Prs Is Isosceles With Rp

Here we are providing Triangles Class 9 Extra Questions Maths Chapter 7 with Answers Solutions, Actress Questions for Class nine Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-nine-maths/

Extra Questions for Class 9 Maths Triangles with Answers Solutions

Extra Questions for Class ix Maths Chapter vii Triangles with Solutions Answers

Triangles Class 9 Extra Questions Very Short Reply Type

Triangles Course 9 Extra Questions Question one.
Find the mensurate of each exterior bending of an equilateral triangle.
Solution:
We know that each interior angle of an equilateral triangle is sixty°.
∴ Each exterior angle = 180° – 60° = 120°

Class nine Triangles Extra Questions Question ii.
If in ∆ABC, ∠A = ∠B + ∠C, then write the shape of the given triangle.
Solution:
Here, ∠A = ∠B + ∠C
And in ∆ABC, by angle sum belongings, we take
∠A + ∠B + C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = xc°
Hence, the given triangle is a right triangle.

Triangles Extra Questions Course 9 Question iii.
In ∆PQR, PQ = QR and ∠R = 50°, then find the measure of ∠Q.
Solution:
Here, in ∆PQR, PQ = QR
⇒ ∠R = ∠P = 50° (given)
At present, ∠P + ∠Q  + ∠R = 180°
50° + ∠Q + 50° = 180°
⇒ ∠Q = 180° – l° – fifty°
= 80°

Course 9 Maths Chapter 7 Extra Questions With Solutions Question 4.
If ∆Heaven ≅ ∆MON by SSS congruence rule, then write three equalities of corresponding angles.
Solution:
Since ∆SKY ≅ ∆Monday by SSS congruence dominion, then three equalities of respective angles
are ∠S = ∠M, ∠One thousand = ∠O and ∠Y = ∠Northward.

Triangles Form nine Actress Questions With Solutions Pdf Question v.
Is ∆ABC possible, if AB = 6 cm, BC = 4 cm and AC = one.5 cm ?
Solution:
Since 4 + 1.v = five.5 ≠ half dozen
Thus, triangle is not possible.

Triangle Class ix Actress Questions Question 6.
In ∆MNO, if ∠N = 90°, then write the longest side.
Solution:
We know that, side opposite to the largest bending is longest.
∴ Longest side = MO.

Actress Questions On Triangles Class nine Question seven.
In ∆ABC, if AB = Ac and ∠B = 70°, find ∠A.
Solution:
Here, in ∆ABC AB = AC ∠C = ∠B [∠s opp. to equal sides of a ∆)
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 70° = 180° [∵ ∠B = 70°]
⇒ ∠A = 180° – 70° – 70° = xl°

Class nine Maths Triangles Extra Questions Question eight.
In ∆ABC, if Ad is a median, so show that AB + AC > 2AD.
Solution:

Produce AD to East, such that Advertizement = DE.
In ∆ADB and ∆EDC, we accept
BD = CD, AD = DE and ∠1 = ∠2
∆ADB ≅ ∆EDC
AB = CE
At present, in ∆AEC, we have
AC + CE > AE
Air conditioning + AB > Advertisement + DE
AB + AC > 2AD [∵ AD = DE]

Triangles Class 9 Extra Questions Short Answer Blazon 1

Extra Questions On Congruence Of Triangles Class nine Question 1.
In the given figure, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.
Solution:

In ∆DAB and ∆CBA, we take
AD = BC [given]
BD = AC [given]
AB = AB [common]
∴ ∆DAB ≅ ∆CBA [by SSS congruence precept]
Thus, ∠DAB =∠CBA [c.p.c.t.]

Extra Questions Of Triangles Class 9 Question 2.
In the given figure, ∆ABD and ABCD are isosceles triangles on the same base BD. Bear witness that ∠ABC = ∠ADC.
Solution:

In ∆ABD, we have
AB = AD (given)
∠ABD = ∠ADB [angles contrary to equal sides are equal] …(i)
In ∆BCD, we take
CB = CD
⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)
Calculation (i) and (ii), we have
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC

Extra Questions For Class 9 Maths Triangles Question 3.
In the given figure, if ∠1 = ∠two and ∠3 = ∠iv, then prove that BC = CD.
Solution:

In ∆ABC and ACDA, we have
∠1 = ∠2 (given)
Air conditioning = Air-conditioning [common]
∠3 = ∠four [given]
So, by using ASA congruence axiom
∆ABC ≅ ∆CDA
Since respective parts of congruent triangles are equal
∴ BC = CD

Class ix Maths Affiliate 7 Actress Questions Question 4.
In the given figure, ∠B < ∠A and ∠C < ∠D. Prove that AD < BC.

Solution:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(two)
[∴ side opposite to greater angle is longer]
Calculation (i) and (ii), we obtain
AO + OD < BO + CO
Advertizing < BC

Triangles Class ix Of import Questions With Solutions Question 5.
In the given figure, Air conditioning > AB and D is a bespeak on AC such that AB = Advertizement. Evidence that BC > CD.
Solution:

Here, in ∆ABD, AB = AD
∠ABD = ∠ADB
[∠south opp. to equal sides of a ∆]
In ∆BAD
ext. ∠BDC = ∠BAD + ∠ABD
⇒ ∠BDC > ∠ABD ….(ii)
As well, in ∆BDC .
ext. ∠ADB > ∠CBD …(3)
From (two) and (3), we take
∠BDC > CD [∵ sides opp. to greater angle is larger]

Class 9 Maths Ch 7 Extra Questions Question 6.
In a triangle ABC, D is the mid-point of side AC such that BD = \(\frac{one}{2}\) AC. Evidence that ∠ABC is a correct angle.
Solution:

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = \(\frac{1}{2}\)Air-conditioning …(i)
Too, BD = \(\frac{1}{2}\)Air conditioning… (ii) [given]
From (i) and (ii), we obtain
Advertizement = BD and CD = BD
⇒ ∠2 = ∠four and ∠ane = ∠3 …..(3)
In ∆ABC, we take
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠iv = 180°
⇒ ∠1 + ∠two + ∠i + ∠two = 180° [using (iii)]
⇒ 2(∠one + ∠2) = 180°
⇒ ∠1 + ∠2 = ninety°
Hence, ∠ABC = xc°

Triangles Class 9 Extra Questions Short Respond Blazon 2

Chapter 7 Maths Grade ix Actress Questions Question 1.
ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Solution:

In ∆ABQ and ∆ACP, we have
AB = Air conditioning (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ Past SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

Questions On Triangles For Class 9 Question 2.
In the given figure, ∆ABC and ∆DBC are ii isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP

Solution:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
Advertising = AD [common)]
∴ By SSS congruence precept, we have
∆ABD ≅ ∆ACD
(2) In ∆ABP and ∆ACP
AB = Air conditioning [given]
∠BAP = ∠CAP [c.p.cit. every bit ∆ABD ≅ ∆ACD]
AP = AP [mutual]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP

Ch vii Maths Class nine Actress Questions Question 3.
In the given figure, it is given that AE = Advertizing and BD = CE. Show that ∆AEB ≅ ∆ADC.

Solution:
We have AE = AD … (i)
and CE = BD … (two)
On calculation (i) and (2),
nosotros have AE + CE = Ad + BD
⇒ Air-conditioning = AB
At present, in ∆AEB and ∆ADC,
we have AE = Advertizement [given]
AB = AC [proved above]
∠A = ∠A [mutual]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Triangles Grade 9 Actress Questions With Solutions Question iv.
In the given figure, in ∆ABC, ∠B = thirty°, ∠C = 65° and the bisector of ∠A meets BC in X. Accommodate AX, BX and CX in ascending order of magnitude.

Solution:
Here, AX bisects ∠BAC.
∴ ∠BAX = ∠CAX = x (say)
Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]
⇒ 2x + 30° + 65° = 180°
⇒ 2x + 95 = 180°
⇒ 2x = 180° – 95°
⇒ 2x = 85°
⇒ ten = \(\frac{85^{\circ}}{ii}\) = 42.59
In ∆ABX, we accept x > xxx°
BAX > ∠ABX
⇒ BX > AX (side opp. to larger angle is greater)
⇒ AX < BX
Also, in ∆ACX, we have 65° > x
⇒ ∠ACX > ∠CAX
⇒ AX > CX [side opp. to larger angle is greater]
⇒ CX > AX … (two)
Hence, from (i) and (ii), we have
CX < AX < BX

Question v.
In effigy, 'South' is any point on the side QR of APQR. Prove that PQ + QR + RP > 2PS.

Solution:
In ∆PQS, we take
PQ + QS > PS …(i)
[∵ sum of any 2 sides of a triangle is greater than the third side]
In ∆PRS, we take
RP + RS > PS …(ii)
Calculation (i) and (ii), we have
PQ + (QS + RS) + RP > 2PS
Hence, PQ + QR + RP > 2PS. [∵ QS + RS = QR]

Question 6.
If two isosceles triangles have a common base of operations, prove that the line joining their vertices bisects them at correct angles.
Solution:
Here, two triangles ABC and BDC having the mutual
base BC, such that AB = Ac and DB = DC.
Now, in ∆ABD and ∆ACD

AB = AC [given]
BD = CD [given]
Advertizing = Ad [mutual]
∴ ΔABD ≅ ΔΑCD [past SSS congruence axiom]
⇒ ∠ane = ∠2 [c.p.c.t.]
Once again, in ∆ABE and ∆ACE, we have
AB = AC [given]
∠1 = ∠2 [proved in a higher place]
AE = AE [common]
∆ABE = ∆ACE [by SAS congruence axiom]
BE = CE [c.p.c.t.]
and ∠3 = ∠four [c.p.c.t.]
But ∠3 + ∠four = 180° [a linear pair]
⇒ ∠3 = ∠iv = 90°
Hence, AD bisects BC at right angles.

Triangles Class 9 Extra Questions Long Answer Type

Question 1.
In the given figure, AP and DP are bisectors of ii side by side angles A and D of quadrilateral ABCD. Bear witness that ii ∠APD = ∠B + 2C.

Solution:
Here, AP and DP are angle bisectors of ∠A and ∠D
∴ ∠DAP = \(\frac{1}{2}\)∠DAB and ∠ADP = \(\frac{one}{ii}\)∠ADC ……(i)
In ∆APD, ∠APD + ∠DAP + ∠ADP = 180°
⇒ ∠APD + \(\frac{1}{2}\) ∠DAB + \(\frac{1}{2}\)∠ADC = 180°
⇒ ∠APD = 180° – \(\frac{ane}{two}\)(∠DAB + ∠ADC)
⇒ ii∠APD = 360° – (∠DAB + ∠ADC) ……(ii)
Also, ∠A + ∠B + C + ∠D = 360°
∠B + 2C = 360° – (∠A + ∠D)
∠B + C = 360° – (∠DAB + ∠ADC) ……(three)
From (two) and (three), nosotros obtain
2∠APD = ∠B + ∠C

Question 2.
In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that
(i) DF = Be (i) AM bisects ∠BAD.

Solution:
(i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠south]
Also, ∠2 = ∠4
⇒ ∠i = ∠iii
⇒ CE = CF (sides opp. to equals ∠s of a ∆]
∴ DF = Be [∵ BC – CE = CD – CF)

(ii) In ∆ADF and ∆ABE, we have
Ad = AB [sides of a square]
DF = Exist [proved to a higher place]
∠D = ∠B = 90°
⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom]
⇒ AF = AE and ∠5 = ∠half-dozen … (i) [c.p.c.t.]
In ∆AMF and ∆AME
AF = AE [proved above]
AM = AM [common]
FM = EM (given)
∴ ∆AMF ≅ ∆AME [by SSS congruence axiom]
∴ ∠7 = ∠8 …(ii) [c.p.c.t.]
Calculation (i) and (ii), we take
∠five + ∠vii = ∠6 + ∠8
∠DAM = ∠BAM
∴ AM bisects ∠BAD.

Question iii.
In correct triangle ABC, correct-angled at C, Chiliad is the mid-point of hypotenuse AB. C is joined to M and produced to a indicate D such that DM = CM. Signal D is joined to point B (see fig.). Bear witness that : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (ii) ∆DBC ≅ ∆ACB (4) CM = \(\frac{1}{ii}\)AB

Solution:
Given : ∆ACB in which 4C = 90° and M is the mid-point of AB.
To Prove :
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC = 90°
(iii) ∆DBC ≅ ∆ACB
(4) CM = \(\frac{1}{2}\)AB
Proof : Consider ∆AMC and ∆BMD,
we have AM = BM [given]
CM = DM [past construction]
∠AMC = ∠BMD [vertically opposite angles]
∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom]
⇒ AC = DB …(i) [by c.p.c.t.]
and ∠1 = ∠two [past c.p.c.t.]
But ∠1 and ∠ii are alternating angles.
⇒ BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180°
⇒ ninety° + CBD = 180°
⇒ ∠CBD = 90°
In ∆DBC and ∆ACB,
nosotros have CB = BC [common]
DB = AC [using (i)]
∠CBD = ∠BCA
∴ ∆DBC ≅ ∆ACB
⇒ DC = AB
⇒ \(\frac{1}{2}\)AB = \(\frac{one}{two}\)DC
⇒ \(\frac{i}{2}\)AB = CM or CM = \(\frac{1}{two}\)AB (∵ CM = \(\frac{1}{two}\)DC)

Question 4.
In figure, ABC is an isosceles triangle with AB = Ac. D is a bespeak in the interior of ∆ABC such that ∠BCD = ∠CBD. Bear witness that AD bisects ∠BAC of ∆ABC.

Solution:
In ∆BDC, nosotros take ∠DBC = ∠DCB (given).
⇒ CD = BD (sides opp. to equal ∠s of ∆DBC)
Now, in ∆ABD and ∆ACD,
we have AB = Ac [given]
BD = CD [proved above]
Advertising = AD [common]
∴ By using SSS congruence precept, we obtain
∆ABD ≅ ∆ACD
⇒ ∠BAD = ∠CAD [c.p.ç.t.]
Hence, Advert bisects ∠BAC of ∆ABC.

Question five.
Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to whatsoever two angles and the included side of the other triangle.
Solution:

Given : Ii Every bit ABC and DEF in which
∠B = ∠E,
∠C = ∠F and BC = EF
To Evidence : ∆ABC = ∆DEF
Proof : We take 3 possibilities
Case I. If AB = DE,
nosotros have AB = DE,
∠B = ∠East and BC = EF.
So, by SAS congruence axiom, nosotros have ∆ABC ≅ ∆DEF

Case II. If AB < ED, then take a bespeak Monday ED
such that EM = AB.
Join MF.
At present, in ∆ABC and ∆MEF,
we have
AB = ME, ∠B = ∠E and BC = EF.
So, by SAS congruence axiom,
we take ΔΑΒC ≅ ΔΜEF
⇒ ∠ACB = ∠MFE
Simply ∠ACB = ∠DFE
∴ ∠MFE = ∠DFE

Which is possible but when FM coincides with B FD i.east., M coincides with D.
Thus, AB = DE
∴ In ∆ABC and ∆DEF, nosotros have
AB = DE,
∠B = ∠Due east and BC = EF
So, by SAS congruence axiom, we have
∆ABC ≅ ∆DEF
Case 3. When AB > ED
Accept a point M on ED produced
such that EM = AB.
Join MF
Proceeding as in Example II, we tin can prove that
∆ABC = ∆DEF
Hence, in all cases, nosotros have
∆ABC = ∆DEF.

Question 6.
In the given figure, side QR is produced to the indicate S. If the bisectors of ∠PQR and ∠PRS meet at T,
prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Solution:
Hither, QT is angle bisector of ∠PQR

Triangles Class 9 Extra Questions HOTS

Question 1.
Show that the difference of any ii sides of a triangle is less than the 3rd side.
Solution:

Consider a triangle ABC
To Prove :
(i) Ac – AB < BC
(2) BC – AC < AB
(three) BC – AB < AC
Construction : Take a signal D on AC
such that AD = AB.
Join BD.
Proof : In ∆ABD, we have ∠3 > ∠1 …(i)
[∵ exterior ∠ is greater than each of interior contrary angle in a ∆]
Similarly, in ∆BCD, nosotros have
∠2 > ∠4 …..(ii) [∵ ext. ∠ is greater then interior opp. angle in a ∆]
In ∆ABD, we have
Advertisement = AB [by construction]
∠i = ∠2 …(iii) [angles opp. to equal sides are equal in a triangle]
From (i), (two) and (iii), we have
⇒ ∠3 > ∠4 =
⇒ BC > CD
⇒ CD < BC
AC – Advert < BC
Air conditioning – AB < BC [∵ Advertisement = AB]
Hence, AC – AB < BC
Similarly, nosotros can evidence
BC – Ac < AB
and BC – AB < AC

Question 2.
In the figure, O is the interior point of ∆ABC. BO meets Air-conditioning at D. Testify that OB + OC < AB + Air-conditioning.

Solution:
In ∆ABD, AB + Advertisement > BD …(i)
∵ The sum of any 2 sides of a triangle is greater than the third side. Also, we have
BD = BO + OD
AB + Advertizing > BO + OD ….(ii)
Similarly, in ∆COD, nosotros take
OD + DC > OC … (iii)
On adding (2) and (iii), we have
AB + AD + OD + DC > BO + OD + OC
⇒ AB + AD + DC > BO + OC
⇒ AB + AC > OB + OC
or OB + OC < AB + Air-conditioning
Hence, proved.

Triangles Form 9 Actress Questions Value Based (VBQs)

Question 1.
A campaign is started by volunteers of mathematical club to heave school and its surrounding under Swachh India Abhiyan. They made their own logo for this campaign. What values are acquired by mathematical club ?
If it is given that ∆ABC ≅ ∆ECD, BC = AE.
Prove that ∆ABC ≅ ∆CEA.

Solution:
Hither, information technology is given that
∆ABC ≅ ∆ECD
AB = CE [c.p.c.t.]
BC = CD [c.p.c.t.]
Air conditioning = ED [c.p.c.t.]
At present, in ∆ABC and ∆CEA
BC = AE [given]
AB = EC [proved to a higher place]
Air conditioning = Air conditioning [common]
∴ By using SSS congruence axiom, we have
∆ABC ≅ ∆CEA
Value : Cleanliness and social concerning.

Question two.
Rajiv, a skillful pupil and actively involved in applying cognition A of mathematics in daily life. He asked his classmate Rahul to make triangle as shown by choosing one of the vertex equally mutual. Rahul tried only not correctly. Afterward sometime Rajiv hinted Rahul most congruency of triangle. At present, Rahul stock-still vertex C equally mutual vertex and locate point D, E such that AC = CD and BC = CE. Was the triangle made past Rahul is coinciding ? Write the condition satisfying congruence.
What value is depicted past Rajiv'south activity?

Solution:
In ∆ABC and ∆December, we have
Air-conditioning = DC [by construction]
BC = EC [by structure]
∠ACB = ∠ECD [vert. opp. ∠s]
Past using SAS congruence axiom, we have
∆ABC ≅ ∆Dec
Value : Cooperative learning, use of concept and friendly nature.

Prs Is Isosceles With Rp,

Source: https://ncertmcq.com/triangles-class-9-extra-questions/

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