Maharashtra Board Class 9 Math Part 2 Solution Chapter 3 – Triangles
Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 3: Triangles. Marathi or Math Part 2 Medium Students of Class 9 get here Triangles full Exercise Solution.
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Std |
Maharashtra Class 9 |
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Subject |
Math Part 2 Solution |
| Chapter |
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.
Solution:
Given, ∠A = 70°, ∠B =40°
We know, Exterior angle is equal to the sum of remote interior angles.
∴ ∠A + ∠B = ∠ACD
∠ACD = 70° + 40° = 130°
(2) In △PQR, ∠P = 70°, ∠Q = 65° then find ∠R.
Solution:
Sum of angles of a triangle is 180°
∠P + ∠Q + ∠R = 180°
= 70° + 65° + ∠R = 180°
= ∠R = 45°
(3) The measures of angles of a triangle are x°, (x-20)°, (x-40)°.
Find the measure of each angle.
Solution:
Sum of angles of a triangle is 180°
∴ x° + (x – 20)° + (x – 40)° = 180°
= 3x – 60° = 180°
= 3x = 120°
= x = 40°
∴ The angles are: 40°, 20°, 0°
(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.
Solution:
Let, the smallest angle be x°
A/Q, other angles are 2x° and 3x°
Sum of angles of a triangle is 180°
∴ x° + 2x° + 3x° = 180°
= 6x° = 180°
= x = 30°
∴ The angles are: 30°, 60°, 90°
(5) In figure 3.9, measures of some angles are given. Using the measures find the values of x, y, z.
Solution:
Given, ∠TMR = 140°
∵ NR is a straight ray
∴ ∠NME + ∠EMR = 180°
= ∠NME = 180° – 140° = 40°
= 3 = 40°
Also, TM is a straight ray
∴ ∠TEN + ∠NEM = 180°
= ∠NEM = 180° – 100° = 80°
= y = 80°
Sum of angles of a triangle is 180°
∴ x + y + 3 = 180°
= x = 180° – 80° – 40°
= 60°
(6) In figure 3.10, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.
Solution:-
∵ AB || DE and AD transversal
∴ ∠BAD = ∠ADE (Alternate angles)
= ∠ADE = 70°
Sum of angles of △DRE is 180°
∴ ∠D + ∠E + ∠DRE = 180°
= 70° + 40° + ∠DRE = 180°
= ∠DRE = 70°
∵ ARD is straight line
∴ ∠ARE + ∠ERD = 180°
=> ∠ARE = 180° – 70° = 110°
(7) In △ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°. Find measure of ∠AOB.
Solution:
In △ABC,
∠A + ∠B + ∠C = 180°
= 2 ∠OAB + 2 ∠OBA + 70° = 180° (∵ AO and BO are angle bisectors)
= ∠OAB + ∠OBA = 180°-70°/2
= 55°
In △AOB,
∠AOB + ∠OAB + ∠BOA = 180°
= ∠AOB + 55° = 180°
= ∠AOB = 125°
(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 ∠PTQ = 90°.
Solution:
∵ AB || CD and PQ transversal
∴ ∠BPQ + ∠PQD = 180° (Angles on same side of transversal)
= 2 ∠TPQ + 2∠PQT = 180° (∴ PT and QT are angle bisectors)
= ∠TPQ + ∠PQT = 90°
Sum of angles of △PTQ is 180°
∴ ∠PTQ + ∠TPQ + ∠PQT = 180°
= ∠PTQ + 90° = 180°
= ∠PTQ = 90°
(9) Using the information in figure 3.12, find the measures of ∠a, ∠b and ∠c.
Solution:
In figure, ∠DBF = ∠ABC (Vertically opposite angles)
B = 70°
∵ BCE is a straight line
∴ ∠c + 100° = 180°
=> C = 80°
In △ABC, a + b + c = 180°
=> a + 70° + 80° = 180°
=> a = 30°
(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 = ½ ∠EDF (ii) EF = FG.
Solution:
∵ ED || FG and DF transversal
∴ ∠EDF = ∠DFG (Alternate angles) —– (1)
(i) Again, Exterior angle is equal to sum of remote interior angles of a triangle.
∴ ∠DEF + ∠EDF = ∠DFM
= ½ DEF + 1/2 ∠EDF = ½ ∠DFM
= ∠DEG + ½ ∠EDF = ∠DFG (∵ EG & FG are angle bisectors)
= ∠DEG = ∠DFG – 1/2 ∠DFG (By (i))
= ½ ∠DFG = 1/2 ∠EDF (By (i))
(ii) Also, ED || FG and EG transversal
∴ ∠DEG = ∠EGF (Alternate angles)
∵ EG is angle bisector
∴ ∠DEG = ∠GEF
∴ ∠EGF = ∠GEF
∴ △GEF is an isosceles triangle
∴ EF = GF
Practice Set 3.2
(1) In 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) By . . . . . . . . . . test
△ABC ≅ △PQR
Ans:
(i) S-S-S
(ii) By . . . . . . . . . . test
△XYZ ≅ △LMN
Ans:
(ii) S-A-S
(iii) By . . . . . . . . . . test
△PRQ ≅ △STU
Ans:
A-S-A
(iv) By . . . . . . . . . . test
△LMN ≅ △PTR
Ans:
(iv) Hypotenuse
(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 (…. sides of congruent triangles)
Solution:
△ABC ≅ △PQR (A-S-A list)
∴ ∠BAC = ∠QPR (Corresponding angles of congruent triangles)
seg AB = seg PQ (corresponding sides of congruent triangles)
Seg AC = seg PR
(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 (Corresponding ….. angles of congruent triangles)
Seg PQ ≅ —- (corresponding sides of congruent triangles)
Solution:
△PQT ≅ △SRT (S-A-S test)
{∵ PT = TS, QT = TR and ∠PTQ = ∠STR (Vertically opposite angles}
∠PQT = ∠SRT (Corresponding angles of congruent triangle)
∠TPQ = ∠TSR
Seg PQ = Seg SR (Corresponding sides of congruent triangle)
(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.
Solution:
In △ABC and △PQR,
BC = PR and AB = PQ
∠BAC = ∠RQP
∴ △ABC ≅ △PQR [S-A-S test]
AC = QR
∠ABC = ∠QPR and ∠ACB = ∠QRP (Corresponding angles & sides of congruent △S)
(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,
LM = PN, LN = PM (Given)
MN = NM (common)
∴ △LMN ≅ △PNM (S-S-S test)
∠LMN = ∠PNM
∠MLN = ∠NPM
∠LNM = ∠PMN (Corresponding angles of congruent triangle)
(5) In figure 3.24, seg AB ≅ seg CB and seg AD ≅ seg CD.
Prove that
△ABD ≅ △CBD
Solution:
In △ABD and △CBD,
AD = CD, AB = CB (given)
BD = BD (Common)
∴ △ABD ≅ △CBD [S-S-S test]
(6) In figure 3.25, ∠P ≅ ∠R
seg PQ ≅ seg RQ
Prove that,
△PQT ≅ △RQS
Solution:
In △PQT and △RQS
∠P = ∠R, PQ = RQ (Given)
∠Q is common
∴ △PQT ≅ △RQS (A-S-A test)
Practice Set 3.3
(1) Find the values of x and y using the information shown in figure 3.37.
Find the measure of ∠ABD and m ∠ACD.
Solution:
In △ABC, AB = AC (Given)
∴ ∠ABC = ∠ACB (Angles opposite to equal sides are equal)
∴ x = 50°
In △BCD, BD = CD (given)
∴ ∠DBC = ∠DCB (Angles opposite to equal sides are equal)
= y = 60°
∴ m ∠ABD = ∠ABC + ∠DBC
= x + 60°
= 50 + 60° = 110°
m ∠ACD = ∠ACB + ∠DCB
= y + 50° = 60° + 50° = 110°
(2) The length of hypotenuse of a right angled triangle is 15. Find the length of median of its hypotenuse.
Solution:-
We know length of median on hypotenuse is half the length of the hypotenuse.
∵ Hypotenuse = 15
Median = 15/2 = 7.5
(3) In △PQR, ∠Q = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS).
Solution:
∵ PQR is a right angled triangle and PR is a hypotenuse and QS is median on the hypotenuse.
∴ l(QS) = ½ × hypotenuse
= ½ × 12 = 6
(4) In figure 3.38, point G is the point of concurrence of the medians of △PQR. If GT = 2.5, find the lengths of PG and PT.
Solution:
∵ PT is median of △PQR and G is a point of concurrence of the medians.
∴ PG : GT = 2:1
∴ PG = 2×2.5
= 5
∴ PT = PG + GT
= 5+2.5
= 7.5
Practice Set 3.4
(1) In figure 3.48, point A is on the bisector of ∠XYZ. If AX = 2 cm then find AZ.
Solution:
∵ A is a point on the bisector of ∠XYZ
∴ Lines draw from A on both sides of the bisector are equal.
∴ AX = AZ
∴ AZ = 2cm
(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.
Solution:
∵ Seg PR ≅ Seg PT
∴ SP is an angle bisector of ∠TSR
∴ ∠RSP = 1/2 ∠RST = ½ × 56° = 28°
(Lines drawn from P are equal then the lines is an angle bisector)
(3) In △PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.
Solution:
Angle opposite to longer side is greatest and opposite to smaller side is smallest.
∴ ∠P is the greatest angle and ∠Q is the smallest angle.
(4) In △FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.
Solution:
∵ ∠F = 80°, ∠A = 40°
∴ ∠N = 180° – ∠F – ∠A
= 60°
Sides opposite to greatest angle is greatest and opposite to smallest angle is greatest.
∴ AN is the greatest side and FN is the smallest side.
(5) Prove that an equilateral triangle is equiangular.
Solution:
In △ABC,
AB = AC
∴ ∠C = ∠B (Angles opposite to equal sides)
AC = BC
∠B = ∠A (Angles opposite to equal sides)
AB = BC
∠C = ∠A (Angles opposite to equal sides)
∴ ∠A = ∠B = ∠C
(6) Prove that, if the bisector of ∠BAC of △ABC is perpendicular to side BC, then △ABC is an isosceles triangle.
Solution:
Let, AD be the angle bisector of ∠A of △ABC.
In △ABD and △ABD,
∠BAD = ∠DAC (∵AD is bisector)
∠ADC = ∠ADB = 90°
AD is common
∴ △ABD ≅ △ACD (A-S-A test)
∴ AB = AC and ∠ABD = ∠ACD (Corresponding sides & angles of congruent triangles)
∴ ABC is an isosceles triangle.
(7) In figure 3.50, if seg PR ≅ seg PQ, show that seg PS > seg PQ.
Solution:
In △PRS, ∠PRS is an obtuse angle
Hence, ∠PRS is the greatest angle
Side opposite to ∠PRS is PS. which is the greatest side
∴ Seg PS > seg PR = seg PQ
∴ Seg PS > seg PQ
(8) In figure 3.51, in △ABC, seg AD and seg BE are altitudes and AE = BD.
Prove that seg AD ≅ seg BE
Solution:
In △ABD and △ABE,
AE = BD and ∠AEB = ∠ADB = 90°, AB is common
∴ △ABD ≅ △ABE (S-A-S test)
∴ Seg AD ≅ Seg BE (Corresponding side of congruent triangle)
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:
Corresponding angle are:
∠X = ∠L, ∠Y = ∠M, ∠Z = ∠N
Rations of corresponding sides are:
XY/LM = YZ/MN = XZ/LN
(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
∴ XY/PQ = YZ/QR = XZ/PR
= 4/8 = 6/QR = 5/PR
∴ QR = 6 × 8/4 = 12
PR = 5 × 8/4 = 10
(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:
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: (D) 3.4 cm.
(ii) In △PQR, If ∠R > ∠Q then . . . . .
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR
Ans: (B) PQ > PR.
(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: (B) PQ < TQ.
(2) △ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.
Solution:
In △ABC, BD and CE are medians
∴ AE = BE = ½ AB and AD = DC = ½ AC
Now, In △BEC and △BDC,
AB = AC
= ½ AB = ½ AC
= BE = CD
∠B = ∠C (Equal angles of isosceles triangles)
BC is common
∴ △BEC ≅ △BDC (S-A-S test)
∴ BD = CE (Corresponding sides of congruent triangles)
(3) In △PQR, If PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.
Solution:
In △PQR,
PQ > PR
∴ ∠R > ∠Q (Angles opposite to the sides follow the inequality)
∴ ½ ∠R > ½ ∠Q
∠SRQ > ∠SQR (∵ SQ & SR are bisectors of ∠Q & ∠R)
∴ SQ > SR
(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.
Solution:
In △ADE, AD = AE
∴ ∠AED = ∠ADE
Now, ∠AED + ∠AEC = 180°
∠ADE + ∠ADB = 180°
∴ ∠AEC – ∠ADB = 0
= ∠AEC = ∠ADB
∴ in △ADB & △AEC,
∠AEC = ∠ADB
BD = EC
AD = AE
∴ △ABD ≅ △ACE (S-A-S test)
(5) In figure 3.60, point S is any point on side QR of △PQR.
Prove that: PQ + QR + RP > 2PS
Solution:
In △PQS
PQ + QS > PS (Sum of two sides of triangle is greater than third side)
Similarly,
In △PSR,
PR + SR > PS
∴ Adding we get, PQ + QS + PR + SR > PS + PS
= PQ + QR + PR > 2 PS
(6) In figure 3.61, bisector of ∠BAC intersects side BC at point D.
Prove that AB > BD
Solution:
∵ AD is angle bisector
∴ ∠BAD = ∠DAC
In △DAC, ∠DAC < ∠ADB (Exterior angle of △DAC)
∴ ∠BAD < ∠ADB
= AB > BD (Sides opposite to the greater and smaller angles are greater and smaller)
(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.
Solution:
∵ PT || SR, and QS transversal,
∠QPT = ∠PSR (corresponding angles)
∵ PT || SR and PR transversal
∠TPR = ∠PRS (Alternate angles)
∵ PT is bisector of ∠QPR
∴ ∠QPT = ∠TPR
= ∠PSR = ∠PRS
∴ PS = PR (sides opposite to the angles are also equal)
(8) In figure 3.63, seg AD ⊥ seg BC.
seg AE is the bisector of ∠CAB and
C – E – D. Prove that
∠DAE = ½ (∠C – ∠B)
Solution:
∵ AE is bisector of ∠CAB
∴ ∠CAE = ∠BAE —- (i)
∴ In △ADB,
∠B + ∠ADB + ∠DAB = 180°
= ∠B = 90° – ∠DAB
In △ACD,
∠C + ∠CAD + ∠ADC = 180°
= ∠C = 90° – ∠CAD
∴ ∠C – ∠B = ∠DAB – ∠CAD
= (∠DAE + ∠BAE) – (∠CAE – ∠DAE)
= 2 ∠DAE + ∠BAE – ∠BAE (By (i))
= 2 ∠DAE
= ∠DAE = ½ (∠C – ∠B)
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