Chapter 3 – Triangles
Practice set 3.1
1. In figure 3.8, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.
Given:
∠ACD is an exterior angle of ∆ABC
∠B = 40°
∠A = 70°
To find:
∠ACD
Solution:
∠ACD = ∠A + ∠B …[Remote Interior angles theorem]
∴ ∠ACD = 70 + 40 …[Given]
∴ ∠ACD = 110°
Ans: The measure of ∠ACD is 110⁰.
2. In ∆PQR, ∠P = 70°, ∠Q = 65° then find ∠R.
Given:
In ∆PQR,
∠P = 70°
∠Q = 65°
To find:
∠R
Solution:
In ∆PQR,
∠P + ∠Q + ∠R = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ 70 + 65 + ∠R = 180
∴ 135 + ∠R = 180
∴ ∠R = 180 – 135
∴ ∠R = 45º
Ans: The measure of ∠R is 45⁰.
3. The measures of angles of a triangle are x°, (x – 20)°, (x – 40)°. Find the measure of each angle.
Given:
Angles of a triangle are x°, (x – 20)°, (x – 40)
To find:
Measure of each angle
Solution:
xº + (x – 20)º + (x – 40)º = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ x + x – 20 + x – 40 = 180
∴ 3x – 60 = 180
∴ 3x = 180 + 60
∴ 3x = 240
∴ x = \(\large \frac {240}{3}\)
∴ x = 80
∴ Measure of first angle = x = 80º
Measure of second angle
= (x – 20)º
= (80 – 20)º
= 60º
Measure of third angle
= (x – 40)º
= (80 – 40)º
= 40º
Ans: The measures of angles of the triangles are 80º, 60º, 40º.
4. The 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.
To find:
Measures of three angles of a triangle
Solution:
Let the measure of the smallest angle of the triangle be xº
∴ Measure of other two angles will be 2xº and 3xº
x + 2x + 3x = 180 …[Sum of the measures of all angles of triangle is 180º]
∴ 6x = 180
∴ x = \(\large \frac {180}{6}\)
∴ x = 30
∴ Measure of the first angle = x = 30º
Measure of the second angle
= 2x
= 2 × 30
= 60º
Measure of the third angle
= 3x
= 3 × 30
= 90º
Ans: The measures of angles of the triangle are 30º, 60º, 90º.
5. In figure 3.9, measures of some angles are given. Using the measures find the values of x, y, z.
Given:
∠TEN = 100⁰
∠EMR = 140⁰
To find:
x, y, z
Solution:
∠NEM + ∠TEN = 180º …[Angles in linear pair]
∴ y + 100 = 180
∴ y = 180 – 100
∴ y = 80º
∠EMN + ∠EMR = 180º …[Angles in linear pair]
∴ z + 140 = 180
∴ z = 180 – 140
∴ z = 40º
In ∆ENM,
∠ENM + ∠NEM + ∠EMN = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ x + y + z = 180
∴ x + 80 + 40 = 180
∴ x = 180 – 120
∴ x = 60º
Ans: The measure of x, y, and z is 60º, 80º and 40º respectively.
6. In figure 3.10, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.
Given:
line AB || line DE
∠BAD = 70⁰
∠RED = 40⁰
To find:
∠DRE
∠ARE
Solution:
Ray AB || ray DE and AD is the transversal …[Given]
∠BAD ≅ ∠ADE …[Alternate angles theorem]
∵ ∠BAD = 70º
∴ ∠ADE = 70º
i.e. ∠RDE = 70º …(i) [A – R – D]
In ∆RDE,
∠DRE + ∠RDE + ∠DER = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ ∠DRE + 70 + 40 = 180 …[From (i) and Given]
∴ ∠DRE = 180 – 110
∴ ∠DRE = 70º
∴ ∠ARE + ∠DRE = 180º …[Angles in linear pair]
∴ ∠ARE + 70 = 180
∴ ∠ARE = 180 – 70
∴ ∠ARE = 110º
Ans: The measures of ∠DRE is 70º and ∠ARE is 110º.
7. In ∆ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°. Find measure of ∠AOB.
Given:
In ∆ABC,
BO is the angle bisector of ∠A
AO is the angle bisector of ∠B
∠C = 70°
To find:
∠AOB
Solution:
Let ∠CBO ≅ ∠OBA = x … (i)
and ∠CAO ≅ ∠OAB = y … (ii) [Angle bisector theorem]
In ∆ABC,
∠ABC + ∠BAC + ∠ACB = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ ∠CBO + ∠OBA + ∠CAO + ∠OAB + ∠ACB = 180º …[Angle addition property]
∴ x + x + y + y + 70 = 180 …[From (i) and (ii)]
∴ 2x + 2y = 180 – 70
∴ 2(x + y) = 110
∴ x + y = 55 … (iii)
In ∆AOB,
∠AOB + ∠OBA + ∠OAB = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ ∠AOB + x + y = 180 …[From (iii)]
∴ ∠AOB + 55 = 180
∴ ∠AOB = 180 – 55
∴ ∠AOB = 125º
Ans: The measure of ∠AOB is 125⁰.
8. In 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°.
Given:
line AB || line CD and line PQ is the transversal
Ray PT is the bisector of ∠BPQ
Ray QT is the bisector of ∠PQD
To prove:
m∠PTQ = 90°
Proof:
Let ∠BPT = ∠TPQ = x … (i)
∠PQT = ∠TQD = y … (ii) [Angle bisector theorem]
Line AB || line CD and PQ is the transversal,
∠BPQ + ∠PQD = 180º …[Interior angles theorem]
∴ ∠BPT + ∠TPQ + ∠PQT + ∠TQD = 180º …[Angle addition property]
∴ x + x + y + y = 180 …[From (i) and (ii)]
∴ 2x + 2y = 180
∴ 2(x + y) = 180
∴ x + y = 90 … (iii)
In ∆PTQ,
∠TPQ + ∠TQP + ∠PTQ = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ x + y + ∠PTQ = 180
∴ 90 + ∠PTQ = 180 …[From (iii)]
∴ ∠PTQ = 180 – 90
∴ ∠PTQ = 90º
Hence proved.
9. Using the information in figure 3.12, find the measures of ∠a, ∠b and ∠c.
Given:
Two angles each of measure 70⁰ and 100⁰
To find:
∠a, ∠b and ∠c
Solution:
∠b = 70º …[Vertically Opposite angles]
∠c + 100 = 180 …[Angles in linear pair]
∴ ∠c = 180 – 100
∴ c = 180 – 100
∴ ∠c = 80º …(ii)
In the given triangle,
∠a + ∠b + ∠c = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ ∠a + 70 + 80 = 180
∴ ∠a = 180 – 150
∴ ∠a = 30º
Ans: The measures of ∠a, ∠b and ∠c are 30⁰, 70⁰ and 80⁰ respectively.
10. In figure 3.13, line DE || line GF ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively.
Prove that,
(i) ∠DEG = \(\large \frac {1}{2}\) ∠EDF
(ii) EF = FG.
Given:
line DE || line GF
ray EG is the bisector of ∠DEF
ray FG is the bisector of ∠DFM
To prove:
(i) ∠DEG = \(\large \frac {1}{2}\) ∠EDF
(ii) EF = FG
Proof:
Let ∠DEG = ∠GEF = x …(i)
∠DFG = ∠GFM = y …(ii) …[Angle bisector theorem]
∠DEF = ∠DEG + ∠GEF …[Angle addition property]
∠DEF = x + x
∴ ∠DEF = 2x …(iii)
Seg ED || seg FG and DF is the transversal,
∠EDF ≅ ∠DFG …[Alternate angles theorem]
∴ ∠EDF ≅ ∠DFG = y …(iv)
seg ED || seg FG and EM is the transversal,
∠DEF ≅ ∠GFM …[Corresponding angles theorem]
∴ ∠DEF = ∠GFM = y …(v)
∴ ∠DEF = ∠EDF …[From (iii), (iv)]
∴ y = 2x …(vi)
∠GFM is an exterior angle of ∆GEF.
∴ ∠GFM = ∠GEF + ∠EGF …[Remote interior angles theorem]
∴ y = x + ∠EGF
∴ 2x = x + ∠EGF …[From iv]
∴ ∠EGF = 2x – x
∴ ∠EGF = x …(vii)
In ∆EFG,
∠GEF ≅ ∠EGF …[From (i), (vii)]
seg EF ≅ seg FG …[Converse of isosceles triangle theorem]
∴ EF = FG
Hence proved.
Practice set 3.2
1. In figure 3.48, point A is on the bisector of ∠XYZ. If AX = 2 cm then find AZ.
(i)
Ans: ∆ABC ≅ ∆PQR by SSS Test
(ii)
Ans: ∆XYZ ≅ ∆LMN by SAS Test
(iii)
Ans: ∆PRQ ≅ ∆STU by ASA Test
(iv)
Ans: ∆LMN ≅ ∆PTR by Hypotenuse Side Test
2. Observe 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)
From the information shown in the figure, in ∆ABC and ∆PQR
∠ABC ≅ ∠PQR
seg BC ≅ seg QR
∠ACB ≅ ∠PRQ
∴ ∆ABC ≅ ∆PQR …_____ test
∴ ∠BAC ≅ _____ …Corresponding angles of congruent triangles.
seg AB ≅ _____
and _____ ≅ seg PR …Corresponding sides of congruent triangles
Solution:
∠ABC ≅ ∠PQR
seg BC ≅ seg QR
∠ACB ≅ ∠PRQ
∴ ∆ABC ≅ ∆PQR … ASA test
∴ ∠BAC ≅ ∠QPR …Corresponding angles of congruent triangles.
seg AB ≅ seg PQ
and seg AC ≅ seg PR …Corresponding sides of congruent triangles
(ii)
From the information shown in the figure,
In ∆PTQ and ∆STR
seg PT ≅ seg ST
∠PTQ ≅ ∠STR . .Vertically opposite angles
seg TQ ≅ seg TR
∴ ∆PTQ ≅ ∆STR … _____ test
∴ ∠TPQ ≅ _____
and _____ ≅ ∠TRS
seg PQ ≅ _____ …Corresponding sides of congruent triangles.
Solution:
In ∆PTQ and ∆STR
seg PT ≅ seg ST
∠PTQ ≅ ∠STR . .Vertically opposite angles
seg TQ ≅ seg TR
∴ ∆PTQ ≅ ∆STR … SAS test
∴ ∠TPQ ≅ ∠TSR
and ∠TQP ≅ ∠TRS
seg PQ ≅ seg SR …Corresponding sides of congruent triangles.
3. From the information shown in the figure, state the test assuring the congruence of ∆ABC and ∆PQR. Write the remaining congruent parts of the triangles.
Given:
∠BAC = 90⁰
∠PQR = 90⁰
seg AB ≅ seg PQ
seg BC ≅ seg PR
To find:
Remaining congruent parts of the triangles.
Solution:
In ∆ABC and ∆QPR,
BAC = ∠PQR = 90º ...[Given]
Hypotenuse BC ≅ Hypotenuse PR …[Given]
side AB ≅ side QP …[Given]
∴ ∆ABC ≅ ∆QPR …[By Hypotenuse – side test]
∴ side AC ≅ side QR …[c.s.c.t.]
∠ABC ≅ ∠QPR …[c.a.c.t.]
∠ACB ≅ ∠QRP …[c.a.c.t.]
Ans: Remaining congruent parts are;
side AC ≅ side QR
∠ABC ≅ ∠QPR
∠ACB ≅ ∠QRP
4. As shown in the following figure, in ∆LMN and ∆PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.
Solution:
In ∆LMN and ∆PNM,
side LM ≅ side PN …[Given]
side LN ≅ side PM …[Given]
side MN ≅ side NM …[Common side]
∆LMN ≅ ∆PNM …[By SSS test]
∠LMN ≅ ∠PNM …[c.a.c.t.]
∠LNM ≅ ∠PMN …[c.a.c.t.]
∠NLM ≅ ∠MPN …[c.a.c.t.]
Ans: Remaining congruent parts are;
∠LMN ≅ ∠PNM
∠LNM ≅ ∠PMN
∠NLM ≅ ∠MPN
5. In figure 3.24, seg AB ≅ seg CB and seg AD ≅ seg CD. Prove that ∆ABD ≅ ∆CBD
Given:
seg AB ≅ seg CB
Seg AD ≅ seg CD
To prove:
∆ABD ≅ ∆CBD
Proof:
In ∆ABD and ∆CBD,
side AB ≅ side CB …[Given]
side AD ≅ side CD …[Given]
side BD ≅ side BD …[Common side]
∴ ∆ABD ≅ ∆CBD …[SSS test]
Hence proved.
6. In figure 3.25, ∠P ≅ ∠R, seg PQ ≅ seg RQ. Prove that, ∆PQT ≅ ∆RQS
Given:
∠P ≅ ∠R
seg PQ ≅ seg RQ
To prove:
∆PQT ≅ ∆RQS
Proof:
In ∆PQT and ∆RQS,
∠P ≅ ∠R …[Given]
side PQ ≅ side RQ …[Given]
∠Q ≅ ∠Q …[Common angle]
∴ ∆PQT ≅ ∆RQS …[By ASA test]
Hence proved.
Practice set 3.3
1. Find the values of x and y using the L information shown in figure 3.37. Find the measure of ∠ABD and m∠ACD.
Given:
side AB ≅ side AC
side DB ≅ side DC
∠ACB = 50º
∠DBC = 60º
To find:
∠ABD and ∠ACD
Solution:
In ∆ABC,
side AB ≅ side AC …[Given]
∴ ∠ABC ≅ ∠ACB …[Isosceles triangle theorem]
∵ ∠ACB = 50º
∴ ∠ABC = x = 50º
In ∆DBC,
side DB ≅ side DC …[Given]
∴ ∠DBC ≅ ∠DCB …[Isosceles triangle theorem]
∵ ∠DBC = 60º
∴ ∠DCB = y = 50º
∠ABD = ∠ABC + ∠DBC …[Angle addition property]
∴ ∠ABD = 50 + 60
∴ ∠ABD = 110º
∠ACD = ∠ACB + ∠DCB …[Angle addition property]
∴ ∠ACD = 50 + 60
∴ ∠ACD = 110º
Ans: The measure of ∠ABD is 110º and ∠ACD is 110º.
2. The length of the hypotenuse of a right angled triangle is 15. Find the length of the median of its hypotenuse.
Given:
In ∆ABC,
∠ABC = 90º
AC = 15 units
Seg BD is median on hypotenuse AC
To find:
BD
Solution:
In ∆ABC,
∠ABC = 90º …[Given]
Seg BD is median on hypotenuse AC …[Given]
∴ BD = \(\large \frac {1}{2}\) AC …[In a right angled triangle the median drawn on the hypotenuse is half of the hypotenuse]
∴ BD = \(\large \frac {1}{2}\) × 15
∴ BD = 7.5 units
Ans: The measure of BD is 7.5 units.
3. In ∆ PQR, ∠Q = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS).
Given:
In ∆ PQR,
∠Q = 90°
PQ = 12
QR = 5
QS is a median
To find:
l(QS)
Solution:
In ∆PQR,
∠PQR = 90º …[Given]
∴ PR² = PQ² + QR² …[By Pythagoras theorem]
∴ PR² = 12² + 5² …[Given]
∴ PR² = 144 + 25
∴ PR² = 169
∴ PR = 13 units …(i) [Taking square roots on both sides]
Seg QS is median on hypotenuse PR.
∴ QS = \(\large \frac {1}{2}\) PR …[In a right angled triangle the median drawn on the hypotenuse is half of the hypotenuse]
∴ QS = \(\large \frac {1}{2}\) × 13 …[From (i)]
∴ QS = 6.5 units
Ans: Length of QS is 6.5 units.
4. In figure 3.38, point G is the point of concurrence of the medians of D PQR. If GT = 2.5, find the lengths of PG and PT.
Given:
Point G is the point of concurrence of the medians of ∆PQR.
GT = 2.5
To find:
PG and PT
Solution:
In ∆PQR,
Point G is centroid …[Given]
\(\large \frac {PG}{GT}\) = \(\large \frac {2}{1}\) …[Centroid divides the median in the ratio 2:1]
∴ \(\large \frac {PG}{2.5}\) = \(\large \frac {2}{1}\)
∴ PG = 2 × 2.5
∴ PG = 5 cm
∴ PT = PG + GT …[P – G – T]
∴ PT = 2.5 + 5
∴ PT = 7.5 cm
Ans: The measure of PG is 5 cm and that of PT is 7.5 cm.
Practice set 3.4
1. In figure 3.48, point A is on the bisector of ∠XYZ. If AX = 2 cm then find AZ.
Given:
AY is the bisector of ∠XYZ
AX = 2
To find:
AZ
Solution:
AY is the bisector of ∠XYZ …[Given]
Point A is equidistant from ray YX and ray YZ …[Angle bisector theorem]
∴ AX = AZ
∵ AX = 2 cm
∴ AZ = 2 cm
Ans: The length of AZ is 2 cm.
2. In 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.
Given:
∠RST = 56°
seg PT ⊥ ray ST
seg PR ⊥ ray SR
seg PR ≅ seg PT
To find:
∠RSP
Solution:
PR = PT …[Given]
∴ Point P lies on bisector of ∠RST …[Angle bisector theorem]
∴ ∠RSP = \(\large \frac {1}{2}\) × ∠RST
∴ ∠RSP = \(\large \frac {1}{2}\) × 56
∴ ∠RSP = 28º
Ans: The measure of ∠RSP is 28⁰.
3. In ∆PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.
Given:
In ∆PQR
PQ = 10 cm
QR = 12 cm
PR = 8 cm
To find:
Greatest and the smallest angle of the triangle
Solution:
∴ QR > PQ > PR …[From the given measurements]
∴ ∠P > ∠R > ∠Q …[Angle opposite to greater side is greater]
∴ ∠P is the greatest angle and ∠Q is the smallest angle of ∆PQR
Ans: The greatest angle is ∠P and the smallest angle is ∠Q.
4. In ∆ FAN, ∠F = 80°, ∠A = 40° . Find out the greatest and the smallest side of the triangle. State the reason.
Given:
In ∆ FAN,
∠F = 80°
∠A = 40°
To find:
Greatest and the Smallest side
Solution:
In ∆FAN,
∠F + ∠A + ∠N = 180º …[Sum of the measures of all angles of a triangle is 180º]
∴ 80 + 40 + N = 180° …[Given]
∴ ∠N = 180 – 120
∴ ∠N = 60º
∴ ∠F > ∠N > ∠A
∴ AN > FA > FN …[Side opposite to greater angle is greater]
∴ Side AN is greatest side and side FN is smallest side of ∆FAN.
Ans: The greatest side is AN and the smallest side is FN.
5. Prove that an equilateral triangle is equiangular.
Given:
In ∆PQR,
PQ = QR = PR.
To prove:
∠P ≅ ∠Q ≅ ∠R
Proof:
In ∆PQR,
side PQ ≅ side PR …[Given]
∴ ∠Q ≅ ∠R …(i) [Isosceles triangle theorem]
In ∆PQR,
side PR ≅ side QR …[Given]
∴ ∠Q ≅ ∠P …(ii) [Isosceles triangle theorem]
∴ ∠P ≅ ∠Q ≅ ∠R …[From (i) and (ii)]
∴ ∆PQR is an equiangular triangle
Hence proved.
6. Prove that, if the bisector of ∠BAC of ∆ ABC is perpendicular to side BC, then ∆ ABC is an isosceles triangle.
Given:
In ∆ABC,
AD bisects ∠BAC
B – D – C
seg AD ⊥ side BC
To prove:
∆ABC is an isosceles triangle
Proof:
In ∆ADB and ∆ADC
∠BAD ≅ ∠CAD …[Given]
side AD ≅ side AD …[Common side]
∠ADB ≅ ∠ADC …[Each measures 90°]
∴ ∆ADB ≅ ∆ADC …[ASA test]
∴ side AB ≅ side AC …[c.s.c.t.]
∴ ∆ABC is an isosceles triangle …[By definition]
Hence proved.
7. In figure 3.50, if seg PR ≅ seg PQ, show that seg PS > seg PQ.
Given:
seg PR ≅ seg PQ
To prove:
seg PS > seg PQ
Proof:
In ∆PQR,
side PQ ≅ side PR …[Given]
∴ ∠PQR ≅ ∠PRQ …(i) [Isosceles triangle theorem]
∠PRQ is exterior angle of ∆PRS …[Definition]
∴ ∠PRQ > ∠S …[Exterior angle property]
∴ ∠PQR > ∠S …(ii) [From (i)]
In ∆PQS,
∠PQS > ∠S …[From (ii) and Q – R – S]
∴ PS > PQ …[In a triangle side opposite to greater angle is greater]
Hence proved.
8. In figure 3.51, in ∆ ABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD ≅ seg BE
Given:
In ∆ ABC,
seg AD and seg BE are altitudes
AE = BD
To prove:
seg AD ≅ seg BE
Proof:
In ∆AEB and ∆BDA,
∠AEB = ∠BDA = 90° …[Given]
Hypotenuse AB ≅ Hypotenuse BA
…[Common side]
seg AE ≅ seg BD …[Given]
∴ ∆AEB ≅ ∆BDA …[Hypotenuse side test]
∴ seg BE ≅ seg AD …[c.s.c.t.]
i.e. seg AD ≅ seg BE
Hence proved.
Practice set 3.5
1. If ∆ XYZ ~ ∆ LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.
Solution:
∆XYZ ~ ∆LMN …[Given]
∴ \(\large \frac {XY}{LM}\) = \(\large \frac {YZ}{MN}\) = \(\large \frac {XZ}{LN}\) …[c.s.s.t.]
∠X ≅ ∠L …[c.a.s.t]
∠Y ≅ ∠M …[c.a.s.t]
∠Z ≅ ∠N …[c.a.s.t]
Ans: The ratio is \(\large \frac {XY}{LM}\) = \(\large \frac {YZ}{MN}\) = \(\large \frac {XZ}{LN}\) and corresponding angles of the triangles are ∠X ≅ ∠L, ∠Y ≅ ∠M and ∠Z ≅ ∠N.
2. In ∆ XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If ∆ XYZ ~ ∆ PQR and PQ = 8 cm then find the lengths of remaining sides of ∆ PQR.
Solution:
∆XYZ ~ ∆PQR …[Given]
∴ \(\large \frac {XY}{PQ}\) = \(\large \frac {YZ}{QR}\) = \(\large \frac {XZ}{PR}\) …[c.s.s.t.]
∴ \(\large \frac {4}{8}\) = \(\large \frac {6}{QR}\) = \(\large \frac {5}{PR}\)
∴ \(\large \frac {4}{8}\) = \(\large \frac {6}{QR}\)
∴ 4 × QR = 6 × 8
∴ 4 × QR = 48
∴ QR = \(\large \frac {48}{4}\)
∴ QR = 12
∴ \(\large \frac {4}{8}\) = \(\large \frac {5}{PR}\)
∴ 4 × PR = 5 × 8
∴ 4 × PR = 40
∴ PR = \(\large \frac {40}{4}\)
∴ PR = 10
Ans: The length of QR is 12 cm and PR is 10 cm.
3. Draw 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.
Solution:
∆ABC ~ ∆DEF
∴ \(\large \frac {AB}{DE}\) = \(\large \frac {BC}{EF}\) = \(\large \frac {AC}{DF}\) …[c.s.s.t.]
∠A ≅ ∠D …[c.a.s.t]
∠B ≅ ∠E …[c.a.s.t]
∠C ≅ ∠F …[c.a.s.t]
Ans: The ratio is \(\large \frac {AB}{DE}\) = \(\large \frac {BC}{EF}\) = \(\large \frac {AC}{DF}\) and corresponding angles of the triangles are ∠A ≅ ∠D, ∠B ≅ ∠E and ∠C ≅ ∠F.
Problem Set 3
1. Choose 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 ______
(A) 3.7 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm
Ans: Option (D) : 3.4 cm
Solution:
If the length of the third side is 3.4 cm, then, 3.4 + 1.5 = 4.9 cm < 5 cm
But, we know that, in a triangle, the sum of the length of two sides should be greater than the third side.
Thus, the length of the third side of the triangle cannot be 3.4 cm.
(ii) In ∆ PQR, If ∠R > ∠Q then ______
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR
Ans: Option (B) : PQ > PR
Explanation: Side opposite to greater angle is longer.
(iii) In ∆ TPQ, ∠T = 65°, ∠P = 95° which of the following is a true statement ?
(A) PQ < TP
(B) PQ < TQ
(C) TQ < TP < PQ
(D) PQ < TP < TQ
Ans: Option (B) : PQ < TQ
Solution:
∠Q = 180° – (95° + 65°)
∴ ∠Q = 20°
∴ ∠Q < ∠T < ∠P
∴ PT < PQ < TQ
2. ∆ ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.
Given:
∆ ABC is an isosceles triangle.
AB = AC
Seg BD and seg CE are medians
To prove:
BD = CE
Proof:
AE = BE = \(\large \frac {1}{2}\) AB …(i) [CE is median on side AB]
AD = CD = \(\large \frac {1}{2}\) AC …(ii) [BD is median on side AC]
∵ AB = AC …(iii) [Given]
∴ AE = BE = AD = CD …(iv) [From (i), (ii) and (iii)]
In ∆ABD and ∆ACE,
side AB ≅ side AC …[Given]
∠A ≅ ∠A …[Common angle]
side AD ≅ side AE …[From (iv)]
∆ABD ≅ ∆ACE …[SAS test]
∴ side BD ≅ side CE …[c.s.c.t.]
i.e. BD = CE
Hence proved.
3. In ∆ PQR, If PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.
Given:
In ∆ PQR,
PQ > PR and bisectors of ∠Q and ∠R intersect at S.
To prove:
SQ > SR
Proof:
In ∆PQR,
PQ > PR …[Given]
∴ ∠PRQ > ∠PQR …[Angle opposite to greater side is greater]
∴ 2∠SRQ > 2∠SQR …[Rays QS and RS bisect ∠PQR and ∠PRQ respectively]
∴ ∠SRQ > ∠SQR …(i)
In ∆SQR,
∠SRQ > ∠SQR …[From (i)]
∴ SQ > SR …[Side opposite to greater angle is greater]
Hence proved.
4. In figure 3.59, point D and E are on side BC of ∆ ABC, such that BD = CE and AD = AE. Show that ∆ ABD ≅ ∆ ACE.
Given:
BD = CE
AD = AE
To prove:
∆ ABD ≅ ∆ ACE
Proof:
In ∆ADE,
side AD ≅ side AE …[Given]
∴ ∠ADE ≅ ∠AED …(i) [Isosceles triangle Theorem]
∠ADB + ∠ADE = 180° …[Angles in Linear pair]
∴ ∠ADB = 180° – ∠ADE …(ii)
∴ ∠AEC + ∠AED = 180° …[Angles in Linear pair]
∴ ∠AEC = 180⁰ – ∠AED …(iii)
∴ ∠ADB ≅ ∠AEC …(iv) [From (i), (ii), (iii)]
In ∆ABD and ∆AEC,
side AD ≅ side AE …[Given]
∠ADB ≅ ∠AEC …[From (iv)]
side BD ≅ side CE …[Given]
∴ ∆ABD ≅ ∆ACE …[SAS test]
Hence proved.
5. In figure 3.60, point S is any point on side QR of ∆ PQR. Prove that : PQ + QR + RP > 2PS
Given:
In ∆PQS,
PQ + QS > PS … (i) [Sum of two sides of a triangle is greater than the third side]
To prove:
PQ + QR + RP > 2PS
Proof:
In ∆PRS,
PR + SR > PS … (ii) [Sum of two sides of a triangle is greater than the third side]
∴ PQ + QS + SR + PR > 2PS …[Adding (i) and (ii)]
∴ PQ + QR + RP > 2PS …[Q – S – R]
Hence proved.
6. In figure 3.61, bisector of ∠BAC intersects side BC at point D. Prove that AB > BD.
Given:
AD intersects side BC at point D
To prove:
AB > BD
Proof:
∠BAD ≅ ∠DAC … (i) [AD bisects ∠BAC]
∠ADB is an exterior angle of ∆ADC
∴ ∠ADB > ∠DAC …[Exterior angle property]
∴ ∠ADB > ∠BAD …. (ii) [From (i)]
In ∆ABD
∠ADB > ∠BAD ...[From (ii)]
∴ AB > BD …[In a triangle, side opposite to greater angle is greater]
Hence proved.
7. In 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.
Given:
Seg PT is the bisector of ∠QPR.
Line PT || line SR
SR intersects ray QP at point S.
To Prove:
PS = PR
Proof:
In DPQR,
ray PT bisects ∠QPR …[Given]
∴ ∠QPT ≅ ∠TPR … (i)
Line PT || line RS and QS is the transversal,
∠QPT ≅ ∠PSR … (ii) [Corresponding angles theorem]
Line PT || line RS and PR is the transversal,
∠TPR ≅ ∠PRS … (iii) [Alternate angles theorem]
In ∆PSR,
∠PSR ≅ ∠PRS …[From (i), (ii), (ii)] [Converse of isosceles triangle theorem]
∴ side PS ≅ side PR
∴ PS = PR
Hence proved.
8. In figure 3.63, seg AD ⊥ seg BC. seg AE is the bisector of ∠CAB and C – E – D. Prove that ∠DAE = \(\large \frac {1}{2}\)(∠B – ∠C)
Given:
seg AD ⊥ seg BC
seg AE is the bisector of ∠CAB
C – E – D
To prove:
∠DAE = \(\large \frac {1}{2}\)(∠B – ∠C)
Proof:
∠BAE ≅ ∠CAE …(i) [Ray AE bisects ∠BAC]
∠BAD + ∠DAE = ∠BAE …[Angle addition property]
∴ ∠DAE = ∠BAE – ∠BAD …(ii)
∠CAE + ∠DAE = ∠CAD …[Angle addition property]
∴ ∠DAE = ∠CAD – ∠CAE ….(iii)
Now,
∠DAE + ∠DAE = ∠BAE – ∠BAD + ∠CAD – ∠CAE …[Adding (ii) and (ii)]
∴ 2∠DAE = ∠BAE – ∠BAD + ∠CAD – ∠BAE …[From (i)]
∴ 2∠DAE = ∠CAD – ∠BAD …(iv)
In ∆ BAD,
∠BAD + ∠ADB + ∠ABD = 180° …[Sum of the measures of all angles of a triangle is 180°]
∴ ∠BAD + 90 + ∠B = 180
∴ ∠BAD = 180 – 90 – ∠B
∴ ∠BAD = 90 – ∠B …(v)
In ∆DAC,
∠CAD + ∠ADC + ∠ACD = 180° …[Sum of the measures of all angles of a triangle is 180°]
∴ ∠CAD + 90 + ∠C = 180
∴ ∠CAD = 180 – 90 – ∠C
∴ ∠CAD = 90 – ∠C …(vi)
Now,
2 ∠DAE = (90 – ∠C) – (90 – ∠B) … [From (iv), (v), (vi)]
∴ 2 ∠DAE = 90 – ∠C – 90 + ∠B
∴ 2 ∠DAE = ∠B – ∠C
∴ ∠DAE = \(\large \frac {1}{2}\) (∠B – ∠C)
Hence proved.