MAK LABS — Std 9 Maths Part 2 Chapter 3 Triangles — Practice Sets 3.1 to 3.5 & Problem Set 3 Solutions

1In figure 3.8, ∠ACD is an exterior angle of ΔABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.

Fig. 3.8

∠ACD is an exterior angle of ΔABC, and ∠A and ∠B are its remote interior angles.

∠ACD = ∠A + ∠B ...... theorem of remote interior angles

= 70° + 40° = 110°

∴ ∠ACD = 110°

2In ΔPQR, ∠P = 70°, ∠Q = 65° then find ∠R.

∠P + ∠Q + ∠R = 180° ...... sum of the angles of a triangle

70° + 65° + ∠R = 180°

135° + ∠R = 180°

∴ ∠R = 180° − 135° = 45°

∴ ∠R = 45°

3The measures of angles of a triangle are x°, (x − 20)°, (x − 40)°. Find the measure of each angle.

x + (x − 20) + (x − 40) = 180 ...... sum of the angles of a triangle

3x − 60 = 180

3x = 240 ∴ x = 80

x − 20 = 60 and x − 40 = 40

∴ The angles are 80°, 60° and 40°.

4The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.

Let the smallest angle be x°. Then the other two angles are 2x° and 3x°.

x + 2x + 3x = 180

6x = 180 ∴ x = 30

2x = 60 and 3x = 90

∴ The angles are 30°, 60° and 90°.

5In figure 3.9, measures of some angles are given. Using the measures find the values of x, y, z.

Fig. 3.9 — T lies on ray ME produced, R on ray NM produced

∠TEN + ∠NEM = 180° ...... angles in a linear pair
100° + y = 180° ∴ y = 80°

∠EMR + ∠EMN = 180° ...... angles in a linear pair
140° + z = 180° ∴ z = 40°

∠EMR is an exterior angle of ΔENM.
∴ ∠EMR = x + y ...... theorem of remote interior angles
140° = x + 80° ∴ x = 60°

∴ x = 60°, y = 80°, z = 40°

Check: 60° + 80° + 40° = 180°.

6In figure 3.10, line AB ∥ line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

Fig. 3.10 — A, R, D are collinear

Line AB ∥ line DE and seg AD is a transversal.
∴ ∠ADE = ∠BAD = 70° ...... alternate angles

In ΔDRE, ∠DRE + ∠RDE + ∠DER = 180°
∠DRE + 70° + 40° = 180°
∴ ∠DRE = 70°

∠ARE + ∠DRE = 180° ...... angles in a linear pair (A − R − D)
∠ARE = 180° − 70° = 110°

∴ ∠DRE = 70°, ∠ARE = 110°

7In ΔABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°, find the measure of ∠AOB.

∠A + ∠B + ∠C = 180° ∴ ∠A + ∠B = 180° − 70° = 110°

∴ ½∠A + ½∠B = 55°

∠OAB = ½∠A and ∠OBA = ½∠B ...... rays AO and BO are bisectors
∴ ∠OAB + ∠OBA = 55°

In ΔAOB, ∠AOB = 180° − (∠OAB + ∠OBA) = 180° − 55° = 125°

∴ ∠AOB = 125°

8In figure 3.11, line AB ∥ line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m∠PTQ = 90°.

Fig. 3.11

Given: Line AB ∥ line CD; ray PT and ray QT bisect ∠BPQ and ∠PQD.

To prove: m∠PTQ = 90°

∠BPQ + ∠PQD = 180° ...... interior angles, line AB ∥ line CD

∴ ½∠BPQ + ½∠PQD = 90°

∴ ∠TPQ + ∠TQP = 90° ...... (I) rays PT and QT are bisectors

In ΔPTQ, ∠PTQ + ∠TPQ + ∠TQP = 180° ...... sum of the angles of a triangle

∠PTQ + 90° = 180° ...... from (I)

∴ m∠PTQ = 90°

9Using the information in figure 3.12, find the measures of ∠a, ∠b and ∠c.

Fig. 3.12

∠b = 70° ...... vertically opposite angles

∠c + 100° = 180° ...... angles in a linear pair ∴ ∠c = 80°

∠a + ∠b + ∠c = 180° ...... sum of the angles of a triangle
∠a + 70° + 80° = 180° ∴ ∠a = 30°

∴ ∠a = 30°, ∠b = 70°, ∠c = 80°

10In figure 3.13, line DE ∥ line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that (i) ∠DEG = ½∠EDF (ii) EF = FG.

Fig. 3.13

Given: Line DE ∥ line GF; ray EG bisects ∠DEF; ray FG bisects ∠DFM.

To prove: (i) ∠DEG = ½∠EDF (ii) EF = FG

(i) Line DE ∥ line GF and line EM is a transversal.
∴ ∠GFM = ∠DEF ...... (I) corresponding angles

∠GFM = ½∠DFM ...... ray FG bisects ∠DFM
∴ ∠DEF = ½∠DFM, that is, ∠DFM = 2∠DEF ...... (II)

∠DFM is an exterior angle of ΔDEF.
∴ ∠DFM = ∠DEF + ∠EDF ...... (III) theorem of remote interior angles

From (II) and (III), 2∠DEF = ∠DEF + ∠EDF ∴ ∠DEF = ∠EDF

∠DEG = ½∠DEF ...... ray EG bisects ∠DEF
∴ ∠DEG = ½∠EDF

(ii) Line DE ∥ line GF and seg EG is a transversal.
∴ ∠FGE = ∠DEG ...... alternate angles

But ∠DEG = ∠GEF ...... ray EG bisects ∠DEF
∴ ∠FGE = ∠FEG

In ΔEFG, the sides opposite to these congruent angles are congruent.
∴ EF = FG ...... converse of isosceles triangle theorem

1In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

(i) ΔABC ≅ ΔPQR

(ii) ΔXYZ ≅ ΔLMN

(iii) ΔPRQ ≅ ΔSTU

(iv) ΔLMN ≅ ΔPTR

Three pairs of sides are congruent: ΔABC ≅ ΔPQR by S-S-S test
Two pairs of sides and the angles included between them are congruent: ΔXYZ ≅ ΔLMN by S-A-S test
Two pairs of angles and the sides included between them are congruent: ΔPRQ ≅ ΔSTU by A-S-A test
Both are right angled triangles with congruent hypotenuses and one more pair of congruent sides: ΔLMN ≅ ΔPTR by hypotenuse-side test

2Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

(i)

Fig. 3.20

In ΔABC and ΔPQR,
∠ABC ≅ ∠PQR, seg BC ≅ seg QR, ∠ACB ≅ ∠PRQ

∴ ΔABC ≅ ΔPQR ...... A-S-A test

∴ ∠BAC ≅ ∠QPR ...... corresponding angles of congruent triangles

seg AB ≅ seg PQ and seg AC ≅ seg PR ...... corresponding sides of congruent triangles

(ii)

Fig. 3.21

In ΔPTQ and ΔSTR,
seg PT ≅ seg ST, ∠PTQ ≅ ∠STR (vertically opposite angles), seg TQ ≅ seg TR

∴ ΔPTQ ≅ ΔSTR ...... S-A-S test

∴ ∠TPQ ≅ ∠TSR and ∠TQP ≅ ∠TRS ...... corresponding angles of congruent triangles

seg PQ ≅ seg SR ...... corresponding sides of congruent triangles

3From the information shown in the figure, state the test assuring the congruence of ΔABC and ΔPQR. Write the remaining congruent parts of the triangles.

Fig. 3.22

In ΔABC and ΔPQR,
∠BAC = ∠PQR = 90° ...... given
hypotenuse BC ≅ hypotenuse PR ...... given
seg AB ≅ seg PQ ...... given

∴ ΔBAC ≅ ΔPQR, that is, ΔABC ≅ ΔQPR ...... hypotenuse-side test

Remaining congruent parts: seg AC ≅ seg QR, ∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP

The right angles are at A and Q, so the correspondence is A ↔ Q, B ↔ P, C ↔ R. Writing the congruence as ΔABC ≅ ΔPQR would pair the wrong vertices.

4As shown in the following figure, in ΔLMN and ΔPNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

Fig. 3.23

In ΔLMN and ΔPNM,
seg LM ≅ seg PN ...... given
seg LN ≅ seg PM ...... given
seg MN ≅ seg NM ...... common side

∴ ΔLMN ≅ ΔPNM ...... S-S-S test

Remaining congruent parts: ∠LMN ≅ ∠PNM, ∠LNM ≅ ∠PMN and ∠MLN ≅ ∠NPM ...... c.a.c.t.

5In figure 3.24, seg AB ≅ seg CB and seg AD ≅ seg CD. Prove that ΔABD ≅ ΔCBD.

Fig. 3.24

Given: seg AB ≅ seg CB and seg AD ≅ seg CD

To prove: ΔABD ≅ ΔCBD

In ΔABD and ΔCBD,

seg AB ≅ seg CB ...... given

seg AD ≅ seg CD ...... given

seg BD ≅ seg BD ...... common side

∴ ΔABD ≅ ΔCBD ...... S-S-S test

6In figure 3.25, ∠P ≅ ∠R, seg PQ ≅ seg RQ. Prove that ΔPQT ≅ ΔRQS.

Fig. 3.25 — S on side PQ, T on side QR

Given: ∠P ≅ ∠R and seg PQ ≅ seg RQ

To prove: ΔPQT ≅ ΔRQS

In ΔPQT and ΔRQS,

∠QPT ≅ ∠QRS ...... given

seg PQ ≅ seg RQ ...... given

∠PQT ≅ ∠RQS ...... common angle

∴ ΔPQT ≅ ΔRQS ...... A-S-A test

1Find the values of x and y using the information shown in figure 3.37. Find the measure of ∠ABD and m∠ACD.

Fig. 3.37 — AB = AC and DB = DC

In ΔABC, seg AB ≅ seg AC ...... given
∴ ∠ABC ≅ ∠ACB ...... isosceles triangle theorem
∴ x = 50°

In ΔDBC, seg DB ≅ seg DC ...... given
∴ ∠DCB ≅ ∠DBC ...... isosceles triangle theorem
∴ y = 60°

∠ABD = ∠ABC + ∠CBD = 50° + 60° = 110°

∠ACD = ∠ACB + ∠BCD = 50° + 60° = 110°

∴ x = 50°, y = 60°, ∠ABD = 110°, m∠ACD = 110°

2The length of hypotenuse of a right angled triangle is 15. Find the length of median of its hypotenuse.

In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse.

Median = ½ × 15 = 7.5

∴ The length of the median is 7.5.

3In ΔPQR, ∠Q = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS).

In ΔPQR, ∠Q = 90°.
PR² = PQ² + QR² ...... Pythagoras theorem
= 12² + 5² = 144 + 25 = 169
∴ PR = 13

Seg QS is the median on hypotenuse PR.
∴ QS = ½ PR ...... property of median on the hypotenuse
= ½ × 13 = 6.5

∴ l(QS) = 6.5

4In figure 3.38, point G is the point of concurrence of the medians of ΔPQR. If GT = 2.5, find the lengths of PG and PT.

Fig. 3.38

The centroid divides each median in the ratio 2 : 1.
∴ PG : GT = 2 : 1

PG = 2 × GT = 2 × 2.5 = 5

PT = PG + GT = 5 + 2.5 = 7.5

∴ PG = 5, PT = 7.5

1In figure 3.48, point A is on the bisector of ∠XYZ. If AX = 2 cm then find AZ.

Fig. 3.48

Seg AX ⊥ ray YX and seg AZ ⊥ ray YZ, so AX and AZ are the distances of A from the sides of ∠XYZ.

Every point on the bisector of an angle is equidistant from the sides of the angle.

∴ AZ = AX = 2 cm

∴ AZ = 2 cm

2In figure 3.49, ∠RST = 56°, seg PT ⊥ ray ST, seg PR ⊥ ray SR and seg PR ≅ seg PT. Find the measure of ∠RSP. State the reason for your answer.

Fig. 3.49

Point P is equidistant from the sides of ∠RST, since PR = PT and both are perpendicular distances.

Any point equidistant from the sides of an angle is on the bisector of the angle.
∴ Ray SP bisects ∠RST.

∴ ∠RSP = ½∠RST = ½ × 56° = 28°

∴ ∠RSP = 28°

3In ΔPQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.

The angle opposite to the greater side is greater.

The greatest side is QR = 12 cm; the angle opposite to it is ∠P.

The smallest side is PR = 8 cm; the angle opposite to it is ∠Q.

∴ Greatest angle: ∠P; smallest angle: ∠Q

4In ΔFAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.

∠N = 180° − (80° + 40°) = 60°

So ∠F = 80° is the greatest angle and ∠A = 40° is the smallest angle.

The side opposite to the greater angle is greater.
Side opposite to ∠F is AN; side opposite to ∠A is FN.

∴ Greatest side: AN; smallest side: FN

5Prove that an equilateral triangle is equiangular.

Given: ΔABC is equilateral, that is, AB = BC = CA.

To prove: ∠A = ∠B = ∠C

In ΔABC, seg AB ≅ seg AC ...... given
∴ ∠C ≅ ∠B ...... (I) isosceles triangle theorem

Seg AB ≅ seg BC ...... given
∴ ∠C ≅ ∠A ...... (II) isosceles triangle theorem

From (I) and (II), ∠A ≅ ∠B ≅ ∠C

∴ ΔABC is equiangular.

6Prove that, if the bisector of ∠BAC of ΔABC is perpendicular to side BC, then ΔABC is an isosceles triangle.

Given: In ΔABC, ray AD bisects ∠BAC and meets side BC at D; seg AD ⊥ side BC.

To prove: ΔABC is isosceles.

In ΔABD and ΔACD,

∠BAD ≅ ∠CAD ...... ray AD bisects ∠BAC

seg AD ≅ seg AD ...... common side

∠ADB ≅ ∠ADC ...... each is a right angle

∴ ΔABD ≅ ΔACD ...... A-S-A test

∴ seg AB ≅ seg AC ...... c.s.c.t.

∴ ΔABC is an isosceles triangle.

7In figure 3.50, if seg PR ≅ seg PQ, show that seg PS > seg PQ.

Fig. 3.50 — Q − R − S

Given: seg PR ≅ seg PQ; Q − R − S

To prove: seg PS > seg PQ

In ΔPQR, seg PQ ≅ seg PR ...... given
∴ ∠PQR ≅ ∠PRQ ...... (I) isosceles triangle theorem

∠PRQ is an exterior angle of ΔPRS.
∴ ∠PRQ > ∠PSR ...... exterior angle property

∴ ∠PQR > ∠PSR ...... from (I)
that is, in ΔPQS, ∠PQS > ∠PSQ

The side opposite to the greater angle is greater.
∴ PS > PQ

8In figure 3.51, in ΔABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD ≅ seg BE.

Fig. 3.51

Given: Seg AD ⊥ side BC, seg BE ⊥ side AC, AE = BD

To prove: seg AD ≅ seg BE

In ΔAEB and ΔBDA,

∠AEB ≅ ∠BDA ...... each is a right angle

hypotenuse AB ≅ hypotenuse BA ...... common side

seg AE ≅ seg BD ...... given

∴ ΔAEB ≅ ΔBDA ...... hypotenuse-side test

∴ seg BE ≅ seg AD ...... c.s.c.t.

1If ΔXYZ ∼ ΔLMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.

Corresponding angles: ∠X ≅ ∠L, ∠Y ≅ ∠M, ∠Z ≅ ∠N

Ratios of corresponding sides: XY/LM = YZ/MN = XZ/LN

2In ΔXYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm. If ΔXYZ ∼ ΔPQR and PQ = 8 cm then find the lengths of remaining sides of ΔPQR.

ΔXYZ ∼ ΔPQR ∴ XY/PQ = YZ/QR = XZ/PR

4/8 = 6/QR = 5/PR

4/8 = 6/QR ∴ QR = (6 × 8)/4 = 12 cm

4/8 = 5/PR ∴ PR = (5 × 8)/4 = 10 cm

∴ QR = 12 cm, PR = 10 cm

3Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same signs. Show the lengths of corresponding sides by numbers in proportion.

ΔABC ∼ ΔPQR

In the sketch, ∠A ≅ ∠P, ∠B ≅ ∠Q = 90° and ∠C ≅ ∠R (shown by the same signs).

AB/PQ = 3/6, BC/QR = 4/8, AC/PR = 5/10, and each ratio equals ½. The corresponding sides are in proportion, so ΔABC ∼ ΔPQR.

Any pair of triangles with equal angles will do; the sides of the second triangle must be the same multiple of the sides of the first.

1Choose the correct alternative answer for the following questions.

(i) If two sides of a triangle are 5 cm and 1.5 cm, the length of its third side cannot be — (D) 3.4 cm
The third side must be greater than 5 − 1.5 = 3.5 cm and less than 5 + 1.5 = 6.5 cm. 3.4 cm is less than 3.5 cm.
(ii) In ΔPQR, if ∠R > ∠Q then — (B) PQ > PR
The side opposite to ∠R is PQ and the side opposite to ∠Q is PR; the side opposite to the greater angle is greater.
(iii) In ΔTPQ, ∠T = 65°, ∠P = 95°. Which of the following is a true statement? — (B) PQ < TQ
∠Q = 20°. The sides opposite to ∠Q, ∠T and ∠P are TP, PQ and TQ, so TP < PQ < TQ. Only option (B) agrees with this.

2ΔABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

D is the midpoint of AC, E is the midpoint of AB

Given: AB = AC; seg BD and seg CE are medians.

To prove: BD = CE

AD = ½ AC and AE = ½ AB ...... D and E are midpoints

But AB = AC ...... given ∴ AD = AE

In ΔABD and ΔACE,

seg AB ≅ seg AC ...... given

∠BAD ≅ ∠CAE ...... common angle

seg AD ≅ seg AE ...... proved above

∴ ΔABD ≅ ΔACE ...... S-A-S test

∴ BD = CE ...... c.s.c.t.

3In ΔPQR, if PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.

Given: PQ > PR; rays QS and RS bisect ∠Q and ∠R.

To prove: SQ > SR

In ΔPQR, PQ > PR ...... given
∴ ∠PRQ > ∠PQR ...... angle opposite to the greater side is greater

∴ ½∠PRQ > ½∠PQR

∴ ∠SRQ > ∠SQR ...... rays RS and QS are bisectors

In ΔSQR, the side opposite to the greater angle is greater.
∴ SQ > SR

4In figure 3.59, point D and E are on side BC of ΔABC, such that BD = CE and AD = AE. Show that ΔABD ≅ ΔACE.

Fig. 3.59

Given: BD = CE and AD = AE

To prove: ΔABD ≅ ΔACE

In ΔADE, seg AD ≅ seg AE ...... given
∴ ∠ADE ≅ ∠AED ...... (I) isosceles triangle theorem

∠ADB = 180° − ∠ADE and ∠AEC = 180° − ∠AED ...... angles in linear pairs
∴ ∠ADB ≅ ∠AEC ...... from (I)

In ΔABD and ΔACE,
seg BD ≅ seg CE ...... given
∠ADB ≅ ∠AEC ...... proved above
seg AD ≅ seg AE ...... given

∴ ΔABD ≅ ΔACE ...... S-A-S test

5In figure 3.60, point S is any point on side QR of ΔPQR. Prove that PQ + QR + RP > 2PS.

Fig. 3.60

Given: S is a point on side QR of ΔPQR.

To prove: PQ + QR + RP > 2PS

In ΔPQS, PQ + QS > PS ...... (I) sum of two sides of a triangle is greater than the third side

In ΔPSR, PR + SR > PS ...... (II)

Adding (I) and (II),
PQ + QS + SR + PR > 2PS

But QS + SR = QR ...... Q − S − R
∴ PQ + QR + RP > 2PS

6In figure 3.61, bisector of ∠BAC intersects side BC at point D. Prove that AB > BD.

Fig. 3.61

Given: Ray AD bisects ∠BAC; B − D − C

To prove: AB > BD

∠ADB is an exterior angle of ΔADC.
∴ ∠ADB > ∠DAC ...... exterior angle property

But ∠DAC = ∠BAD ...... ray AD bisects ∠BAC
∴ ∠ADB > ∠BAD

In ΔABD, the side opposite to ∠ADB is AB and the side opposite to ∠BAD is BD.
∴ AB > BD ...... side opposite to the greater angle is greater

7In figure 3.62, seg PT is the bisector of ∠QPR. A line parallel to seg PT and passing through R intersects ray QP at point S. Prove that PS = PR.

Fig. 3.62

Given: Seg PT bisects ∠QPR; line RS ∥ seg PT; S is on ray QP.

To prove: PS = PR

Seg PT ∥ seg SR and line QS is a transversal.
∴ ∠QPT ≅ ∠PSR ...... (I) corresponding angles

Seg PT ∥ seg SR and seg PR is a transversal.
∴ ∠TPR ≅ ∠PRS ...... (II) alternate angles

∠QPT ≅ ∠TPR ...... (III) seg PT bisects ∠QPR

From (I), (II) and (III), ∠PSR ≅ ∠PRS

In ΔPSR, the sides opposite to these angles are congruent.
∴ PS = PR ...... converse of isosceles triangle theorem

8In figure 3.63, seg AD ⊥ seg BC, seg AE is the bisector of ∠CAB and C − E − D. Prove that ∠DAE = ½(∠C − ∠B).

Fig. 3.63 — as drawn, the order of points on BC is C − D − E − B

Given: Seg AD ⊥ seg BC; seg AE bisects ∠CAB.

To prove: ∠DAE = ½(∠C − ∠B)

In ΔADC, ∠ADC = 90°
∴ ∠CAD = 90° − ∠C ...... (I)

∠CAE = ½∠A ...... seg AE bisects ∠CAB
= ½(180° − ∠B − ∠C) ...... sum of the angles of ΔABC
= 90° − ½∠B − ½∠C ...... (II)

∠DAE = ∠CAE − ∠CAD ...... D lies between C and E
= (90° − ½∠B − ½∠C) − (90° − ∠C) ...... from (I) and (II)
= ½∠C − ½∠B

∴ ∠DAE = ½(∠C − ∠B)

The question says C − E − D, but figure 3.63 shows D between C and E (C − D − E). The proof follows the figure. If E were between C and D, ∠B would be greater than ∠C and the result would read ½(∠B − ∠C).

3. Triangles

Practice Set 3.1

Practice Set 3.2

Practice Set 3.3

Practice Set 3.4

Practice Set 3.5

Problem Set 3