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Given bisector og the base angles of a triangle enclose an angle of 135°

i.e. ∠BOC = 135°

But,

135° = 90° + \(\frac{1}{2}\)∠A

\(\frac{1}{2}\)∠A = 135° – 90°

∠A = 45° (2)

= 90°

Therefore, is right angled triangle right angled at A.

(i) No,

Two right angles would up to 180° so the third angle becomes zero. This is not possible. Therefore, the triangle cannot have two right angles.

(ii) No,

A triangle can’t have two obtuse angles as obtuse angle means more than 90°. So, the sum of the two sides exceeds more than 180° which is not possible. As the sum of all three angles of a triangle is 180°.

(iii) Yes,

A triangle can have two acute angle as acute angle means less than 90°.

(iv) No,

Having angles more than 60° make that sum more than 180° which is not possible as the sum of all angles of a triangle is 180°.

(v) No,

Having all angles less than 60° will make that sum less than 180° which is not possible as the sum of all angles of a triangle is 180°.

(vi) Yes,

A triangle can have three angles equal to 60° as in this case the sum of all three is equal to 180° which is possible. This type of triangle is known as equilateral triangle.

A triangle where one of the internal angles is obtuse (greater than 90 degrees) is called an obtuse triangle. The sum of angles of obtuse triangle is also 180°.

Let,

∠1 = 2k, ∠2 = k and ∠3 = 3k

∠1 + ∠2 + ∠3 = 180°

6k = 180°

k = 30°

Therefore, minimum angle be ∠2 = k = 30°.

Given,

In ΔABC

∠ ABC = ∠ ACB

Divide both sides by 2, we get

\(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB

∠OBC = ∠OCB [Therefore, OB, OC bisects ∠B and ∠C]

Now,

∠BOC = 90° + \(\frac{1}{2}\)∠A

120° – 90° = \(\frac{1}{2}\)∠A

30° x 2 = ∠A

∠A = 60°

Now in ΔABC

∠A + ∠ABC + ∠ACB = 180°[Sum of all angles of a triangle]

60° + 2∠ABC = 180°[Therefore, ∠ABC = ∠ACB]

2∠ABC = 180° – 60°

2∠ABC = 120°

∠ABC = 60°

Therefore,

∠ABC = ∠ACB = 60°

Hence, proved

In ΔABC

∠A + ∠B + ∠C = 180°

60° + ∠B + ∠C = 180°

∠B + ∠C = 120°

\(\frac{1}{2}\)∠B + \(\frac{1}{2}\)∠C = 60°(i)

∠BOC + ∠OBC + ∠OCB = 180°

∠BOC + \(\frac{1}{2}\)∠B + \(\frac{1}{2}\)∠C = 180°

∠BOC + \(\frac{1}{2}\)(∠B + ∠C) = 180°

∠BOC + 60° = 180°[From (i)]

∠BOC = 120°

In BOC,

∠BOC + ∠OCB + ∠OBC = 180°

∠BOC + 1/2 × (80) + 1/2 × (40) = 180°

∠BOC = 180° - 70

∠BOC = 110°

(b) ∠1 = ∠4

From the figure, M is the mid-point of both AC and BD.

By the corresponding parts of congruent triangles, ∠1 = ∠4.

Correct answer is 6m

We have, 

AB = AC = 25cm 

BC = 14cm 

In ⊿ ACD and ⊿ ABD 

∠ADB =∠ADB = 90 

AB = AC = 25cm 

AD = AD (Common) 

⊿ ABD ≅∠ACD 

∴BD = CD = 7cm (By c.p.c.t) 

In ⊿ACD 

AB² = AD² + BD² 

25² = AD² + 72 

625 = AD² + 49 

AD² = 625 - 49 

AD² = 576 

AD = \(\sqrt{576}\)

AD = 24 cm

False

Two right angled triangles are congruent, if the hypotenuse and a side of one of the triangle are equal to the hypotenuse and one of the side of the other triangle.

We have ΔPAB and ΔPQR

∠P = ∠P(common)

∠PAB = ∠PQR (Corresponding angles) 

Then, ΔPAB ~ ΔPQR (BY AA similarity) 

So, \(\frac{PB}{PR}\) =\(\frac{AB}{QR}\) (Corresponding parts of similar triangle area proportion)

Or , \(\frac{PB}6\) = \(\frac{3}9\)

Or PB = \(\frac{3}9\) x 6 

Or PB= 2 cm

We have , XY || BC 

In Δ AXY and ΔABC

∠A = ∠A (Common)

∠AXY = ∠ABC (Corresponding angles) 

Then, Δ AXY ~ΔABC (By AA Similarity) 

So,\(\frac{AX}{BY}\) = \(\frac{XY}{BC}\) (Corresponding parts of similar triangle area proportion) 

Or \(\frac{1}4\) = \(\frac{XY}6\)

Or XY = 6/4

Or XY = 1.5cm

Given, 

∠ABC = 90° and BD⊥AC 

BD = 8 cm 

AD = 4 cm 

Required to find: CD.

We know that, 

ABC is a right angled triangle and BD⊥AC. 

Then, ΔDBA∼ΔDCB  [By AA similarity] 

\(\frac{BD}{CD}\) = \(\frac{AD}{BD}\)

BD² = AD x DC 

(8)² = 4 x DC 

DC = \(\frac{64}{4}\) = 16 cm 

Therefore, CD = 16 cm

It is given that ABC is a right angled triangle 

and BD is the altitude drawn from the right angle to the hypotenuse. 

In ∆ DBA and ∆ DCB, we have : 

∠BDA = ∠CDB

∠DBA = ∠DCB = 90°

Therefore, by AA similarity theorem, we get : 

∆DBA - ∆ DCB 

⇒ BD/CD = AD/BD

⇒ CD = BD^2/AD 

CD = 8×8/4 = 16 cm

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. 

In ∆ BDC and ∆ ABC, we have : 

∠ABC = ∠BBC = 90° (given) 

∠C = ∠C (common)

By AA similarity theorem, we get : 

∆ BDC- ∆ ABC 

AB/BD = BC/DC 

⇒ 5.7/3.8 = BC/5.4 

⇒  = 5.7/3.8 ×5.4 

= 8.1 

Hence, BC = 8.1 cm

Given 

ABC and DEF are two similar triangles, ∠A = 57° and ∠E = 73° 

We know that SAS similarity criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. 

In ΔABC and ΔDEF

If \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) and ∠A = ∠D, then ΔABC ~ ΔDEF

So, 

∠A = ∠D 

⇒ ∠D = 57° … (1) 

Similarly, ∠B = ∠E 

⇒ ∠B = 73° … (2) 

We know that the sum of all angles of a triangle is equal to 180°. 

⇒ ∠A + ∠B + ∠C = 180° 

⇒ 57° + 73° + ∠C = 180° 

⇒ 130° + ∠C = 180° 

⇒ ∠C = 180° - 130° = 50° 

∴ ∠C = 50°

In ΔAEB and ΔAFC,

∠AEB and ∠AFC (Each 90º)

∠A = ∠A (Common angle)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)

∴ BE = CF (By CPCT)

Data: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively 

To Prove: Altitude BE = Altitude CF. 

Proof: 

In ∆ABC, 

AB = AC and CF ⊥ AB, 

BE ⊥ AC. 

∴ ∠BEC = ∠CFB = 90° (Data) 

Angles opposite to equal sides of an isosceles triangles are equal. 

BC is common. 

∴ ∆BEC ≅ ∆CFB (ASA postulate) 

∴ BE = CF.

Data : BE and CF are two equal altitudes of a triangle ABC. 

To Prove: 

ABC is an isosceles triangle. 

Proof : BE = CF (data) 

In ∆BCF and ∆CBE, 

∠BFC = ∠CEB = 90° 

(data) BC is common hypotenuse. 

As per Right angle, 

hypotenuse, side postulate, 

∴ ∆BCF ≅ ∆CBE 

∴ ∠CBF = ∠BCE 

∴ ∠CBA = ∠BCA 

∴ AB = AC 

∴ ∆ABC is an isosceles triangle.

False

If two angles of a triangle are equal then third angle may or may not be equal to each of the other two angles.

True

The congruent figures superimpose each other completely.

Correct answer is 

(a) Δ ABD ≅ Δ ACD 

(b) Δ XYZ ≅ Δ UZY

(c) Δ ACE ≅ Δ BDE 

(d) Δ ABC ≅ Δ CDE

(e) not possible 

(f) Δ LOM ≅ Δ CDE

 A right-angled triangle may have all sides equal.

False

False

Since, given triangles are congruent by SAS.

Two figures are congruent, if they have the same shape.

True

If three angles of two triangles are equal, triangles are congruent.

False

If two sides and one angle of a triangle are equal to the two sides and angle of another triangle, then the two triangles are congruent.

False

Two right angles are congruent.

True

Two acute angles are congruent.

False

False

Since, two triangles are congruent if two sides and included angle of one triangle are equal to the two sides and included angle of another triangle.

If two triangles are congruent, then the corresponding angles are equal.

True

Correct answer is (b) ASA congruence criterion

Correct answer is 

(i) Δ PQR ≅ Δ STU (ii) Not congruent

Correct option is (C) 42 cm

\(\because\) Tangents from an outer point to a circle are of equal length.

\(\therefore ER=ES=5\,cm,\)    (Tangents of circle from point E & ES = 5 cm)

\(SF=PF=7\,cm,\)       (Tangents of circle from point F & PF = 7 cm)

\(GP=QG=5\,cm\)       (Tangents of circle from point G & QG = 5 cm)

and \(QD=DR=4\,cm\)   (Tangents of circle from point D & DR = 4 cm)

\(\therefore\) Perimeter of quadrilateral DEFG

\(=(DR+ER)+(ES+SF)\) \(+(PF+GP)+(QG+QD)\)

\(=(4+5)+(5+7)+(7+5)+(5+4)\)

\(=9+12+12+9\)

\(=18+24=42\,cm\)

Correct option is: (C) 42 cm

Two line segments are congruent, if they are of Equal lengths.

Two line segments are congruent, if they are of equal lengths.

Correct option is B) same length

(i) All circles are similar
(ii) All squares are similar
(iii)All equilateral triangles are similar
(iv) Two triangles are similar, if their corresponding angles are equal
(v) Two triangles are similar, if their corresponding sides are proportional
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Correct option is  D) Both A & B

(b) trapezium 

Diagonals of a trapezium divide each other proportionally.

(i) All circles are similar. 

(ii) All squares are similar. 

(iii) All equilateral triangles are similar. 

(iv) Two triangles are similar, if their corresponding angles are equal. 

(v) Two triangles are similar, if their corresponding sides are proportional. 

(vi) Two polygons of the same number of sides are similar, if (a) equal their corresponding angles are and their corresponding sides are (b) proportional.

(i)  congruent 

(ii) similar 

(iii)  equilateral 

(iv) (a) equal 

(b) proportional