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(b) ASA congruence criterion

Given that □ABCD is a square. 

ΔAPB is an equilateral triangle. 

Now in ΔAPD and ΔBPC 

AP = BP ( ∵ sides of an equilateral triangle) 

AD = BC (∵ sides of a square) 

∠PAD = ∠PBC [ ∵ 90° – 60°] 

∴ ΔAPD ≅ ΔBPC (by SAS congruence)

Correct option is  C) S.A.S

Correct option is B) S.A.S

Given: 

AD = AE …(i) 

AB = AC …(ii) 

Subtracting AD from both sides, we get: 

⇒ AB – AD = AC – AD 

⇒ AB – AD = AC - AE (Since, AD = AE) 

⇒ BD = EC …(iii) 

Dividing equation (i) by equation (iii), we get: 

AD/DB = AE/EC 

Applying the converse of Thales’ theorem, DE‖BC 

⇒ ∠DEC + ∠ECB = 180° (Sum of interior angles on the same side of a Transversal Line is 0°.) 

⇒ ∠DEC + ∠CBD = 180° (Since, AB = AC ⇒ ∠B = ∠C) 

Hence, quadrilateral BCED is cyclic. 

Therefore, B,C,E and D are concylic points.

Correct option is   D) BD = CE

Correct option is  A) 45 units

Correct option is  B) 2CF = CE

Given: ∠ACB = 90° And CD⊥AB

To Prove; BC²/AC² = BD/AD

Proof: In ∆ ACB and ∆ CDB 

∠ACB = ∠COB = 90° (Given) 

∠ = ∠ (Common) 

By AA similarity-criterion ∆ ACB ~ ∆CDB 

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional. 

∴ BC/BD = AB/BC 

⇒ BC² = BD. AB … (1) 

In ∆ ACB and ∆ ADC 

∠ACB = ∠ADC = 90° (Given) 

∠CAB = ∠DAC (Common) 

By AA similarity-criterion ∆ ACB ~ ∆ADC

When two triangles are similar, then the ratios of their corresponding sides are proportional.

∴ AC/AD = AB/AC 

⇒ AC² = AD. AB ….(2) 

Dividing (2) by (1), we get 

BC²/AC² = BD/AD

In ΔABC,

AB = AC (Given)

∴ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

AC = AD

∴ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal) In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)

∴ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º

∴ 2(∠ACB + ∠ACD) = 180º

∴ 2(∠BCD) = 180º

∴ ∠BCD = 90º

Corresponding altitudes of two similar triangles are 6 cm and 9 cm (given)

We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.

Ratio in the areas of two similar triangles = (6)² : (9)² = 36 : 81 = 4 : 9

The longest side of a right angled triangle is called its hypotenuse .

Median is also called altitude in an equilateral triangle.

The sum of an exterior angle of a triangle and its adjacent angle is always a right angle.

The longest side of a right angled triangle is called its hypotenuse.

Answer is:

(b) Any triangle

(d) BC > CA and CA < CD.

∠B = ∠C = 55º ⇒ AC = AB 

⇒ ∠BAC = 180º – (∠B + ∠C) = 180º – 110º = 70º 

Now in ΔACD, ∠ACD = 180º – 55º = 125º 

∴ ∠CAD = 180º – (125º + 25º) = 30º 

⇒ ∠CAD > ∠CDA ⇒ CD > AC 

(∵ In a given triangle, the greater angle has greater side opposite to it) Also ∠BAC > ∠ABC ⇒ BC > AC 

∴ BC > CA and CA < CD.

The angles are \(\frac{8}{12}\times180°,\frac{3}{12}\times180°\) i.e., 120º, 45º and 15º. 

Also we know that the longest side is opposite the greatest angle and so on. 

∴ Let the longest side opposite the greatest angle 120º be \(x\) and let the next longest side opposite angle 45º be y. Then, by the sine rule

\(\frac{sin\,120°}{x}\) = \(\frac{sin\,45°}{y}\)

⇒ \(\frac{x}{y}=\frac{sin\,120°}{sin\,45°}=\frac{\frac{\sqrt3}{2}}{\frac{1}{\sqrt2}}=\frac{\frac{\sqrt3}{2}}{\frac{1}{\sqrt2}}=\frac{\sqrt6}{2}.\)

(i) In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given) 

AD = AD (Common)

∴ ΔABD ≅ ΔACD (By SSS congruence rule)

∴ ∠BAD = ∠CAD (By CPCT)

∴ ∠BAP = ∠CAP …. (1)

(ii) In ΔABP and ΔACP,

AB = AC (Given)

∠BAP = ∠CAP [From equation (1)]

AP = AP (Common)

∴ ΔABP ≅ ΔACP (By SAS congruence rule)

∴ BP = CP (By CPCT) … (2)

(iii) From equation (1),

∠BAP = ∠CAP 

Hence, AP bisects ∠A. 

In ΔBDP and ΔCDP, 

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule) ∴

∠BDP = ∠CDP (By CPCT) … (3)

Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP

∴ ∠BPD = ∠CPD (By CPCT) …. (4)

∠BPD + ∠CPD = 180 (Linear pair angles)

∠BPD + ∠BPD = 180

2∠BPD = 180 [From equation (4)]

∠BPD = 90^o … (5)

From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

∴ AM and DN are medians of ∆ABC & ∆DEF respectively 

∴ BM = MC & EN = NF 

⇒ BM = 1/2 BC & EN = 1/2 EF 

But, BC = EF ∴BM = EN ...(i) 

In ∆ABM & ∆DEN we have 

 AB = DE [Given] 

 AM = DN [Given] 

 BM = EN [By (i)] 

∴ By SSS criterion of congruence we have 

 ∆ABM ≅ ∆DEN ⇒ ∠B = ∠E ...(ii) [By cpctc] 

 Now, In ∆ABC & ∆DEF 

 AB = DE [Given] 

 ∠B = ∠E [By (ii)] 

 BC = EF [Given] 

∴ By SAS criterion of congruence we get ∆ABC ≅ ∆DEF

(c) \(\frac{ac}{b+c}\)

In Δs LMK and ONK, 

∠KML = ∠ONK = 46° 

∠K = ∠K            (Common) 

∴ ΔLMK ~ ΔONK       (AA similarity)

⇒ \(\frac{KM}{KN}=\frac{LM}{ON}⇒\frac{b+c}{c} =\frac{a}{x}⇒x=\frac{ac}{b+c}\).

In ∆BAC and ∆PQR, 

seg BA ≅ seg PQ

seg BC ≅ seg PR 

∠BAC ≅ ∠PQR = 90° [Given] 

∴ ∆BAC ≅ ∆PQR [Hypotenuse side test] 

∴ seg AC ≅ seg QR [c.s.c.t.] 

∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP [c.a.c.t.]

Length of hypotenuse = 15 [Given] 

Length of median on the hypotenuse = 1/2

x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse] 

= (1/2) x 15 = 7.5 

∴ The length of the median on the hypotenuse is 7.5 units.

(B) The answer is PQ > PR

We observe that the length of the side opposite to 30° is half the length of the hypotenuse.

(D) 3.4 cm

Sum of the lengths of two sides of a triangle > length of the third side 

Here, 1.5 cm + 3.4 cm = 4.9 cm < 5 cm 

∴ Third side ≠ 3.4 cm

i. By SSS test 

∆ABC ≅ ∆PQR

ii. By SAS test 

∆ XYZ ≅ ∆LMN

iii. By ASA test 

∆PRQ ≅ ∆STU

iv. By hypotenuse side test 

∆LMN ≅ ∆PTR

From the activity we observe that, when corresponding angles of two triangles are equal, the ratios of their corresponding sides are also equal i.e., their corresponding sides are in the same proportion.

∆XYZ ~ ∆LMN [Given] 

∴ ∠X ≅ ∠L 

∠Y ≅ ∠M > 

∠Z ≅ ∠N [Corresponding angles of similar triangles]

XY/LM = YZ/MN = XZ/LN [Corresponding sides of similar triangles]

Correct answer is (b) 3

Correct answer is (d) 6 cm

Correct answer is (c) 5 cm

Correct answer is (d) Difference of any two sides of a triangle is greater than the third side

We have, 

In ∆BAC, by Pythagoras theorem, we have 

BC² = AB² + AC² 

⇒ BC² = c² + b² 

⇒ BC = √(c² + b²) 

In ∆ABD and ∆CBA 

∠B = ∠B  [Common] 

∠ADB = ∠BAC  [Each 90°] 

Then, ∆ABD ͏~ ∆CBA   [By AA similarity] 

Thus, \(\frac{AB}{CB} = \frac{AD}{CA}\) [Corresponding parts of similar triangles are proportional] 

\(\frac{c}{\sqrt{(c^2 + b^2)}}\) = \(\frac{AD}{b }\)

∴ AD = \(\frac{bc}{\sqrt{(c^2 + b^2)}}\)

Correct option is (C) S.S.S.

In triangle \(\triangle PQR\;and\;\triangle SQR,\)

PR = SQ     (Given),

PR = SR      (Given),

QR = QR     (Common side)

\(\therefore\) \(\triangle PQR\cong \triangle SQR\)     (By SSS congruence criteria)

Correct option is  C) S.S.S.

Correct option is (A) similar

Any two circles are always similar but congruent if they have same radii.

Correct option is  A) similar

Correct option is (D) S.A.S.

In triangles \(\triangle POR\;and\;\triangle QOS,\)

PO = OQ

and RO = OS   (PQ and RS bisect at O)

\(\angle POR=\angle QOS\)   (Vertically opposite angles)

\(\therefore\) \(\triangle POR\cong \triangle QOS\)   (By S.A.S. congruence rule)

Correct option is  D) S.A.S.

Correct option is (D) S.S.S.

In \(\triangle ABC\) & \(\triangle CDA\),

AB = CD,    (Given)

BC = DA,    (Given)

AC = AC    (Common side)

\(\therefore\) \(\triangle ABC\) \(\cong\) \(\triangle CDA\)  (By SSS congruence rule)

Hence, both triangles are congruent by SSS rule.

Correct option is  D) S.S.S.