Triangles — Chapter Test

In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.

Answer

Given,

In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF.

The two triangles will not be congruent as AB and EF are not corresponding sides.

In the adjoining figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°. Prove that

(a) △PBQ ≅ △QCR

(b) PQ = QR

(c) ∠PRQ = 45°

Answer

(a) Given, ABCD is a square.

∴ AB = BC.

Given, AP = BQ.

∴ AB - AP = BC - BQ .......(i)

In △PBQ and △QCR,

BQ = CR (Given)

From (i) we get,

PB = QC

∠PBQ = ∠QCR (Both are equal to 90°, as each angle in square = 90°.)

Hence, proved △PBQ ≅ △QCR by SAS axiom.

(b) We know that corresponding parts of congruent triangles are equal.

∴ PQ = QR.

Hence, proved that PQ = QR.

(c) Considering △PQR.

We know PQ = QR and ∠Q = 90°.

Hence, △PQR is an isosceles triangle with ∠P = ∠R = x.

Sum of angles of triangle = 180°.

⇒ ∠P + ∠Q + ∠R = 180°

⇒ x + 90° + x = 180°

⇒ 2x = 90°

⇒ x = 45°.

Hence, proved that ∠PRQ = 45°.

In the adjoining figure, OA ⊥ OD, OC ⊥ OB, OD = OA and OB = OC. Prove that AB = CD.

Answer

From figure,

∠AOD = ∠COB (Each 90°)

Adding ∠AOC to both sides,

⇒ ∠AOD + ∠AOC = ∠AOC + ∠COB

⇒ ∠COD = ∠AOB.

Now, in △AOB and △DOC

OA = OD (Given)

OB = OC (Given)

∠AOB = ∠COD (Proved)

∴ △AOB ≅ △DOC (SAS axiom)

We know that corresponding parts of congruent triangles are equal.

∴ AB = CD.

Hence, proved that AB = CD.

In the adjoining figure, PQ || BA and RS || CA. If BP = RC, prove that:

(i) △BSR ≅ △PQC

(ii) BS = PQ

(iii) RS = CQ.

Answer

Given, BP = RC

⇒ BR - PR = PC - PR

⇒ BR = PC.

Now, in △BSR and △PQC,

∠B = ∠P (Corresponding angles)

∠R = ∠C (Corresponding angles)

BR = PC (Proved)

Hence, proved △BSR ≅ △PQC by ASA axiom.

(ii) We know that corresponding parts of congruent triangles are equal.

∴ BS = PQ.

(iii) We know that corresponding parts of congruent triangles are equal.

∴ RS = CQ.

In the adjoining figure, AB = AC, D is a point in the interior of △ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of △ABC.

Answer

Given, AB = AC.

∴ ∠ABC = ∠ACB (As angles opposite to equal sides of isosceles triangle are equal.)

Given, ∠DBC = ∠DCB

∴ DB = DC. (Sides opposite to equal angles are equal.)

In △ABD and △ACD,

AB = AC (Given)

DB = DC (Proved)

AD = AD (Common)

Thus, △ABD ≅ △ACD by SSS axiom.

We know that corresponding parts of congruent triangle are equal.

∴ ∠BAD = ∠CAD.

Hence, proved that AD bisects ∠BAC.

In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.

Answer

Given,

DE bisects ∠D.

∴ ∠EDA = ∠EDC = $\dfrac{∠D}{2}$.

AB || CD and DE cuts these parallel lines.

∠DEA = ∠EDC = $\dfrac{∠D}{2}$.

In △ADE,

∠ADE = ∠DEA = $\dfrac{∠D}{2}$.

Hence, AD = AE.

Given,

CE bisects ∠C.

∴ ∠ECB = ∠ECD = $\dfrac{∠C}{2}$.

AB || CD and CE cuts these parallel lines.

∠CEB = ∠ECD = $\dfrac{∠C}{2}$.

In △BCE,

∠BEC = ∠BCE = $\dfrac{∠C}{2}$

Hence, BE = BC.

⇒ AB = AE + BE = AD + BC

⇒ AB = AD + BC.

Hence, proved that AB = AD + BC.

In △ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that △CAE is isosceles.

Answer

From figure,

∠DAC= ∠ACE (Alternate angles)

∠BAD = ∠CEA (Corresponding angles)

But, ∠BAD = ∠DAC (as AD is bisector of ∠BAC)

∴ ∠ACE = ∠CEA

AE = AC (Sides opposite to equal angles are equal.)

∴ △CAE is isosceles triangle.

Hence, proved that △CAE is isosceles triangle.

In the adjoining figure, ABC is a right angled triangle at B. ADEC and BCFG are squares. Prove that AF = BE.

Answer

Given, ADEC and BCFG are squares.

Considering △BCE and FCA we get,

From figure,

∠BCE = ∠BCA + 90

∠ACF = ∠BCA + 90

∠BCE = ∠ACF

AC = CE (Sides of squares are equal)

BC = CF (Sides of squares are equal)

△BCE ≅ △FCA (By SAS axiom).

We know that corresponding parts of congruent triangles are equal.

∴ AF = BE.

Hence, proved that AF = BE.

In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.

Answer

∠1 = ∠4 (Vertically opposite angles)

⇒ 2∠2 = 2∠3

⇒ ∠2 = ∠3.

TS = TR (Given)

⇒ ∠TRS = ∠TSR (As angle opposite to equal side are equal)

⇒ ∠TRS - ∠2 = ∠TSR - ∠3

⇒ ∠ARB = ∠BSA.

∠RTB = ∠STA (Common angle)

△RBT ≅ △SAT (By ASA axiom.)

We know that corresponding sides of congruent triangles are equal.

∴ RB = SA.

Hence, proved that RB = SA.

In the figure (1) given below, find the value of x.

Answer

From figure,

∠ABD = ∠BAD = 36° (As angles opposite to equal sides are equal.)

∠BDA = 180° - (36° + 36°) = 180° - 72° = 108°.

From figure,

∠BDA + ∠ADC = 180°

108° + ∠ADC = 180°

∠ADC = 72°.

In △ADC,

As AD = AC,

∴ ∠ADC = ∠ACD = 72° (As angles opposite to equal sides are equal.)

∠ADC + ∠ACD + ∠DAC = 180°

72° + 72° + ∠DAC = 180°

∠DAC = 180° - 144° = 36°.

From figure,

∠BAD + ∠DAC + x = 180°

36° + 36° + x = 180°

72° + x = 180°

x = 108°.

Hence, x = 108°.

In the figure (2) given below, AB = AC and DE || BC. Calculate

(i) x

(ii) y

(iii) ∠BAC

Answer

(i) Since, AB = AC

∠ABC = ∠ACB

⇒ 2x° = (y - 2)°

⇒ y° = 2x° + 2° .......(i)

From figure,

∠ADE = ∠ABC (Corresponding angles)

⇒ x° + y° - 36° = 2x°

⇒ y° - 36° = x°

Substituting value of y from (i) in above equation,

⇒ 2x° + 2° - 36° = x°

⇒ 2x° - x° = 34°

⇒ x° = 34°

⇒ x = 34.

Hence, x = 34.

(ii) Substituting value of x in (i),

⇒ y° = 2(34)° + 2°

⇒ y° = 68° + 2° = 70°.

⇒ y = 70.

Hence, y = 70.

(iii) From figure,

∠BAC = 180° - (2x° + y° - 2°)

= 180° - (2 × 34° + 70° - 2°)

= 180° - (68° + 68°)

= 180° - 136° = 44°.

Hence, ∠BAC = 44°.

In the figure (3) given below, calculate the size of each lettered angle.

Answer

From figure,

∠AEB = ∠DEC = 80° (Vertically opposite angles are equal).

In △AEB,

⇒ x + 54° + 80° = 180°

⇒ x + 134° = 180°

⇒ x = 46°.

AB = BC

⇒ ∠BAC = ∠ACB (As angles opposite to equal sides are equal.)

∴ ∠ACB = 54°

In △ABC,

∠ABC + ∠ACB + ∠BAC = 180°

⇒ (x + y)° + 54° + 54° = 180°

⇒ y° + 46° + 108° = 180°

⇒ y = 180° - 154° = 26°.

From figure,

∠ECD = ∠BAE = 54° (Alternate angles)

∠BCA + ∠ECD + z = 180°

54° + 54° + z = 180°

108° + z = 180°

z = 72°.

Hence, x = 46°, y = 26° and z = 72°.

In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that

(i) AC > DC

(ii) AB > AD.

Answer

From figure,

∠DAC = ∠ACD = 35° (As angles opposite to equal sides are equal.)

∠ADC = 180° - (35° + 35°) = 180° - 70° = 110°.

Since, ∠ADC > ∠DAC

∴ AC > DC (Side opposite to greater angle is greater.)

Hence, proved that AC > DC.

(ii) From figure,

∠ADB = 180° - ∠ADC = 180° - 110° = 70°.

Considering △ABD,

AD = BD.

∴ ∠BAD = ∠ABD = a.

⇒ a + a + ∠ADB = 180°

⇒ 2a + 70° = 180°

⇒ 2a = 110°

⇒ a = 55°.

Since, ∠ADB > ∠ABD

∴ AB > AD. (Side opposite to greater angle is greater.)

Hence, proved that AB > AD.

In the figure (2) given below, prove that

(i) x + y = 90°

(ii) z = 90°

(iii) AB = BC.

Answer

(i) From figure,

∠ACB = x (Alternate angles)

In △ABC,

⇒ x + (y + y) + ∠ACB = 180°

⇒ x + 2y + x = 180°

⇒ 2x + 2y = 180°

⇒ x + y = 90°.

Hence, proved that x + y = 90°.

(ii) Now in △BCD,

⇒ y + z + ∠BCD = 180° (Sum of all angles in a triangle is 180°)

⇒ y + z + x = 180°

⇒ 90° + z = 180° [∵ x + y = 90°]

⇒ z = 90°.

Hence, proved that z = 90°.

(iii) In △ABC,

∠ACB = ∠BAC = x

∴ AB = BC (As sides opposite to equal angles are equal.)

Hence, proved that AB = BC.

In the adjoining figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that

(i) △ABD ≅ △ACD

(ii) △ABP ≅ △ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

Answer

(i) AB = AC (as ABC is an isosceles triangle on base BC)

DB = DC (as DBC is an isosceles triangle on base BC)

AD = AD (Common)

∴ △ABD ≅ △ACD by SSS axiom.

Hence, proved that △ABD ≅ △ACD.

(ii) AB = AC (as ABC is an isosceles triangle on base BC)

AP = AP (Common)

∠ABP = ∠ACP (As angles opposite to equal sides are equal.)

∴ △ABP ≅ △ACP by AAS axiom.

Hence, proved that △ABP ≅ △ACP.

(iii) We know that,

△ABP ≅ △ACP.

As corresponding parts of congruent triangle are equal.

∴ ∠BAP = ∠CAP ......(i)

also ∠APB = ∠APC.

△DPB ≅ △DPC (By SAS axiom)

As corresponding parts of congruent triangle are equal.

∴ ∠BDP = ∠CDP ......(ii)

From (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

Hence, proved that AP bisects ∠A as well as ∠D.

(iv) Since, ∠BPA = ∠CPA.

From figure,

∠BPA + ∠CPA = 180°

2∠BPA = 180°

∠BPA = 90°.

Hence, proved that AP is the perpendicular bisector of BC.

In the adjoining figure, AP ⊥ l and PR > PQ. Show that AR > AQ.

Answer

Take a point S on PR such that PS = PQ.

Join A and S. Mark angles as shown.

PQ = PS

AP = AP (Common)

∠APQ = ∠APS (Both are equal to 90°)

△APQ ≅ △APS (By SAS axiom)

We know that corresponding parts of congruent triangles are equal.

∠1 = ∠2.

In △ARS,

∠2 > ∠3 (As exterior angle is greater than each interior opposite angle.)

∴ ∠1 > ∠3

⇒ AR > AQ (As side opposite to greater angle is greater.)

Hence, proved that AR > AQ.

If O is any point in the interior of a triangle ABC, show that OA + OB + OC > $\dfrac{1}{2}$(AB + BC + CA).

Answer

In △OBC, OB + OC > BC ......(i) (As sum of any two sides of triangle > third side)

Similarly OC + OA > CA .......(ii)

and, OA + OB > AB .......(iii)

On adding (i), (ii) and (iii), we get

⇒ OB + OC + OC + OA + OA + OB > BC + CA + AB

⇒ 2(OA + OB + OC) > AB + BC + CA

⇒ OA + OB + OC > $\dfrac{1}{2}$(AB + BC + CA).

Hence, proved that OA + OB + OC > $\dfrac{1}{2}$(AB + BC + CA).