Mathematics Part ii Solutions Solutions for Class 9 Maths Chapter 3 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among class 9 students for Maths Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 9 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 27:
Question 1:
In the given figure, ACD is an exterior angle of ABC. B = 40°, A = 70°.
Find the measure of ACD.
Answer:
In ABC,
∠ACD = ∠A + ∠B (Exterior angle property)
= 70â + 40â
= 110â
Hence, the measure of ACD is 110ââ.
Page No 27:
Question 2:
Answer:
In PQR,
∠P + ∠Q + ∠R = 180â (Angle sum property)
⇒ 70â + 65â + ∠R = 180â
⇒ 135â + ∠R = 180â
⇒ ∠R = 180â − 135â
= 45â
Hence, the measure of R is 45â.
Page No 27:
Question 3:
The measures of angles of a triangle are x°, ( x20)°, (x40)°. Find the measure of each angle.
Answer:
Let us suppose the angles ∠P, ∠Q, ∠Rof a PQR be x°, (x 20)°, (x 40)° respectively.
∠P + ∠Q + ∠R = 180â (Angle sum property)
⇒ xâ + (x 20)° + (x 40)° = 180â
⇒ 3x 60 = 180
⇒ 3x = 240
⇒ x â= 80
Therefore,
∠P = 80â
∠R = (80 20)°
= 60â
∠R = (80 40)°
= 40â
Hence, the measure of each angle is 80â, 60â and 40ârespectively.
Page No 27:
Question 4:
Answer:
Let us suppose the angles of a PQR such that ∠P < ∠Q < ∠R.
A.T.Q,
∠Q = 2∠P
∠R = 2∠P
Now, ∠P + ∠Q + ∠R = 180â (Angle sum property)
⇒ ∠P + 2∠P + 3∠P = 180â
⇒ 6∠P = 180â
⇒ ∠P = 30â
Therefore,
∠P = 30â
∠R = 60â
∠R = 90â
Hence, the measure of each angle is 30â, 60â and 90ârespectively.
Page No 28:
Question 5:
In the given figure, measures of some angles are given. Using the measures find the values of x, y, z.
Answer:
∠NEM + ∠NET = 180â (Linear angle property)
⇒ y + 100â = 180â
⇒ y = 80â
Also, ∠NME + ∠EMR = 180â (Linear angle property)
⇒ z + 140â = 180â
⇒ z = 40â
Now, In â³NEM
∠N + ∠E + ∠M = 180â (Angle sum property)
⇒ x + y + z = 180â
⇒ x + 80â + 40â = 180â
⇒ x + 120â = 180â
⇒ x = 60â
Hence, the values of x, y and z are 60â, 80â and 40ârespectively.
Page No 28:
Question 6:
In the given figure, line AB || line DE. Find the measures of DRE and ARE using given measures of some angles.
Answer:
AB || DE and AD is a transversal line.
∠BAR = ∠RDE = 70â (Alternate angles)
In â³DER
∠D + ∠E + ∠DRE = 180â (Angle sum property)
⇒ 70â + 40â + ∠DRE = 180â
⇒ 110â + ∠DRE = 180â
⇒ ∠DRE = 70â
Now, ∠ARE = ∠DRE + ∠RDE (Exterior angle property)
= 70â + 40â
= 110â
Hence, the measures of ∠DRE and ∠ARE are 70â and 110ârespectively.
Page No 28:
Question 7:
In ABC, bisectors of A and B intersect at point O. If C = 70° . Find measure of AOB.
Answer:
If the bisectors of ∠X and ∠Y of a â³XYZ intersect at point O, then
= 90â + 35â
= 125â
Hence, the measure of ∠AOB is 125â.
Page No 28:
Question 8:
In the given Figure, 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 °.
Answer:
AB || CD and PQ is a transversal line.
∠BPQ + ∠DQP = 180â (Angles on the same side of a transversal line are supplementary angles)
⇒ ∠QPT + ∠PQT = 90â
In â³PQT
∠QPT + ∠PQT + ∠PTQ = 180â (Angle sum property)
⇒ 90â + ∠PTQ = 180â
⇒ ∠PTQ = 90â
Hence proved.
Page No 28:
Question 9:
Using the information shown in figure, find the measures of a, b and c.
Answer:
c + 100â = 180â (Linear angle property)
⇒ c = 80â
Now, b = 70â (Vertically opposite angles)
a + b + c = 180â (Angle sum property)
⇒ a + 70â + 80â = 180â
⇒ a + 150â = 180â
⇒ a = 30â
Hence, the values of a, b and c are 30â, 70â and 80ârespectively.
Page No 28:
Question 10:
(i) DEG = EDF (ii) EF = FG.
Answer:
(i) Given: DE || GF
Now, DEF = GFM (Corresponding angles as DM is a transversal line)
⇒ 2DEG = DFG (Ray EG and ray FG are bisectors of DEF and DFM)
⇒ 2DEG = EDF (âµ EDF = DFG, alternate angles as DF is a transversal line)
⇒ DEG = EDF
(ii) Given: DE || GF
DEG = EGF (Alternate angles as EG is a transversal line)
∴ GEF = EGF (âµ DEG = GEF)
∴ EF = FG (Sides opposite to equal angles)
Page No 31:
Question 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)
ABC PQR
(ii)
By . . . . . . . . . . test
XYZ LMN
(iii)
By . . . . . . . . . . test
LMN PTR
(iv)
By . . . . . . . . . . test
LMN PTR
Answer:
(i) SSC Test
(ii) SAS Test
(iii) ASA Test
(iv) Hypotenuse Side Test.
Page No 32:
Question 2:
(i)
(ii)
PTQ STR ....... test
seg PQ corresponding sides of congruent triangles
Answer:
(ii)
PTQ STR ....... test
seg PQ corresponding sides of congruent triangles
Page No 32:
Question 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.
Answer:
âIn â³ABC and â³QPR
AB = PQ (Given)
BC = PR (Given)
∠A = ∠Q = 90â (Given)
By RHS test of congruency
â³ABC ≅ â³QPR
∠B = ∠P (corresponding angles of congruent triangles)
∠C = ∠R (corresponding angles of congruent triangles)
âAC = QR (corresponding sides of congruent triangles)
Page No 32:
Question 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.
Answer:
âIn â³LMNand â³PNM
LM = PN (Given)
LN = PM (Given)
MN = NM (Common)
By SSS test of congruency
â³LMN ≅ â³PNM
∠L = ∠P (corresponding angles of congruent triangles)
∠LMN = ∠PNM (corresponding angles of congruent triangles)
∠MNL = ∠NMP (corresponding angles of congruent triangles)
Page No 32:
Question 5:
In the given figure, seg AB seg CB and seg ADseg CD.
Prove that ABD CBD
Answer:
In â³ABD and â³CBD
AB = CB (Given)
AD = CD (Given)
BD = BD (Common)
By SSS test of congruency
â³ABD ≅ â³CBD
Page No 33:
Question 6:
In the given figure, P R seg PQ seg RQ
Prove that, PQT RQS
Answer:
In â³PQT and â³RQS
∠P = ∠R (Given)
PQ = RQ (Given)
∠Q = ∠Q (Common)
By ASA test of congruency
â³PQT ≅ â³RQS
Page No 38:
Question 1:
Find the values of x and y using the information shown in figure .
Find the measure of ABD and mACD.
Answer:
In â³ABC, AB = AC
∴ ∠ABC = ∠ACB (Angles opposite to equal sides are equal)
⇒ x = 50â
In â³BDC, DB = DC
∴ ∠DBC= ∠DCB (Angles opposite to equal sides are equal)
⇒ y = 60â
Now, ∠ABD = ∠ABC + ∠DBC
= 50â + 60â
= 110â
Also, ∠ACD = ∠ACB + ∠DCB
= 50â + 60â
= 110â
Hence, the values of x and y are 50â and 60ârespectively.
Page No 38:
Question 2:
The length of hypotenuse of a right angled triangle is 15. Find the length of median of its hypotenuse.
Answer:
In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse.
Therefore, the length of median of its hypotenuse is 7.5 units.
Page No 38:
Question 3:
In PQR, Q = 90° , PQ = 12, QR = 5 and QS is a median. Find l(QS).
Answer:
In â³PQR,
PR2 = PQ2 + QR2 (By Pythagoras theorem)
⇒ PR2 = 122 + 52
⇒ PR2 = 144 + 25
⇒ PR2 = 132
⇒ PR = 13
In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse.
= 6.5 units.
Hence, the length of QR is 6.5 units.
Page No 38:
Question 4:
In the given figure, P point G is the point of concurrence of the medians of PQR . If GT = 2.5, find the lengths of PG and PT.
Answer:
The point of concurrence of medians of a triangle divides each median in the ratio 2 : 1.
Let us suppose the length of PG and GT be 2x and x.
Therefore, the value of x is 2.5
Now, PG = 2x
= 2(2.5)
= 5 units
Now, PT = PG + GT
= 5 + 2.5
= 7.5 units
Hence, the lenght of PG and GT is 5 and 7.5units respectively.
Page No 43:
Question 1:
In the given figure, point A is on the bisector of XYZ. If AX = 2 cm then find AZ.
Answer:
By Part I of the angle bisector theorem, every point on the bisector of an angle is equidistant from the sides of the angle.
∴ AZ = AX
= 2 cm
Hence, the length of AZ is 2 cm.
Page No 43:
Question 2:
In the given figure, 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.
Answer:
By Part II of the angle bisector theorem, any point equidistant from sides of an angle is on the bisector of the angle
∠TSP = ∠RSP
Now, ∠TSP + ∠RSP = ∠TSR
⇒ ∠RSP + ∠RSP = 56â
⇒ 2∠RSP = 56â
⇒ ∠RSP = 28â
Hence, the measure of ∠RSP is 28â.
Page No 43:
Question 3:
In PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.
Answer:
In â³PQR, QR and PR are the longest and smallest side.
The angle opposite to the longest side is the largest angle and the angle opposite to the smallest side is the smallest angle .
Hence, the greatest and the smallest angle of the triangle are ∠QPR and ∠PQR respectively.
Page No 43:
Question 4:
In FAN, F = 80° , A = 40° . Find out the greatest and the smallest side of the triangle. State the reason.
Answer:
In âFAN,
∠F + ∠A + ∠N = 180â (Angle sum property)
⇒ 80â + 40â + ∠N = 180â
⇒ 120â + ∠N = 180â
⇒ ∠N = 180â − 120â
= 60â
Now, in â³FAN, ∠F and ∠A are the largest and smallest angle.
The side opposite to the largest angle is the greatest side and the side opposite to the smallest angle is the smallest side .
Hence, the greatest and the smallest side of the triangle are AN and âFN respectively.
Page No 43:
Question 5:
Prove that an equilateral triangle is equiangular.
Answer:
Consider an equilateral triangle ABC.
In â³ABC, AB = BC
∴ ∠C = ∠A ...(1) (Angles opposite to equal sides)
In â³ABC, AB = CA
∴ ∠C = ∠B ...(2) (Angles opposite to equal sides)
From (1) and (2), we get
∠A = ∠B = ∠C
Hence, an equilateral triangle is equiangular.
Page No 44:
Question 6:
Answer:
In â³ADB and â³ADC
∠BAD = ∠CAD (AD is the bisector of ∠BAC)
∠BDA = ∠CDA = 90â (Given)
AD = DA (Common)
By ASA test of congruency
â³ADB ≅ â³ADC
∴ ∠B = ∠C (corresponding angles of congruent triangles)
If two angles of a traingle are equal, then the traingle is said to be a isosceles triangle.
Page No 44:
Question 7:
Answer:
In â³PRS, ∠PRS is an obtuse angle.
Since, a traingle can have maximum one obtuse angle.
Hence, ∠PRS is the greatest angle.
Therefore, the side opposite to ∠PRS i.e., PS is the longest side.
Now, seg PS > seg PR > seg RS
∴ seg PS > seg PQ [âµseg PRseg PQ]
Page No 44:
Question 8:
In the given figure, in ABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD seg BE
Answer:
In â³ADB and â³BEA
∠ADB = ∠BEA = 90â (Given)
BD = AE (Given)
AB = BA (Common)
By RHS test of congruency
â³ADB ≅ â³BEA
∴ AD = BE (corresponding angles of congruent triangles)
Hence proved.
Page No 47:
Question 1:
If XYZ ~ LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.
Answer:
Consider, XYZ ~ LMN
Page No 47:
Question 2:
In XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If XYZ ~ â PQR and PQ = 8 cm then find the lengths of remaining sides of PQR.
Answer:
Consider, XYZ ~ PQR
Hence, the lengths of remaining sides of PQR are 12 and 10 cm.
Page No 47:
Question 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.
Answer:
The pair of similar triangles is given below:
Page No 49:
Question 1:
(i) If two sides of a triangle are 5 cm and 1.5 cm, the lenght of its third side cannot be . . . . . . . .
(ii) In PQR, If R > Q then . . . . . . . .
(A) QR > PR
(D) QR < PR
(iii) In TPQ, T = 65° , P = 95° which of the following is a true statement ?
(C) TQ < TP < PQ
Answer:
(i)
According to the triangle inequality "the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side".
Therefore, the lenght of the third side cannot be greater than 3.4 cm.
Hence, the correct option is (D).
(ii) In PQR, If R > Q
∴ PQ > PR (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).
(iii) In TPQ, T = 65° , P = 95°
A triangle can have maximum one obtuse angle.
T < P
PQ < TQ (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).
Page No 49:
Question 2:
ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.
Answer:
âIn â³ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
Dividing both sides by 2, we get
⇒ ∠EBC = ∠DCB ...(2)
In â³BEC and â³CDB
∠EBC = ∠DCB [From (2)]
BE = CD (E and D are the mid points of AB & AC respectively and AB = AC)
BC = CB (Common)
By SAS test of congruency
â³BEC ≅ â³CDB
∴ BD = CE (corresponding sides of congruent triangles)
Page No 49:
Question 3:
Answer:
In â³PQR, PQ > PR
Now, the angle opposite to the greater side is greater than angle opposite to the smaller side.
∴ ∠R > ∠Q
Dividing both sides by 2, we get
⇒ ∠SRQ > ∠SQR
⇒ SQ > SR (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.
Page No 49:
Question 4:
Answer:
âIn â³ADE, AD = AE
∴ ∠ADE = ∠AED (Angles opposite to equal sides)
Now,
∠ADE + ∠ADB = 180â ...(1)
∠AED + ∠AEC = 180â ...(2)
Subtracting (2) from (1), we get
∠ADB − ∠AEC = 0
⇒ ∠ADB = ∠AEC ...(3)
In â³ABD and â³ACE
∠ADE = ∠AEC [From (3)]
BD = CE (Given)
AD = AE (Given)
By SAS test of congruency
â³ABD ≅ â³ACE
Page No 49:
Question 5:
In the given figure, point S is any point on side QR of PQR Prove that : PQ + QR + RP > 2PS
Answer:
In â³PQS
PQ + QS > PS ...(1)
(Sum of two sides of a traingle is greater than the third side)
In â³PRS
RP + RS > PS ...(2)
â (Sum of two sides of a traingle is greater than the third side)
Adding (1) and (2), we get
PQ + QS + RP + RS > PS + PS
⇒ PQ + QR + RP > 2PS
Page No 50:
Question 6:
In the given figure, bisector of BAC intersects side BC at point D. Prove that AB > BD
Answer:
Given: AD is the bisector of ∠A
∴ ∠DAB = ∠DAC
Now, ∠BDA is the exterior angle of the âDAC
∴ ∠BDA > ∠DAC
⇒ ∠BDA > ∠DAB (âµ ∠DAC = ∠DAB)
⇒ AB > BD (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.
Page No 50:
Question 7:
In the given figure, seg PT is the bisector of QPR. A line through R intersects ray QP at point S. Prove that PS = PR
Answer:
According to the figure, we have
PT || SR
If PT || SR and PS is a transversal line, then
QPT = PSR (Corresponding angles) ....(1)
If PT || SR and PR is a transversal line, then
TPR = PRS (Alternate angles) ....(2)
We are given that seg PT is the bisector of QPR.
∴ QPT = TPR ...(3)
From (1), (2) and (3)
PSR = PRS
The sides opposite to equal angles are also equal in a triangle.
∴ PS = PR
Page No 50:
Question 8:
In the given figure, seg AD seg BC. seg AE is the bisector of CAB and C - E - D. Prove that
Answer:
In âADB,
∠ADB + ∠DAB + ∠B = 180° [Angle sum property]
⇒ 90° + ∠DAB + ∠B = 180°
⇒∠B = 90° − ∠DAB .....(2)
In âADC,
∠ADC+∠DAC+∠C = 180° [Angle sum property]
⇒ 90° + ∠DAC + ∠C = 180°
⇒∠C = 90° − ∠DAC .....(3)
Subtracting (2) from (3), we get
∠C − ∠B = ∠DAC − ∠DAB
⇒ ∠C − ∠B = (∠AEC + ∠DAE) − (∠BAE − ∠DAE)
⇒ ∠C − ∠B = ∠AEC + ∠DAE − ∠BAE + â∠DAE
â⇒ ∠C − ∠B = 2∠DAE [From (1)]
â⇒ DAE =
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