Frank Solutions for Chapter 12 Isosceles Triangles Class 9 Mathematics ICSE
Exercise 12.1
1. Find the angles of an isosceles triangle whose equal angles and the non-equal angles are in the ratio 3:4.
Answer:
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,
Sum of angles of a triangle = 180°
Hence,
3x + 3x + 4x = 180°
⇒ 10x = 180°
We get,
x = 18°
Therefore,
3x = 3 × 18° = 54° and
4x = 4 × 18° = 72°
Hence,
The angles of a triangle are 54°, 54° and 72°
2. Find the angles of an isosceles triangle which are in the ratio 2:2:5
Answer
Let equal angles be 2x each
So, non-equal angle is 5x
We know that,
Sum of angles of a triangle = 180°
2x + 2x + 5x = 180°
9x = 180°
x = 20°
Therefore,
2x = 2 × 20° = 40°
5x = 5 × 20° = 100°
Hence, the angles of a triangle are 40°, 40° and 100°
3. Each equal angle of an isosceles triangle is less than the third angle by 15°. Find the angles.
Answer
Therefore, non-equal angle is x + 15°
We know that,
Sum of angles of a triangle = 180°
x + x + (x + 15°) = 180°
⇒ 3x + 15° = 180°
⇒ 3x = 180° – 15°
⇒ 3x = 165°
We get,
x = 55°
So,
(x + 15°) = 55° + 15° = 70°
Hence, the angles of a triangle are 55°, 55° and 70°
4. Find the interior angles of the following triangles
(a)
(a)
∠A = 110°
AB = AC
∠C = ∠B (angles opposite to two equal sides are equal)
Now,
By angle sum property,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + ∠B = 180°
⇒ 110° + 2∠B = 180°
⇒ 2∠B = 180° – 110°
⇒ 2∠B = 70°
We get,
∠B = 35°
∠C = 35°
Hence,
The interior angles are ∠B = 35° and ∠C = 35°
(b)
In △ABC,
AB = AC
∠ACB = ∠ABC ...(1) [∵ angles opposite to two equal sides are equal]
Now,
∠ACB + ∠ACD = 180° [linear pair]
∠ACB = 180° – ∠ACD
∠ACB = 180° – 105°
∠ACB = 75°
So,
∠ABC = 75° [from equation (1)]
Now, in △ABC,
By angle sum property,
∠ABC + ∠ACB + ∠BAC = 180°
75° + 75° + ∠BAC = 180°
150° + ∠BAC = 180°
∠BAC = 180° – 150°
We get,
∠BAC = 30°
Hence,
In △ABC,
∠A = 30°
∠B = 75°
∠C = 75°
(c)
In △ABD,
Given that,
AD = BD
∠ABD = ∠BAD …(angles opposite to two equal sides are equal)
Now,
∠ABD = 37° …(given)
Hence,
∠BAD = 37°
By exterior angle property,
∠ADC = ∠ABD + ∠BAD
∠ADC = 37° + 37°
We get,
∠ADC = 74°
In △ADC,
AC = DC …(given)
∠ADC = ∠DAC …(angles opposite to two equal sides are equal)
∠DAC = 74°
Now,
∠BAC = ∠BAD + ∠DAC
∠BAC = 37° + 74°
We get,
∠BAC = 111°
In △ABC,
∠BAC + ∠ABC + ∠ACB = 180°
111° + 37° + ∠ACB = 180°
∠ACB = 180° – 111° – 37°
We get,
∠ACB = 32°
Therefore,
The interior angles of △ABC are 37°, 111° and 32°
(d)
AD = CD …(given)
∠ACD = ∠CAD …(angles opposite to two equal sides are equal)
Now,
∠ACD = 50° …(given)
∠CAD = 50°
By exterior angle property,
∠ADB = ∠ACD + ∠CAD
∠ADB = 500 + 500
∠ADB = 1000
In △ADB,
AD = BD …(given)
∠DBA = ∠DAB …(angles opposite to two equal sides are equal)
Also,
∠ADB + ∠DBA + ∠DAB = 180°
⇒ 100° + 2∠DBA = 180°
⇒ 2∠DBA = 180° – 100°
⇒ 2∠DBA = 80°
We get,
∠DBA = 40°
∠DAB = 40°
∠BAC = ∠DAB + ∠CAD
⇒ ∠BAC = 40° + 50°
⇒ ∠BAC = 90°
Therefore, the interior angles of △ABC are 50°, 90° and 40°
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.
Let ∠ABC = x
Hence,
∠BCA = x (since AB = AC)
In △ABC,
∠ABC + ∠BCA + ∠BAC = 180° …(1)
But
∠BAC + ∠DAC = 180° …(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 △ADC,
∠ADC + ∠DCA + ∠DAC = 180° …(3)
But ∠BAC + ∠DAC = 180° …(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 = 180°
⇒ x + y = 90°
⇒ ∠BCA + ∠DCA = 90°
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.
Join PO and produce to meet QR at point S
In △PQS and △PRS
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.
Answer
Let D and E be the mid points of AB and AC respectively
Now,
Join BE and CD
Then BE and CD become the medians of this isosceles triangle
In △ABE and △ACD
AB = AC (given)
AD = AE (D and E are mid points 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
In △QDP,
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 = 50° and ∠R = 110°, find ∠PSR.
In △PQS,
PQ = PS
Therefore,
∠PQS = ∠PSQ (angles opposite to two equal sides are equal)
∠P + ∠PQS + ∠PSQ = 180°
⇒ 50° + 2∠PQS = 180°
⇒ 2∠PQS = 180° – 50°
We get,
2∠PQS = 130°
⇒ ∠PQS = 65°
So,
∠PQS = ∠PSQ = 65° …(1)
In △SRQ,
SR = RQ
Hence,
∠RQS = ∠RSQ (angles opposite to two equal sides are equal)
∠R + ∠RQS + ∠RSQ = 180°
⇒ 110° + 2∠RQS = 180°
⇒ 2∠RQS = 180° – 110°
We get,
2∠RQS = 70°
∠RQS = 35°
So,
∠RQS = ∠RSQ = 35° ….(2)
Adding equations (1) and (2), we get,
∠PSQ + ∠RSQ = 65° + 35°
∠PSR = 100°
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 70°, find the measures of ∠DBC and ∠BAC.
In △BDC,
∠BDC = 70°
BD = BC
Hence,
∠BDC = ∠BCD (angles opposite to two equal sides are equal)
∠BCD = 70°
Now,
∠BCD + ∠BDC + ∠DBC = 180°
⇒ 70° + 70° + ∠DBC = 180°
⇒ ∠DBC = 180° – 140°
We get,
∠DBC = 40°
∠DBC = ∠ABC (BC is the angle bisector)
Hence,
∠ABC = 40°
In △ABC,
Since AB = AC, ∠ABC = ∠ACB
Hence,
∠ACB = 40°
⇒ ∠ACB + ∠ABC + ∠BAC = 180°
⇒ 40° + 40° + ∠BAC = 180°
⇒ ∠BAC = 180° – 80°
⇒ ∠BAC = 100°
Therefore, the measure of ∠BAC = 100° and ∠DBC = 40°
11. △PQR 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
(i) In △PQR,
PQ = PR
Hence,
∠R = ∠Q …(1)
Now,
In △QNT and △RMT
∠QNT = ∠RMT = 90°
∠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 △PNT and △PMT
TN = TM (proved)
PT = PT (common)
∠PNT = ∠PMT = 90°
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
Answer
PQ = QR (given)
∠PRQ = ∠QPR …(1)
In △PRS,
PR = RS (given)
∠PSR = ∠RPS …(2)
Now,
Adding equations (1) and (2), we get,
∠QPR + ∠RPS = ∠PRQ + ∠PSR
∠QPS = ∠PRQ + ∠PSR …(3)
Now,
In △PRS,
∠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.
In △KTM,
KT = TM (given)
Hence,
∠TKM = ∠TMK …(1)
Now,
∠KTL = ∠TKM + ∠TMK
⇒ 80° = ∠TKM + ∠TKM …[from (1)]
⇒ 80° = 2∠TKM
We get,
∠TKM = 40° = ∠TMK = ∠LMK …(2)
But,
∠TKM = ∠TKL (KT is the angle bisector)
Hence,
∠TKL = 40°
In △KTL,
∠TKL + ∠KTL + ∠KLT = 180°
⇒ 40° + 80° + ∠KLT = 180°
⇒ ∠KLT = 180° – 40° – 80°
We get,
∠KLT = 600 = ∠KLM
Therefore,
∠KLM = 60° and ∠LMK = 40°
14. Equal sides QP and RP of an isosceles △PQR are produced beyond P to S and T such that △PST is an isosceles triangle with PS = PT. Prove that TQ = SR.
In △PTQ and △PSR
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.
Answer
AB = AC (given)
AD = AD (common)
∠BAD = ∠CAD (AD bisects ∠BAC)
Hence,
△ADB ≅ △ADC
Therefore,
BD = DC and ∠BDA = ∠CDA
But,
∠BDA + ∠CDA = 180°
∠BDA = ∠CDA = 90°
Therefore,
AD bisects BC perpendicularly
Hence, proved
16. In the figure △ABC is isosceles with AB = AC. Prove that:
(i) ∠A : ∠B = 1 : 3
(ii) ∠ADE = ∠BCD
In △DEC,
∠DEC = ∠ADE + ∠A = 2a (ext. Angle to △ADE)
DE = DC
⇒ ∠DEC = ∠DEC = 2a …(ii)
In △BDC, let ∠B = b
DC = BC
⇒ ∠BDC = ∠B = b …(iii)
In △ABC,
∠ADB = ∠ADE + ∠EDC + ∠BDC
180° = a + ∠EDC + b (from (i) and (iii))
∠EDC = 180° - a – b …(iv)
Now again in △DEC
180° = ∠EDC + ∠DCE + ∠DEC (from (ii))
⇒ 180° = ∠EDC + 2a + 2a
⇒ ∠EDC = 180° - 4a …(v)
Equating (iv) and (v)
180° - a – b = 180° - 4a
⇒ 3a = b …(vi)
⇒ a/b = 1/3 = ∠A/∠B
Hence, ∠A : ∠B = 1 : 3
(ii) In △ADE,
∠ADE = a (from (i))
In △BCD,
∠BCE = b (AB = AC ⇒ ∠B = ∠C) …(vii)
⇒ ∠BCD = ∠BCE - ∠DCE
⇒ ∠BCD = b – 2a (from (vii) and (ii))
But b = 3a (from (vi))
Therefore, ∠BCD = 3a – 2a = a
Hence, ∠ADE = ∠BCD
17. In △ABC, D is mid-point of BC, AD is equal to AC. AC is produced to E, such that CE = AC. Prove that :
(i) ∠ADB = ∠DCE
(ii) AB = CE
(i) In △ADC,
AD = AC (given)
Therefore, ∠ADC = ∠ACD …(i)
But ∠ADB + ∠ADC = 180° …(ii)
And ∠ACD + ∠DCE = 180° …(iii)
From (i), (ii) and (iii)
∠ADB = ∠DCE
(ii) In △ABD and △DCE
BD = CD (D is mid-point of BC)
∠ADB = ∠DCE (proved)
AD = CE (since AC = AC and AC = CE)
Therefore, △ABD ≅ △DCE
Hence, AB = CE
18. In △XYZ, AY and AZ are the bisectors of ∠Y and ∠Z respectively. The perpendicular bisectors of AY and AZ cut YZ at B and C respectively. Prove that line segment YZ is equal to the perimeter of △ABC.
In △ABM and △BMY
MY = MA (BM bisects AY)
BM = BM (common)
∠BMY = ∠BMA
Therefore, △ABM ≅ △BMY
Hence, YB = AB …(i)
In △ACN and △CNZ
NZ = NA (CN bisects AZ)
CN = CN (common)
Therefore, △ACN ≅ △CNZ
Hence, CZ = AC …(ii)
YZ = YB + BC + CZ
Substituting from (i) and (ii)
YZ = YB + BC + AC
Hence, YZ is equal to the perimeter of △ABC
19. △PQR is an isosceles triangle with PQ = PR. QR is extended to S and ST is drawn perpendicular to QP produced, and SN is perpendicular to PR produced. Prove that QS bisects ∠TSN.
In △PQR, let ∠PQR = x
PQ = PR
⇒ ∠PQR = ∠PRQ = x …(i)
In △RNS,
∠NRS = ∠PRQ = x (vertically opposite angles)
∠RNS = 90° (given)
⇒ ∠NSR + ∠RNS + ∠NRS = 180°
⇒ ∠NSR = 90° - x …(ii)
Now in Quadrilateral PTRS
∠PTS = 90° (given)
⇒ ∠TPR = ∠PQR + ∠PRQ = 2x (exterior angle to triangle PQR)
⇒ ∠PRS = 180° - ∠PRQ = 180° - x (QRS is a st. Line)
⇒ ∠PTS + ∠TPR + ∠PRS + ∠TSR = 360° (angles of a quad. = 360°)
⇒ 90° + 2x + 180° - x + ∠TSR = 360°
⇒ ∠TSR = 90° - x …(iii)
From (ii) and (iii)
∠TSR = ∠NSR
Therefore, QS bisects ∠TSN.
20. In the given figure, D and E are points on AB and AC respectively. AE and CD intersect at P such that AP = CP. If ∠BAE = ∠BCD, prove that DBDE is isosceles.
In △APD and △EPC,
∠DAP = ∠ECP ...(∵ ∠BAE = ∠BCD)
AP = CP …(given)
∠APD = ∠EPC …(vertically opposite angles)
∴ △APD ≅ △EPC …(By ASA Congruence criterion)
⇒ AD = EC …(c.p.c.t)
In △APC,
AP = CP …(given)
⇒ ∠PAC = ∠PCA …(angles opposite to two equal sides are equal)
Now, ∠BAE = ∠BCD and ∠PAC = ∠PCA
⇒ ∠BAC = ∠BCA
⇒ BC = BA ...(sides opposite to two equal angles are equal)
⇒ BE + EC = BD + DA
⇒ BE = BD (∵ EC = DA)
⇒ ∠BDE = ∠BED …(angles opposite to two equal sides are equal)
⇒ △BDE is an isosceles triangle.
21. In DPQR as shown, ∠PQS = ∠RQS and QS ⊥ PR. Find the value of x and y, if PQ = 3x + 1; QR = 5y – 2; PS = x + 1 and SR = y + 2.
QS = QS …[common]
∠QSP = ∠QSR …[each = 90°]
∠PQS = ∠RQS …[given]
∴ △PQS ≅ △SQR …[By ASA criterion]
⇒ PS = RS
⇒ x + 1 = y + 2
⇒ x = y + 1 …(i)
And PQ = SQ
⇒ 3x + 1 = 5y – 2
⇒ 3(y + 1) + 1 = 5y – 2 …[From (i)]
⇒ 3y + 3 + 1 = 5y – 2
⇒ 2y = 6
⇒ y = 3
Putting y = 3 in (i),
x = y + 1 = 3 + 1 = 4
22. If ∠PQS = 60°,
a. find ∠QPR.
b. show that PQ = PS = QS = SR
Answer
(a) In △PQS,
PQ = QS …(given)
⇒ ∠QSP = ∠QPS …(angles opposite to two equal sides are equal)
Now, ∠PQS + ∠QSP + ∠QPS = 180°
⇒ 60° + 2∠QSP = 180°
⇒ 2∠QSP = 120 °
⇒ ∠QSP = 60°
⇒ QPS = 60°
In △PRS,
PS = SR …(given)
⇒ ∠PRS = ∠RPS …(angles opposite to two equal sides are equal)
By exterior angle property,
∠QSP = ∠RPS + ∠PRS
⇒ 60° = 2∠RPS
⇒ ∠RPS = 30°
Now ∠QPR = ∠QPS + ∠RPS = 60° + 30° = 90°
(b) In △PQS,
∠PQS = 60°, ∠QPS = 60° and ∠QSP = 60°
⇒ △PQS is an equilateral triangle.
⇒ PQ = QS = PS
And, PS = SR
⇒ PQ = PS = QS = SR
23. In the given figure, if DPQR is an isosceles triangle, prove that : ∠QSR = exterior ∠PRT.
In △PQR,
∠QPR + ∠PQR + ∠PRQ = 180°
⇒ ∠QPR + 2x + 2y = 180°
⇒ ∠QPR + 2x + 2y = 180°
⇒ ∠QPR = 180° - 2x – 2y …(i)
Since PQ = PR,
∠PRQ = ∠PQR …(angles opposite to two equal sides are equal)
⇒ 2x = 2y
⇒ x = y
Now, ∠PRT = ∠PQR + ∠QPR …(by exterior angle property)
⇒ ∠PRT = 2x + 180° - 2x – 2y …[From (i)]
⇒ ∠PRT = 180° - 2y …(ii)
In △SQR,
∠QSR + ∠SQR + ∠SRQ = 180
⇒ ∠QSR + x + y = 180°
⇒ ∠QSR = 180° - x – y
⇒ ∠QSR = 180° - y – y ….[∵ x = y (proved)]
⇒ ∠QSR = 180° - 2y …(iii)
From (ii) and (iii),
∠QSR = ∠PRT
24. In the given figure, if DABC is an isosceles triangle and ∠PAC = 110°, find the base angle and vertex angle of the DABC.
To find:
Base angles : ∠ABC and ∠ACB
Vertex angle: ∠BAC
In quadrilateral APQC,
∠APQ + ∠PQC + ∠ACQ + ∠PAC = 360°
⇒ 90° + 90° + ∠QCA + 110° = 360°
⇒ ∠ACQ = 360° - 290°
⇒ ∠ACQ = 70°
⇒ ∠ACB = 70° …(i)
In △ABC,
AB = AC …(given)
⇒ ∠ACB = ∠ABC ...(angles opposite to two equal sides are equal)
⇒ ∠ABC = 70° …[From (i)]
In △ABC,
∠ABC + ∠ACB + ∠BAC = 180° …(angle sum property)
⇒ 70° + 70° + ∠BAC = 180°
⇒ ∠BAC = 180° - 140°
⇒ ∠BAC = 40°