]
⇒ y = 77
In △EHK
∠EHK + ∠EKH + ∠HEK = 180°         [Angle sum property of triangle]
⇒ 56° + 77° + z° = 180°
⇒ z = 47
Hence, the correct answer is option (d).

Question 20:

In Fig. 72, if AB || CD and AE || BD, then the value of x is

(a) 38
(b) 48
(c) 58
(d) 68

Answer 20:

In the given figure,
$AE ∥ BD (given) \Rightarrow ∠ A + ∠ B = 180^{\circ} (co - interior angles)$
Sum of all the angles of a polygon is given by $(n - 2) \times 180^{°}$ , where n is the number of sides of the polygon.
The given polygon has number of sides(n) = 5
So, sum of all the angles = $(n - 2) \times 180^{°} = (5 - 2) \times 180^{°} = 3 \times 180^{°} = 540^{°}$
$\Rightarrow ∠ A + ∠ B + ∠ C + ∠ D + ∠ E = 540^{°} \Rightarrow 180 ° + 132 ° + 84 ° + 3 x = 540 ° (since, ∠ A + ∠ B = 180 °) \Rightarrow 3 x + 396 ° = 540 ° \Rightarrow 3 x = 540 ° - 396 ° = 144 °$
Divide both sides by 3, we get
$x = 48 °$

Page-15.36

Question 21:

If the exterior angles of a triangle are (2x + 10)°, (3x − 5)° and (2x + 40)°, then x =
(a) 25
(b) 35
(c) 45
(d) 55

Answer 21:

Sum of the exterior angles of a triangle is 360°
∴(2x + 10)°+ (3x − 5)° + (2x + 40)° = 360°
⇒ 2x + 10 + 3x − 5 + 2x + 40 = 360
⇒ 7x + 45 = 360
⇒ 7x = 315
⇒ x = 45
Hence, the correct answer is option (c)

Question 22:

In Fig. 73, the value of x is
(a) 20
(b) 30
(c) 40
(d) 25

Answer 22:

∠TRS + ∠TRQ = 180° [Linear angles]
⇒ 5x° + ∠TRQ = 180°
⇒ ∠TRQ = 180° − 5x°
Now, ∠QTR + ∠TRQ = ∠PQT [Exterior angle property of triangle]
⇒ 3x° + 180° − 5x° = 120°
⇒ 2x° = 60°
⇒ x = 30
Hence, the correct answer is option (b).

Question 23:

In Fig. 74, if AB || CO, ∠CAB = 49°, ∠CBD = 27° and∠BDC = 112°, then the values of x and y are
(a) x = 41, y = 90
(b) x = 41,y = 63
(c) x = 63, y = 41
(d) x = 90, y = 41

Answer 23:

Since, AB || CD
∠ABD + ∠CDB = 180° [Angles on the same side of a transversal line are supplementary]
⇒ x° + 27° + 112° = 180°
⇒ x° = 41°
⇒ x = 41
Now, In △ABC
∠A + ∠B + ∠C = 180° [Angle sum property of triangle]
⇒ 49° + 41°+ y° = 180°
⇒ y° = 90°
⇒ y = 90
Hence, the correct answer is option (a).

Question 24:

Which of the following is the set of measures of the sides of a triangle?
(a) 8 cm, 4 cm, 20 cm
(b) 9 cm, 17 cm, 25 cm
(c) 11 cm, 16 cm, 28 cm
(d) None of these

Answer 24:

We knwno that Triangle Inequality Theorem states that the sum of two side lengths of a triangle is always greater than the third side.
Using this in (a), we get
8 + 4 ≯ 20
⇒ 12 ≯ 20
So, triangle is not possible
Using this in (b), we get
9 + 17 > 25
⇒ 26 > 25
and
9 + 25 > 7
⇒ 34 > 7
and
17 + 25 > 9
⇒ 42 > 9
So, triangle is possible.
Using this in (c), we get
11 + 16 ≯ 28
⇒ 27 ≯ 28
So, triangle is not possible.
Hence, the correct answer is option (b).

Question 25:

In which of the following cases, a right triangle cannot be constructed?
(a) 12 cm, 5 cm, 13 cm
(b) 8 cm, 6 cm, 10 cm
(c) 5 cm, 9 cm, 11 cm
(d) None of these

Answer 25:

In (a)
12² + 5² = 13²
⇒ 144 + 25 = 169
⇒ 169 = 169
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, a right triangle can be constructed.

In (b)
8² + 6² = 10²
⇒ 44 + 36 = 100
⇒ 100 = 100
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, a right triangle can be constructed.

In (c)
5² + 9² ≠ 11²
⇒ 25 + 81 ≠ 121
⇒ 106 ≠ 121
Since, the sum of the square of two smallest side is not equal to the square of largest side.
Hence, a right triangle can not be constructed.
Hence, the correct answer is option (c).

Question 26:

Which of the following is/are not Pythagorean triplet (s)?
(a) 3,4,5
(b) 8,15,17
(c) 7,24,25
(d) 13,26,29

Answer 26:

In (a)
3² + 4² = 5²
⇒ 9 + 16 = 25
⇒ 25 = 25
Since, the sum of the square of two smallest number is equal to the square of largest number.
Hence, it is a Pythagorean triplet.
In (b)
8² + 15² = 17²
⇒ 64 + 225 = 289
⇒ 289 = 289
Since, the sum of the square of two smallest number is equal to the square of largest number.
Hence, it is a Pythagorean triplet.
In (c)
7² + 24² = 25²
⇒ 49 + 576 = 625
⇒ 625 = 625
Since, the sum of the square of two smallest number is equal to the square of largest number.
Hence, it is a Pythagorean triplet.
In (d)
13² + 26² ≠ 29²
⇒ 169 + 676≠ 841
⇒ 845 ≠ 841
Since, the sum of the square of two smallest number is not equal to the square of largest number.
Hence, it is not a Pythagorean triplet.
Hence, the correct answer is option (d).

Question 27:

In a right triangle, one of the acute angles is four times the other. Its measure is
(a) 68°
(b) 84°
(c) 80°
(d) 72°

Answer 27:

Let the smallest angle be x, then the other angle be 4x.
Now,
x + 4x + 90° = 180°
⇒ 5x = 90°
⇒ x = 18°
Thus, the measure of the angles are 18°, and 4(18)° = 72°
Hence, the correct answer is option (d).

Page-15.37

Question 28:

In which of the following cases can a right triangle ABC be constructed?
(a) AB = 5 cm, BC = 7 cm, AC = 10 cm
(b) AB = 7 cm, BC = 8 cm, AC = 12 cm
(c) AB = 8 cm, BC = 17 cm, AC = 15 cm
(d) None of these

Answer 28:

In (c)
BC² = AC² + AB²
⇒ (17)² = (15)² + (8)²
⇒ 289 =225 + 64
⇒ 289 = 289
Since, the sum of the square of two smallest side is equal to the square of largest side.
Hence, ABC is a right angle triangle at A.
Hence, the correct answer is option (c).

Question 29:

△ ABC is a right triangle 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 29:

In right traingle ABC,
BC² = AC² + AB²
⇒ BC² = (7)² + (24)²
⇒ BC² =49 + 576
⇒ BC² = 625
⇒ BC² = (25)² 2√)2
⇒ BC = 25 cm
Hence, the correct answer is option (c).

Question 30:

If △ABC is an isosceles right-triangle right angled at C such that AC = 5 cm. Then, AB =
(a) 2.5cm
(b) $5 \sqrt{2}$ cm
(c) 10 cm
(d) 5 cm

Answer 30:

Suppose BC is the ladder which is placed againts the wall OA. The foot of the ladder C is 15 m away from the foot O of the wall and its top reaches the window which is 20 m above the ground.

In right traingle ABC,
AB² = BC² + AC²
⇒ AB² = (5)² + (5)²
⇒ AB² =25 + 25
⇒ AB² = 50
⇒ AB² = ${(5 \sqrt{2})}^{2}$
⇒ AB = $5 \sqrt{2}$ cm
Hence, the correct answer is option (b).

Question 31:

Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, the distance between their tops is
(a) 13 m
(b) 14 m
(c) 15 m
(d) 12.8 m

Answer 31:

Suppose AB and CD are two poles.The is distance between AB and CD is 12 m.

In right traingle BDE,
BD² = DE² + BE²
⇒ BD² = (5)² + (12)²
⇒ BD² =25 + 144
⇒ BD² = 169
⇒ BD² = (13)²(52√)2
⇒ BD = 13 m
Hence, the correct answer is option (a).

Question 32:

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 32:

Suppose BC is the ladder which is placed againts the wall OA. The foot of the ladder C is 15 m away from the foot O of the wall and its top reaches the window which is 20 m above the ground.

In right traingle BOC,
BC² = OC² + OB²
⇒ BC² = (15)² + (20)²
⇒ BC² =225 + 400
⇒ BC² = 625
⇒ BC² = (25)²
⇒ BC = 25 m
Hence, the correct answer is option (b).

Question 33:

The hypotenuse of a right triangle is 26 cm long. If one of the remaining two sides is 10 cm long, the length of the other side is
(a) 25 cm
(b) 23 cm
(c) 24 cm
(d) 22 cm

Answer 33:

In right traingle BOC,
BC² = OC² + OB²
⇒ (26)² = (10)² + OB²
⇒ 676 =100 + OB²
⇒ OB² = 576
⇒ OB² = (24)²
⇒ OB = 24 cm
Hence, the correct answer is option (c).

Question 34:

A 15 m long ladder is placed against a wall in such away that the foot of the ladder is 9 m away from the wall. Up to what height does the ladder reach the wall?
(a) 13 m
(b) 10 m
(c) 8 m
(d) 12 m

Answer 34:

Suppose BC is the ladder having length 15 m is placed againts the wall OA. The foot of the ladder C is 9 m away from the foot of the wall O.

In right traingle BOC,
BC² = OC² + OB²
⇒ (15)² = (9)² + OB²
⇒ 225 =81 + OB²
⇒ OB² = 144
⇒ OB² = (12)²
⇒ OB = 12 m
Hence, the correct answer is option (d).