Frank Solutions for Class 9 Maths Chapter 12 Isosceles Triangle provide comprehensive answers with the intention of helping students in solving the most tricky questions with ease. These solutions are helpful in clearing their doubts quickly while revising for the final examination. To obtain in-depth conceptual knowledge, students can practise Frank Solutions on a daily basis.
Chapter 12 consists of problems based on the Isosceles Triangle. Students who want to come out with flying colours in the annual exams are suggested to practise Frank Solutions regularly. The main aim of preparing these solutions is to boost the self-confidence of students in solving complex problems effortlessly. Download Frank Solutions for Class 9 Maths Chapter 12 Isosceles Triangle from the link provided below.
Frank Solutions for Class 9 Maths Chapter 12 Isosceles Triangle Download PDF
Access Frank Solutions for Class 9 Maths Chapter 12 Isosceles Triangle
1. Find the angles of an isosceles triangle whose equal angles and the non-equal angles are in the ratio 3:4.
Solution:
Given that,
The equal angles and the non-equal angles are in the ratio 3:4
Let the equal angles be 3x each,
So, non- equal angle is 4x
We know that,
The sum of angles of a triangle = 1800
Hence,
3x + 3x + 4x = 1800
10x = 1800
We get,
x = 180
Therefore,
3x = 3 × 180 = 540 and
4x = 4 × 180 = 720
Hence,
The angles of a triangle are 540, 540 and 720
2. Find the angles of an isosceles triangle which are in the ratio 2:2:5
Solution:
The equal angles and the non-equal angle are in the ratio 2:2:5
Let equal angles be 2x each
So, the non-equal angle is 5x
We know that,
The sum of angles of a triangle = 1800
2x + 2x + 5x = 1800
9x = 1800
x = 200
Therefore,
2x = 2 × 200 = 400
5x = 5 × 200 = 1000
Hence, the angles of a triangle are 400, 400 and 1000
3. Each equal angle of an isosceles triangle is less than the third angle by 150. Find the angles.
Solution:
Let equal angles of the isosceles triangle be x each
Therefore, the non-equal angle is x + 150
We know that,
The sum of angles of a triangle = 1800
x + x + (x + 150) = 1800
3x + 150 = 1800
3x = 1800 – 150
3x = 1650
We get,
x = 550
So,
(x + 150) = 550 + 150 = 700
Hence, the angles of a triangle are 550, 550 and 700
4. Find the interior angles of the following triangles
(a)
(b)
(c)
(d)
Solution:
(a)
In â–³ABC,
∠A = 1100
AB = AC
∠C = ∠B (angles opposite to two equal sides are equal)
Now,
By angle sum property,
∠A + ∠B + ∠C = 1800
∠A + ∠B + ∠B = 1800
1100 + 2∠B = 1800
2∠B = 1800 – 1100
2∠B = 700
We get,
∠B = 350
∠C = 350
Hence,
The interior angles are ∠B = 350 and ∠C = 350
(b)
In â–³ABC,
AB = AC
∠ACB = ∠ABC ………(1) [∵ angles opposite to two equal sides are equal]
Now,
∠ACB + ∠ACD = 1800 [linear pair]
∠ACB = 1800 – ∠ACD
∠ACB = 1800 – 1050
∠ACB = 750
So,
∠ABC = 750 [from equation (1)]
Now, in â–³ABC,
By angle sum property,
∠ABC + ∠ACB + ∠BAC = 1800
750 + 750 + ∠BAC = 1800
1500 + ∠BAC = 1800
∠BAC = 1800 – 1500
We get,
∠BAC = 300
Hence,
In â–³ABC,
∠A = 300
∠B = 750
∠C = 750
(c)
In â–³ABD,
Given that,
AD = BD
∠ABD = ∠BAD … (angles opposite to two equal sides are equal)
Now,
∠ABD = 370 …. (given)
Hence,
∠BAD = 370
By exterior angle property,
∠ADC = ∠ABD + ∠BAD
∠ADC = 370 + 370
We get,
∠ADC = 740
In â–³ADC,
AC = DC …. (given)
∠ADC = ∠DAC …. (angles opposite to two equal sides are equal)
∠DAC = 740
Now,
∠BAC = ∠BAD + ∠DAC
∠BAC = 370 + 740
We get,
∠BAC = 1110
In â–³ABC,
∠BAC + ∠ABC + ∠ACB = 1800
1110 + 370 + ∠ACB = 1800
∠ACB = 1800 – 1110 – 370
We get,
∠ACB = 320
Therefore,
The interior angles of â–³ABC are 370, 1110 and 320
(d)
In â–³ACD,
AD = CD …… (given)
∠ACD = ∠CAD … (angles opposite to two equal sides are equal)
Now,
∠ACD = 500 …… (given)
∠CAD = 500
By exterior angle property,
∠ADB = ∠ACD + ∠CAD
∠ADB = 500 + 500
∠ADB = 1000
In â–³ADB,
AD = BD …… (given)
∠DBA = ∠DAB ….. (angles opposite to two equal sides are equal)
Also,
∠ADB + ∠DBA + ∠DAB = 1800
1000 + 2∠DBA = 1800
2∠DBA = 1800 – 1000
2∠DBA = 800
We get,
∠DBA = 400
∠DAB = 400
∠BAC = ∠DAB + ∠CAD
∠BAC = 400 + 500
∠BAC = 900
Therefore, the interior angles of â–³ABC are 500, 900 and 400
5. Side BA of an isosceles triangle ABC is produced so that AB = AD. If AB and AC are the equal sides of the isosceles triangle, prove that ∠BCD is a right angle.
Solution:
Let ∠ABC = x
Hence,
∠BCA = x (since AB = AC)
In â–³ABC,
∠ABC + ∠BCA + ∠BAC = 1800 ……. (1)
But
∠BAC + ∠DAC = 1800 ……. (2)
From equations (1) and (2)
∠ABC + ∠BCA + ∠BAC = ∠BAC + ∠DAC
∠DAC = ∠ABC + ∠BCA
∠DAC = x + x
We get,
∠DAC = 2x
Let ∠ADC = y,
Hence,
∠DCA = y (since AD = AC)
Now,
In â–³ADC,
∠ADC + ∠DCA + ∠DAC = 1800 … (3)
But ∠BAC + ∠DAC = 1800 ……. (4)
From equations (3) and (4), we get,
∠ADC + ∠DCA + ∠DAC = ∠BAC + ∠DAC
∠BAC = ∠ADC + ∠DCA
∠BAC = y + y
∠BAC = 2y
Now, substituting the value of ∠BAC and ∠DAC in equation (2)
2x + 2y = 1800
x + y = 900
∠BCA + ∠DCA = 900
Therefore,
∠BCD is a right angle
6. The bisectors of the equal angles of an isosceles triangle PQR meet at O. If PQ = PR, prove that PO bisects ∠P.
Solution:
Join PO and produce to meet QR at point S
In â–³PQS and â–³PRS
PS = PS (common)
PQ = PR (given)
So,
∠Q = ∠R (angles opposite to two equal sides are equal)
Hence,
△PQS ≅ △PRS
Thus,
∠QPS = ∠RPS
Therefore,
PO bisects ∠P
7. Prove that the medians corresponding to equal sides of an isosceles triangle are equal.
Solution:
Let â–³ABC be an isosceles triangle with AB = AC
Let D and E be the midpoints of AB and AC, respectively
Now,
Join BE and CD
Then BE and CD become the medians of this isosceles triangle
In â–³ABE and â–³ACD
AB = AC (given)
AD = AE (D and E are midpoints of AB and AC)
∠A = ∠A (common angle)
Hence,
△ABE ≅ △ACD (SAS criteria)
Therefore,
The medians BE and CD are equal, i.e., BE = CD
8. DPQ is an isosceles triangle with DP = DQ. A straight line CD bisects the exterior ∠QDR. Prove that DC is parallel to PQ
Solution:
In â–³QDP,
DP = DQ
Hence,
∠Q = ∠P (angles opposite to two equal sides are equal)
∠QDR = ∠Q + ∠P
2∠QDC = ∠Q + ∠P (DC bisects angle QDR)
2∠QDC = ∠Q + ∠Q
We get,
2∠QDC = 2∠Q
Hence,
∠QDC = ∠Q
But these angles are alternate angles
Therefore,
DC || PQ
Hence, proved
9. In a quadrilateral PQRS, PQ = PS and RQ = RS. If ∠P = 500 and ∠R = 1100, find ∠PSR.
Solution:
In â–³PQS,
PQ = PS
Therefore,
∠PQS = ∠PSQ (angles opposite to two equal sides are equal)
∠P + ∠PQS + ∠PSQ = 1800
500 + 2∠PQS = 1800
2∠PQS = 1800 – 500
We get,
2∠PQS = 1300
∠PQS = 650
So,
∠PQS = ∠PSQ = 650 …….. (1)
In â–³SRQ,
SR = RQ
Hence,
∠RQS = ∠RSQ (angles opposite to two equal sides are equal)
∠R + ∠RQS + ∠RSQ = 1800
1100 + 2∠RQS = 1800
2∠RQS = 1800 – 1100
We get,
2∠RQS = 700
∠RQS = 350
So,
∠RQS = ∠RSQ = 350 ….. (2)
Adding equations (1) and (2), we get,
∠PSQ + ∠RSQ = 650 + 350
∠PSR = 1000
10. △ABC is an isosceles triangle with AB = AC. Another triangle, BDC is drawn with base BC = BD in such a way that BC bisects ∠B. If the measure of ∠BDC is 700, find the measures of ∠DBC and ∠BAC.
Solution:
In â–³BDC,
∠BDC = 700
BD = BC
Hence,
∠BDC = ∠BCD (angles opposite to two equal sides are equal)
∠BCD = 700
Now,
∠BCD + ∠BDC + ∠DBC = 1800
700 + 700 + ∠DBC = 1800
∠DBC = 1800 – 1400
We get,
∠DBC = 400
∠DBC = ∠ABC (BC is the angle bisector)
Hence,
∠ABC = 400
In â–³ABC,
Since AB = AC, ∠ABC = ∠ACB
Hence,
∠ACB = 400
∠ACB + ∠ABC + ∠BAC = 1800
400 + 400 + ∠BAC = 1800
∠BAC = 1800 – 800
∠BAC = 1000
Therefore, the measure of ∠BAC = 1000 and ∠DBC = 400
11. â–³PQR is isosceles with PQ = PR. T is the mid-point of QR, and TM and TN are perpendiculars on PR and PQ, respectively. Prove that,
(i) TM = TN
(ii) PM = PN and
(iii) PT is the bisector of ∠P
Solution:
(i) In â–³PQR,
PQ = PR
Hence,
∠R = ∠Q …… (1)
Now,
In â–³QNT and â–³RMT
∠QNT = ∠RMT = 900
∠Q = ∠R [from equation (1)]
QT = TR (given)
Hence,
△QNT ≅ △RMT (AAS criteria)
Therefore,
TM = TN
(ii) Since, △QNT ≅ △RMT
NQ = MR ……… (2)
But,
PQ = PR ….. (3) [given]
Now, subtracting (2) from (3), we get,
PQ – NQ = PR – MR
PN = PM
(iii) In â–³PNT and â–³PMT
TN = TM (proved)
PT = PT (common)
∠PNT = ∠PMT = 900
Hence,
△PNT ≅ △PMT
So,
∠NPT = ∠MPT
Therefore,
PT is the bisector of ∠P
12. △PQR is isosceles with PQ = QR. QR is extended to S so that △PRS becomes isosceles with PR = PS. Show that ∠PSR: ∠QPS = 1:3
Solution:
In â–³PQR,
PQ = QR (given)
∠PRQ = ∠QPR …… (1)
In â–³PRS,
PR = RS (given)
∠PSR = ∠RPS …….. (2)
Now,
Adding equations (1) and (2), we get,
∠QPR + ∠RPS = ∠PRQ + ∠PSR
∠QPS = ∠PRQ + ∠PSR …….. (3)
Now,
In â–³PRS,
∠PRQ = ∠RPS + ∠PSR
∠PRQ = ∠PSR + ∠PSR [from equation(2)]
∠PRQ = 2∠PSR …… (4)
Now,
∠QPS = 2∠PSR + ∠PSR [from equation (3) and (4)]
∠QPS = 3∠PSR
∠PSR / ∠QPS = 1 / 3
Therefore,
∠PSR: ∠QPS = 1: 3
Hence, proved
13. In △KLM, KT bisects ∠LKM and KT = TM. If ∠LTK is 800, find the value of ∠LMK and ∠KLM.
Solution:
In â–³KTM,
KT = TM (given)
Hence,
∠TKM = ∠TMK ……. (1)
Now,
∠KTL = ∠TKM + ∠TMK
800 = ∠TKM + ∠TKM …….. [from (1)]
800 = 2∠TKM
We get,
∠TKM = 400 = ∠TMK = ∠LMK …… (2)
But,
∠TKM = ∠TKL (KT is the angle bisector)
Hence,
∠TKL = 400
In â–³KTL,
∠TKL + ∠KTL + ∠KLT = 1800
400 + 800 + ∠KLT = 1800
∠KLT = 1800 – 400 – 800
We get,
∠KLT = 600 = ∠KLM
Therefore,
∠KLM = 600 and ∠LMK = 400
14. Equal sides QP and RP of an isosceles â–³PQR are produced beyond P to S and T such that â–³PST is an isosceles triangle with PS = PT. Prove that TQ = SR.
Solution:
In â–³PTQ and â–³PSR
PQ = PR (given)
PT = PS (given)
∠TPQ = ∠SPR (vertically opposite angles)
Hence,
△PTQ ≅ △PSR
Therefore,
TQ = SR
Hence, proved
15. Prove that the bisector of the vertex angle of an isosceles triangle bisects the base perpendicularly.
Solution:
In â–³ADB and â–³ADC
AB = AC (given)
AD = AD (common)
∠BAD = ∠CAD (AD bisects ∠BAC)
Hence,
△ADB ≅ △ADC
Therefore,
BD = DC and ∠BDA = ∠CDA
But,
∠BDA + ∠CDA = 1800
∠BDA = ∠CDA = 900
Therefore,
AD bisects BC perpendicularly
Hence, proved
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