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It is possible to have a triangle in which two angles are acute.

True

It is possible to have a triangle in which two of the angles are right angles.

False

 It is possible to have a triangle in which each angle is greater than 60°.

False

It is possible to have a triangle in which each angle is less than 60°.

False

It is possible to have a triangle in which each angle is equal to 60°.

True

Correct option is C) 90°

Correct answer is (c) (iii) only

Given that the triangle is acute.

So,

∠1, ∠2 and ∠3 be the angles of the triangle.

∠1 = 90°(Given)

∠2 = ∠3

We know that,

∠1 + ∠2 + ∠3 = 18°

90° + ∠2 + ∠2 = 180°

2∠2 = 180° – 90°

∠2 = 45°

Therefore,

∠2 = ∠3 = 45°

Thus, each acute angle is equal to 45°

Correct answer is (c) Equilateral

(d) (i) and (ii)

From the question it is given that,

One angle of isosceles triangle is 70^o.

We know that, in an isosceles triangle 2 angles are equal corresponding with 2 equal sides,

If 70^o is 3^rd angle of triangle,

70^o + x + x = 180^o

2x + 70^o = 180^o

2x = 180^o – 70^o

x = (110^o/2)

x = 55^o

So, both angles are 55^o

Consider the 70^o as base angle of isosceles triangle

Then, 70^o + 70^o + x = 180^o

x = 180^o – 140^o

x = 40^o

So, one angle is 40^o and another is 70^o

Correct answer is (c) Obtuse angled triangle 

Correct answer is (d) (i) and (ii)

State whether the statements are True or False. 

The top and bottom faces of a kaleidoscope are congruent.

True

(b) a scalene triangle only

A scalene triangle is a triangle that has three unequal sides.

(c) Obtuse angled triangle

We know that, sum of interior angles of triangle is equal to 180^o.

Let us assume the 3^rd angle be Q,

Then, 40^o + 40^o + Q = 180^o

80^o + Q = 180^o

Q = 180 – 80

Q = 100^o

An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles. Since a triangle’s angles must sum to 180^o.

Correct answer is 12m

Correct answer is 100 cm

Correct answer is (b) 5m

Correct answer is 

(i)∠F = 50° (ii) ∠EOF = 120°

Correct option is (C) 130°

Given that \( \triangle ABC \sim \triangle DEF\)

\(\therefore\) \(\angle D=\angle A=50^\circ\)      (Given \(\angle A=50^\circ)\)

In \(\triangle DEF,\) \(\angle D+\angle E+\angle F=180^\circ\)  (Sum of angles in a triangle)

\(\Rightarrow\) \(\angle E+\angle F=180^\circ-\angle D\)

\(=180^\circ-50^\circ=130^\circ\)

Correct option is: (C) 130°

Correct option is (B) ΔABC ~ ΔDEF

ΔABC is similar to ΔDEF is denoted as \(\triangle ABC\sim\triangle DEF.\)

Correct option is: (B) ΔABC ~ ΔDEF

In Fig. 6.24, Δ PQO≅ Δ PQR

Correct answer is 

(c) ΔABC ≅ ΔDCB

Given that AC = AE, AB = AD and 

∠BAD = ∠EAC 

In ΔABC and ΔADE 

AB = AD 

AC = AE 

∠BAD = ∠EAC 

∴ ΔABC ≅ ΔADE (∵ SAS congruence) 

⇒ BC = DE (CPCT)

We know that AB || CD and GE is the transversal

From the figure we know that ∠EGF and ∠GED are interior angles

So we get

∠EGF + ∠GED = 180^o

By substituting the values

∠EGF + 130^o = 180^o

On further calculation

∠EGF = 180^o – 130^o

By subtraction

∠EGF = 50^o

Therefore, ∠EGF = 50^o

Correct option is (C) right

In triangles \(\triangle PAQ\;and\;\triangle PBQ,\)

PA = PB,       (Given)

AQ = BQ,      (Given)

PQ = PQ       (Common side)

\(\therefore\) \(\triangle PAQ\cong\triangle PBQ\)    (By SSS congruence criteria))

\(\therefore\) \(\angle APQ=\angle BPQ\)    (By corresponding property of congruence triangles)

\(\Rightarrow\) \(\angle APO=\angle BPO\)   _______(1)

Now, in triangles \(\triangle APO\;and\;\triangle BPO,\)

AP = BP    (Given)

\(\angle APO=\angle BPO\)       (From (1))

OP = OP      (Common side)

\(\therefore\) \(\triangle APO\cong\triangle BPO\)   (By SAS congruence criteria)

\(\therefore\) \(\angle AOP=\angle BOP\) _______(2)  (By corresponding property of congruence triangles)

Since, \(\angle AOP\;and\;\angle BOP\) forms a linear pair.

\(\therefore\) \(\angle AOP+\angle BOP\) = \(180^\circ\)

\(\Rightarrow\) \(2\angle AOP\) = \(180^\circ\)    \((\because\) \(\angle AOP=\angle BOP\)   (From (2)))

\(\Rightarrow\) \(\angle AOP\) = \(\frac{180^\circ}2=90^\circ\)

\(\Rightarrow\) \(\angle POA\) = \(90^\circ\)

Hence, \(\angle POA\) is a right angle.

Correct option is  C) right

Given: ∆ABC is an equilateral triangle.

 To prove: ∆ABC is equiangular

 i.e. ∠A ≅ ∠B ≅ ∠C …(i) 

[Sides of an equilateral triangle] 

In ∆ABC, 

seg AB ≅ seg BC [From (i)] 

∴ ∠C = ∠A (ii)

 [Isosceles triangle theorem] 

In ∆ABC, 

seg BC ≅ seg AC [From (i)] 

∴ ∠A ≅ ∠B (iii) [Isosceles triangle theorem] 

∴ ∠A ≅ ∠B ≅ ∠C [From (ii) and (iii)] 

∴ ∆ABC is equiangular.

In ∆PQR,

 seg PR ≅ seg PQ [Given] 

∴ ∠PQR ≅ ∠PRQ ….(i) [Isosceles triangle theorem] 

∠PRQ is the exterior angle of ∆PRS. 

∴ ∠PRQ > ∠PSR ….(ii) [Property of exterior angle] 

∴ ∠PQR > ∠PSR [From (i) and (ii)] 

i.e. ∠Q > ∠S ….(iii)

In APQS, 

∠Q > ∠S [From (iii)] 

∴ PS > PQ [Side opposite to greater angle is greater] 

∴ seg PS > seg PQ

Given that D is the mid point of BC of ΔABC. 

DF ⊥ AC; DE = DF 

DE ⊥ AB 

In ΔBED and ΔCFD 

∠BED = ∠CFD (given as 90°) 

BD = CD (∵D is mid point of BC) 

ED = FD (given) 

∴ ΔBED ≅ ΔCFD (RHS congruence)

Given: Seg AD is the bisector of ∠BAC. 

seg AD ⊥ seg BC

 To prove: ∆ABC is an isosceles triangle.

 Proof.

 In ∆ABD and ∆ACD,

 ∠BAD ≅ ∠CAD [seg AD is the bisector of ∠BAC] 

seg AD ≅ seg AD [Common side] 

∠ADB ≅ ∠ADC [Each angle is of measure 90°] 

∴ ∆ABD ≅ ∆ACD [ASA test] 

∴ seg AB ≅ seg AC [c. s. c. t.] 

∴ ∆ABC is an isosceles triangle.

Yes. 

Construction: 

Draw line RM parallel to seg PQ through a point R. 

Proof: 

seg PQ || line RM and seg PR is their transversal. [Construction] 

∴ ∠PRM = ∠QPR ……..(i) [Alternate angles] 

seg PQ || line RM and seg QR is their transversal. [Construction]

∴ ∠SRM = ∠PQR ……..(ii) [Corresponding angles] 

∴ ∠PRM + ∠SRM = ∠QPR + ∠PQR [Adding (i) and (ii)] 

∴ ∠PRS = ∠PQR + ∠QPR [Angle addition property]

Given: 

Seg AD is the bisector of ∠BAC. seg AD ⊥ seg BC 

To prove: AABC is an isosceles triangle.

Proof: 

In ∆ABD and ∆ACD, 

∠BAD ≅ ∠CAD [seg AD is the bisector of ∠BAC] 

seg AD ≅ seg AD [Common side] 

∠ADB ≅ ∠ADC [Each angle is of measure 90°] 

∴ ∆ABD ≅ ∆ACD [ASA test] 

∴ seg AB ≅ seg AC [c. s. c. t.] 

∴ ∆ABC is an isosceles triangle.

Yes 

Proof: 

In ∆ABC and ∆ACB, 

seg AB ≅ seg AC [Given] 

∠BAC ≅ ∠CAB [Common angle] 

seg AC ≅ seg AB [Given] 

∴ ∆ABC ≅ ∆ACB [SAS test] 

∴ ∠ABC ≅ ∠ACB [c. a. c. t.]

Yes 

Proof:

In ∆ABC and ∆ACB, 

seg AB ≅ seg AC [Given] 

∠BAC ≅ ∠CAB [Common angle] 

seg AC ≅ seg AB [Given] 

∴ ∆ABC ≅ ∆ACB [SAS test] 

∴ ∠ABC ≅ ∠ACB [c. a. c. t.]

Yes 

Construction: Draw seg AD ⊥ seg BC. 

Proof: 

In ∆ABD and ∆ACD,

 seg AB≅ seg AC [Given] 

∠ADB ≅ ∠ADC [Each angle is of measure 90°] 

seg AD ≅ seg AD [Common side] 

∴ ∆ABD ≅ ∆ACD [Hypotenuse side test] 

∴ ∠ABD ≅ ∠ACD [c.a.c.t.] 

∴ ∠ABC ≅ ∠ACB [B-D-C]

Yes 

Construction: 

Draw seg AD ⊥ seg BC. 

Proof: 

In ∆ABD and ∆ACD, 

seg AB≅ seg AC [Given] 

∠ADB ≅ ∠ADC [Each angle is of measure 90°] 

seg AD ≅ seg AD [Common side] 

∴ ∆ABD ≅ ∆ACD [Hypotenuse side test] 

∴ ∠ABD ≅ ∠ACD [c.a.c.t.] 

∴ ∠ABC ≅ ∠ACB [B - D - C]

Correct option is (C) 96 cm²

We have \(\triangle PQR\sim\triangle XYZ,\)

QR = 3 cm, YZ = 4 cm and \(ar(\triangle PQR)=54\,cm^2\)

We know that the area of two similar triangles are in the ratio of the squares of their corresponding sides.

\(\because\) \(\triangle XYZ\sim\triangle PQR\)

\(\therefore\) \(\frac{ar(\triangle XYZ)}{ar(\triangle PQR)}=(\frac{XY}{PQ})^2\)

\(=(\frac{YZ}{QR})^2=(\frac{XZ}{PR})^2\)

\(\Rightarrow\) \(\frac{ar(\triangle XYZ)}{ar(\triangle PQR)}=(\frac{YZ}{QR})^2=\frac{YZ^2}{QR^2}\)

\(=\frac{4^2}{3^2}=\frac{16}{9}\)

\(\Rightarrow\) \(ar(\triangle XYZ)=\frac{16}9\,ar(\triangle PQR)\)

\(=\frac{16}9\times54\)       \((\because ar(\triangle PQR)=54\,cm^2)\)

\(=16\times6\) \(=96\,cm^2\)

Correct option is: (C) 96 cm²

(i) Write the side opposite to vertex Y  = \(\overline{XZ}\)

(ii) Write the angle opposite to side \(\overline{PQ}\) = ∠R

i) In ΔABC

∠A + ∠B + ∠C = 180° (angle sum property)

x° + 50° + 60° = 180° 

x° + 110° = 180° 

x° = 180° – 110°

ii) In ΔPQR

∠P + ∠Q + ∠R = 180° (angle – sum property) 

90° + 30° + x° = 180° 

120° + x° – 180° 

x° = 180° – 120° 

∴ x° = 60°

iii) In ΔXYZ,

∠X + ∠Y + ∠Z = 180° (angle – sum property) 

30°+110° + x° =180° 

140° + x° = 180° 

x° = 180° – 140° 

∴ x° = 40°

i) In ΔPQR

x° + 50° = 120° (exterior angle property) 

x°= 120°- 50° 

x°= 70° 

Also ∠P + ∠Q +∠R = 180° (angle – sum property) 

70° + 50° + y° = 180° 

120° + y° = 180° 

y° = 180° – 120° 

y° = 60°

(OR)

y° + 120° = 1800 (linear pair of angles) 

y° = 180°- 120° 

∴ y° = 60°

ii) In the figure ΔRST, 

x° = 80° (vertically opposite angles) 

also ∠R + ∠S + ∠T = 180° (angle – sum property) 

80° + 50°+ y°= 180° 

130° + y° = 180° 

y° = 180° – 130° 

∴ y° = 50°

iii) m ΔMAN, 

x° = ∠M + ∠A (exterior angle property) 

x° = 50° + 60° 

x° = 110° 

Also x° + y° = 180° 

110°+ y°= 180° 

y° = 180° – 110° 

y° = 70°

(OR) 

in ΔMAN, 

∠M + ∠A + ∠N = 180° (angle – sum property )

50° + 60° + y° = 180° 

110° + y° = 180° 

y° = 180°- 110° 

∴ y° = 70°

v) In the figure ΔABC, 

x° = 60° (vertically opposite angles) 

∠A + ∠B + ∠ACB = 180° (angle – sum property) 

y° + 30° + 60° = 180° 

y° + 90° = 180° 

y° = 180° – 90° 

∴ y° = 90°

v) In the figure ΔEFG, 

y° = 90° (vertically opposite angles)

Also in ΔEFG; 

∠F + ∠E + ∠G = 180° (angle – sum property) 

∴ x° + x° + y° = 180 

2x° + 90° = 180° 

2x° = 180°- 90° 

2x° = 90°

x° = 90°/2

∴ x° = 45°

vi) In the figure ΔLET, 

∠L = ∠T = ∠E = x° (vertically opposite angles) 

Also in ΔLET 

∠L + ∠E + ∠T = 180° (angle – sum property) 

x° + x° + x° = 180° 

3x° = 180°

x° = 180°/3

x° = 60°

Correct option is (C) angles are equal (or) sides are in proportion

Two triangles are similar if their corresponding angles are equal or their corresponding sides are in proportion.

Correct option is: (C) angles are equal (or) sides are in proportion

It is given that ∆ ABC - ∆ DEF. 

Therefore, their corresponding sides will be proportional. 

Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides. 

⇒ perimeter of ∆ABC/ perimeter of ∆DEF = BC/EF

Let the perimeter of ∆ABC be X cm 

Therefore, 

X/25 = 9.1/6.5 

⇒ X = 9.1×25/6.5 = 35 

Thus, the perimeter of ∆ABC is 35 cm.

The required two angles are ∠A and ∠E. When ∠A = ∠E, ∆ ABC ≅ ∆ EDF by SAS criterion.

Correct option is (A) ΔABC ≅ ΔDEF

In \(\triangle ABC\;and\;\triangle EFD,\)

AB = EF   (Both hypotenuse are equal)

\(\angle C=\angle D\) \(=90^\circ\)   (Angles opposite to hypotenuse)

AC = ED         (Given)

\(\therefore\) \(\triangle ABC\cong\triangle EFD\)       (By RHS congruence criteria)

Similarly we can say \(\triangle ACB \cong \triangle EDF\)

(B), (C) and (D) are correct.

Correct option is   D) 1 : 4