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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
In the figure, it is given that AB = CD and AD = BC. Prove that ∆ADC ≅ ∆CBA.

Solution:
Given : In the figure, AB = CD, AD = BC

To prove : ∆ADC = ∆CBA
Proof : In ∆ADC and ∆CBA
CD = AB (Given)
AD = BC (Given)
CA = CA (Common)
∴ ∆ADC ≅ ∆CBA (SSS axiom)

Question 2.
In a APQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Given : In ∆PQR, PQ = QR
L, M and N are the mid-points of sides PQ, QR and RP respectively. Join LM, MN and LN

To prove : ∠PNM = ∠PLM
Proof : In ∆PQR,
∵ M and N are the mid points of sides PR and QR respectively
∴ MN || PQ and MN = \(\frac { 1 }{ 2 }\) PQ …(i)
∴ MN = PL
Similarly, we can prove that
LM = PN
Now in ∆NML and ∆LPN
MN = PL (Proved)
LM = PN (Proved)
LN = LN (Common)
∴ ∆NML = ∆LPN (SSS axiom)
∴ ∠MNL = ∠PLN (c.p.c.t.)
and ∠MLN = ∠LNP (c.p.c.t.)
⇒ ∠MNL = ∠LNP = ∠PLM = ∠MLN
⇒ ∠PNM = ∠PLM

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

Mark the correct alternative in each of the following:
Question 1.
If all the three angles of a triangle are equal, then each one of them is equal to
(a) 90°
(b) 45°
(c) 60°
(d) 30°
Solution:
∵ Sum of three angles of a triangle = 180°
∴ Each angle = \(\frac { { 180 }^{ \circ } }{ 3 }\)  = 60° (c)

Question 2.
If two acute angles of a right triangle are equal, then each acute is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
In a right triangle, one angle = 90°
∴ Sum of other two acute angles = 180° – 90° = 90°
∵ Both angles are equal
∴ Each angle will be = \(\frac { { 90 }^{ \circ } }{ 2 }\)  = 45° (b)

Question 3.
An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to
(a) 75°
(b) 80°
(c) 40°
(d) 50°
Solution:
In a triangle, exterior angles is equal to the sum of its interior opposite angles
∴ Sum of interior opposite angles = 100°
∵ Both angles are equal
∴ Each angle will be = \(\frac { { 100 }^{ \circ } }{ 2 }\)  = 50° (d)

Question 4.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:
Let ∠A, ∠B, ∠C be the angles of a ∆ABC and let ∠A = ∠B + ∠C

But ∠A + ∠B + ∠C = 180°
( Sum of angles of a triangle)
∴ ∠A + ∠A = 180° ⇒ 2∠A = 180°
⇒ ∠A = \(\frac { { 180 }^{ \circ } }{ 2 }\)  = 90°
∴ ∆ is a right triangle (d)

Question 5.
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = \(\frac { 1 }{ 2 }\)∠A, then ∠A is equal to
(a) 80°
(b) 75°
(c) 60°
(d) 90°
Solution:
Side BC of ∆ABC is produced to D, then

Ext. ∠ACB = ∠A + ∠B
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)

Question 6.
In ∆ABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
(a) 35°
(b) 90°
(c) 70°
(d) 55°
Solution:
In ∆ABC, ∠B = ∠C
AX is the bisector of ext. ∠CAD
∠DAX = 70°

∴ ∠DAC = 70° x 2 = 140°
But Ext. ∠DAC = ∠B + ∠C
= ∠C + ∠C (∵ ∠B = ∠C)
= 2∠C
∴ 2∠C = 140° ⇒ ∠C = \(\frac { { 140 }^{ \circ } }{ 2 }\) = 70°
∴ ∠ACB = 70° (c)

Question 7.
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is
(a) 55°
(b) 85°
(c) 40°
(d) 9.0°
Solution:
In ∆ABC, BA is produced to D such that ∠CAD = 95°
and let ∠C = 55° and ∠B = x°

∵ Exterior angle of a triangle is equal to the sum of its opposite interior angle
∴ ∠CAD = ∠B + ∠C ⇒ 95° = x + 55°
⇒ x = 95° – 55° = 40°
∴ Other interior angle = 40° (c)

Question 8.
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
(a) 90°
(b) 180°
(c) 270°
(d) 360°
Solution:
In ∆ABC, sides AB, BC and CA are produced in order, then exterior ∠FAB, ∠DBC and ∠ACE are formed

We know an exterior angles of a triangle is equal to the sum of its interior opposite angles
∴ ∠FAB = ∠B + ∠C
∠DBC = ∠C + ∠A and
∠ACE = ∠A + ∠B Adding we get,
∠FAB + ∠DBC + ∠ACE = ∠B + ∠C + ∠C + ∠A + ∠A + ∠B
= 2(∠A + ∠B + ∠C)
= 2 x 180° (Sum of angles of a triangle)
= 360° (d)

Question 9.
In ∆ABC, if ∠A = 100°, AD bisects ∠A and AD⊥ BC. Then, ∠B =
(a) 50°
(b) 90°
(c) 40°
(d) 100°
Solution:
In ∆ABC, ∠A = 100°
AD is bisector of ∠A and AD ⊥ BC

Now, ∠BAD = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50°
In ∆ABD,
∠BAD + ∠B + ∠D= 180°
(Sum of angles of a triangle)
⇒ ∠50° + ∠B + 90° = 180°
∠B + 140° = 180°
⇒ ∠B = 180° – 140° ∠B = 40° (c)

Question 10.
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4:5. The angles of the triangle are
(a) 48°, 60°, 72°
(b) 50°, 60°, 70°
(c) 52°, 56°, 72°
(d) 42°, 60°, 76°
Solution:
In ∆ABC, BC is produced to D and ∠ACD = 108°

Ratio in ∠A : ∠B = 4:5
∵ Exterior angle of a triangle is equal to the sum of its opposite interior angles
∴ ∠ACD = ∠A + ∠B = 108°
Ratio in ∠A : ∠B = 4:5

Question 11.
In a ∆ABC, if ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC =
(a) 60°
(b) 120°
(c) 150°
(d) 30°
Solution:
In ∆ABC, ∠A = 60°, ∠B = 80°
∴ ∠C = 180° – (∠A + ∠B)
= 180° – (60° + 80°)
= 180° – 140° = 40°
Bisectors of ∠B and ∠C meet at O

Question 12.
Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 45° and ∠CDB = 55°, then ∠BOD =
(a) 100°
(b) 80°
(c) 90°
(d) 135°
Solution:
In the figure,

AB and CD intersect at O
and AC || DB, ∠CAB = 45°
and ∠CDB = 55°
∵ AC || DB
∴ ∠CAB = ∠ABD (Alternate angles)
In ∆OBD,
∠BOD = 180° – (∠CDB + ∠ABD)
= 180° – (55° + 45°)
= 180° – 100° = 80° (b)

Question 13.
In the figure, if EC || AB, ∠ECD = 70° and ∠BDO = 20°, then ∠OBD is

(a) 20°
(b) 50°
(c) 60°
(d) 70°
Solution:
In the figure, EC || AB
∠ECD = 70°, ∠BDO = 20°

∵ EC || AB
∠AOD = ∠ECD (Corresponding angles)
⇒ ∠AOD = 70°
In ∆OBD,
Ext. ∠AOD = ∠OBD + ∠BDO
70° = ∠OBD + 20°
⇒ ∠OBD = 70° – 20° = 50° (b)

Question 14.
In the figure, x + y =
(a) 270
(b) 230
(c) 210
(d) 190°

Solution:
In the figure

Ext. ∠OAE = ∠AOC + ∠ACO
⇒ x = 40° + 80° = 120°
Similarly,
Ext. ∠DBF = ∠ODB + ∠DOB
y = 70° + ∠DOB
[(∵ ∠AOC = ∠DOB) (vertically opp. angles)]
= 70° + 40° = 110°
∴ x+y= 120°+ 110° = 230° (b)

Question 15.
If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?
(a) 25°
(b) 30°
(c) 45°
(d) 60°
Solution:
Ratio in the measures of the triangle =3:4:5
Sum of angles of a triangle = 180°
Let angles be 3x, 4x, 5x
Sum of angles = 3x + 4x + 5x = 12x
∴ Smallest angle = \(\frac { 180 x 3x }{ 12x }\) = 45° (c)

Question 16.
In the figure, if AB ⊥ BC, then x =
(a) 18
(b) 22
(c) 25
(d) 32

Solution:
In the figure, AB ⊥ BC
∠AGF = 32°
∴ ∠CGB = ∠AGF (Vertically opposite angles)
= 32°

In ∆GCB, ∠B = 90°
∴ ∠CGB + ∠GCB = 90°
⇒ 32° + ∠GCB = 90°
⇒ ∠GCB = 90° – 32° = 58°
Now in ∆GDC,
Ext. ∠GCB = ∠CDG + ∠DGC
⇒ 58° = x + 14° + x
⇒ 2x + 14° = 58°
⇒ 2x = 58 – 14° = 44
⇒ x = \(\frac { 44 }{ 2 }\) = 22°
∴ x = 22° (b)

Question 17.
In the figure, what is ∠ in terms of x and y?
(a) x + y + 180
(b) x + y – 180
(c) 180° -(x+y)
(d) x+y + 360°

Solution:
In the figure, BC is produced both sides CA and BA are also produced
In ∆ABC,
∠B = 180° -y
and ∠C 180° – x

∴ z = ∠A = 180° – (B + C)
= 180° – (180 – y + 180 -x)
= 180° – (360° – x – y)
= 180° – 360° + x + y = x + y – 180° (b)

Question 18.
In the figure, for which value of x is l₁ || l₂?
(a) 37
(b) 43
(c) 45
(d) 47

Solution:
In the figure, l₁ || l₂
∴ ∠EBA = ∠BAH (Alternate angles)
∴ ∠BAH = 78°

⇒ ∠BAC + ∠CAH = 78°
⇒ ∠BAC + 35° = 78°
⇒ ∠BAC = 78° – 35° = 43°
In ∆ABC, ∠C = 90°
∴ ∠ABC + ∠BAC = 90°
⇒ x + 43° = 90° ⇒ x = 90° – 43°
∴ x = 47° (d)

Question 19.
In the figure, what is y in terms of x?

Solution:
In ∆ABC,
∠ACB = 180° – (x + 2x)
= 180° – 3x …(i)
and in ∆BDG,
∠BED = 180° – (2x + y) …(ii)
∠EGC = ∠AGD (Vertically opposite angles)
= 3y

In quad. BCGE,
∠B + ∠ACB + ∠CGE + ∠BED = 360° (Sum of angles of a quadrilateral)
⇒ 2x+ 180° – 3x + 3y + 180°- 2x-y = 360°
⇒ -3x + 2y = 0
⇒ 3x = 2y ⇒ y = \(\frac { 3 }{ 2 }\)x (a)

Question 20.
In the figure, what is the value of x?
(a) 35
(b) 45
(c) 50
(d) 60

Solution:
In the figure, side AB is produced to D
∴ ∠CBA + ∠CBD = 180° (Linear pair)
⇒ 7y + 5y = 180°
⇒ 12y = 180°
⇒ y = \(\frac { 180 }{ 12 }\) = 15
and Ext. ∠CBD = ∠A + ∠C
⇒ 7y = 3y + x
⇒ 7y -3y = x
⇒ 4y = x
∴ x = 4 x 15 = 60 (d)

Question 21.
In the figure, the value of x is
(a) 65°
(b) 80°
(c) 95°
(d) 120°

Solution:
In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°

Now in ∆ABD,
Ext. ∠DBC = ∠A + ∠D
= 55° + 25° = 80°
Similarly, in ∆BCE,
Ext. ∠DEC = ∠EBC + ∠ECB
= 80° + 40° = 120° (d)

Question 22.
In the figure, if BP || CQ and AC = BC, then the measure of x is
(a) 20°
(b) 25°
(c) 30°
(d) 35°

Solution:
In the figure, AC = BC, BP || CQ

∵ BP || CQ
∴ ∠PBC – ∠QCD
⇒ 20° + ∠ABC = 70°
⇒ ∠ABC = 70° – 20° = 50°
∵ BC = AC
∴ ∠ACB = ∠ABC (Angles opposite to equal sides)
= 50°
Now in ∆ABC,
Ext. ∠ACD = ∠B + ∠A
⇒ x + 70° = 50° + 50°
⇒ x + 70° = 100°
∴ x = 100° – 70° = 30° (c)

Question 23.
In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then
(a) x = 55°, y = 40°
(b) x = 50°, y = 45°
(c) x = 60°, y = 35°
(d) x = 35°, y = 60°

Solution:
In the figure,
∵ AB || CD, EF intersects them at P and Q respectively,
∠APR = 25°, ∠RQC = 30°, ∠CQF = 65°
∵ AB || CD
∴ ∠APQ = ∠CQF (Corresponding anlges)
⇒ y + 25° = 65°
⇒ y = 65° – 25° = 40°
and APQ + PQC = 180° (Co-interior angles)

y + 25° + ∠1 +30°= 180°
40° + 25° + ∠1 + 30° = 180°
⇒ ∠1 + 95° = 180°
∴ ∠1 = 180° – 95° = 85°
Now, ∆PQR,
∠RPQ + ∠PQR + ∠PRQ = 180° (Sum of angles of a triangle)
⇒ 40° + x + 85° = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° = 55°
∴ x = 55°, y = 40° (a)

Question 24.
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94° and 126°. Then, ∠BAC = ?
(a) 94°
(b) 54°
(c) 40°
(d) 44°
Solution:
In ∆ABC, base BC is produced both ways and ∠ACD = 94°, ∠ABE = 126°

Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 94° = ∠BAC + ∠ABC
Similarly, ∠ABE = ∠BAC + ∠ACB
⇒ 126° = ∠BAC + ∠ACB
Adding,
94° + 126° = ∠BAC + ∠ABC + ∠ACB + ∠BAC
220° = 180° + ∠BAC
∴ ∠BAC = 220° -180° = 40° (c)

Question 25.
If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is
(a) 45°
(b) 95°
(c) 135°
(d) 90°
Solution:
In right ∆ABC, ∠A = 90°

Bisectors of ∠B and ∠C meet at O, then 1
∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A
= 90°+ \(\frac { 1 }{ 2 }\) x 90° = 90° + 45°= 135° (c)

Question 26.
The bisects of exterior angles at B and C of ∆ABC, meet at O. If ∠A = .x°, then ∠BOC=

Solution:
In ∆ABC, ∠A = x°
and bisectors of ∠B and ∠C meet at O.

Question 27.
In a ∆ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
(a) 25°
(b) 50°
(c) 100°
(d) 75°
Solution:
In ∆ABC, ∠A = 50°
BC is produced

Bisectors of ∠ABC and ∠ACD meet at ∠E
∴ ∠E = \(\frac { 1 }{ 2 }\) ∠A = \(\frac { 1 }{ 2 }\) x 50° = 25° (a)

Question 28.
The side BC of AABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD =115°,then ∠ALC =
(a) 85°
(b) 72\(\frac { 1 }{ 2 }\) °
(c) 145°
(d) none of these
Solution:
In ∆ABC, BC is produced to D
∠B = 30°, ∠ACD = 115°

Question 29.
In the figure , if l1 || l2, the value of x is
(a) 22 \(\frac { 1 }{ 2 }\)
(b) 30
(c) 45
(d) 60

Solution:
In the figure, l₁ || l₂
EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB

∵ EC and EB are the bisectors of ∠DCB and ∠CBA respectively
∴ ∠CEB = 90°
∴ a + b = 90° ,
and ∠GEB = 90° (∵ ∠CEB = 90°)
2x = 90° ⇒ x = \(\frac { 90 }{ 2 }\) = 45 (c)

Question 30.
In ∆RST (in the figure), what is the value of x?
(a) 40°
(b) 90°
(c) 80°
(d) 100°

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove the segment DE || BC.

Solution:
Given : Sides BA and CA of ∆ABC are produced such that BA = AD are CA = AE. ED is joined.
To prove : DE || BC
Proof: In ∆ABC and ∆DAE AB=AD (Given)
AC = AE (Given)
∠BAC = ∠DAE (Vertically opposite angles)
∴ ∆ABC ≅ ∆DAE (SAS axiom)
∴ ∠ABC = ∠ADE (c.p.c.t.)
But there are alternate angles
∴ DE || BC

Question 2.
In a ∆PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Given : In ∆PQR, PQ = QR
L, M and N are the mid points of the sides PQ, QR and PR respectively

To prove : LM = MN
Proof : In ∆LPN and ∆MRH
PN = RN (∵ M is mid point of PR)
LP = MR (Half of equal sides)
∠P = ∠R (Angles opposite to equal sides)
∴ ALPN ≅ AMRH (SAS axiom)
∴ LN = MN (c.p.c.t.)

Question 3.
Prove that the medians of an equilateral triangle are equal.
Solution:
Given : In ∆ABC, AD, BE and CF are the medians of triangle and AB = BC = CA

To prove : AD = BE = CF
Proof : In ∆BCE and ∆BCF,
BC = BC (Common side)
CE = BF (Half of equal sides)
∠C = ∠B (Angles opposite to equal sides)
∴ ABCE ≅ ABCF (SAS axiom)
∴ BE = CF (c.p.c.t.) …(i)
Similarly, we can prove that
∴ ∆CAD ≅ ∆CAF
∴ AD = CF …(ii)
From (i) and (ii)
BE = CF = AD
⇒ AD = BE = CF

Question 4.
In a ∆ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 120° and AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ 120° + ∠B + ∠B = 180°
⇒ 2∠B = 180° – 120° = 60°
∴ ∠B = \(\frac { { 60 }^{ \circ } }{ 2 }\) = 30°
and ∠C = ∠B = 30°
Hence ∠B = 30° and ∠C = 30°

Question 5.
In a ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Solution:
In ∆ABC, ∠B = 70°
AB =AC
∴ ∠B = ∠C (Angles opposite to equal sides)

But ∠B = 70°
∴ ∠C = 70°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 70° + 70° = 180°
⇒ ∠A + 140°= 180°
∴∠A = 180°- 140° = 40°

Question 6.
The vertical angle of an isosceles triangle is 100°. Find its base angles.
Solution:
In ∆ABC, AB = AC and ∠A = 100°
But AB = AC (In isosceles triangle)

∴ ∠C = ∠B (Angles opposite to equal sides)
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 100° + ∠B + ∠B = 180° (∵ ∠C = ∠B)
⇒ 2∠B = 180° – 100° = 80°
∴ ∠C = ∠B = 40°
Hence ∠B = 40°, ∠C = 40°

Question 7.
In the figure, AB = AC and ∠ACD = 105°, find ∠BAC.
Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ ∠ACB + 105°= 180°
⇒ ∠ACB = 180°-105° = 75°
∴ ∠ABC = ∠ACB = 75°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 75° + 75° = 180°
⇒ ∠A + 150°= 180°
⇒ ∠A= 180°- 150° = 30°
∴ ∠BAC = 30°

Question 8.
Find the measure of each exterior angle of an equilateral triangle.
Solution:
In an equilateral triangle, each interior angle is 60°

But interior angle + exterior angle at each vertex = 180°
∴ Each exterior angle = 180° – 60° = 120°

Question 9.
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Solution:
Given : In an isosceles ∆ABC, AB = AC
and base BC is produced both ways

To prove : ∠ACD = ∠ABE
Proof: In ∆ABC,
∵ AB = AC
∴∠C = ∠B (Angles opposite to equal sides)
⇒ ∠ACB = ∠ABC
But ∠ACD + ∠ACB = 180° (Linear pair)
and ∠ABE + ∠ABC = 180°
∴ ∠ACD + ∠ACB = ∠ABE + ∠ABC
But ∠ACB = ∠ABC (Proved)
∴ ∠ACD = ∠ABE
Hence proved.

Question 10.
In the figure, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.

Solution:
In the given figure,
In ∆ABC,
AB = AC and DB = DC
In ∆ABC,
∵ AB = AC
∴ ∠ACD = ∠ABE …(i) (Angles opposite to equal sides)
Similarly, in ∆DBC,
DB = DC
∴ ∠DCB = ∠DBC .. (ii)
Subtracting (ii) from (i)
∠ACB – ∠DCB = ∠ABC – ∠DBC
⇒ ∠ACD = ∠ABD
∴ Ratio ∠ABD : ∠ACD = 1 : 1

Question 11.
Determine the measure of each of the equal angles of a rightangled isosceles triangle.
OR
ABC is a rightangled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
Given : In a right angled isosceles ∆ABC, ∠A = 90° and AB = AC
To determine, each equal angle of the triangle

∵ ∠A = 90°
∴ ∠B + ∠C = 90°
But ∠B = ∠C
∴ ∠B + ∠B = 90°
⇒ 2∠B = 90°
90°
⇒ ∠B = \(\frac { { 90 }^{ \circ } }{ 2 }\)  = 45°
and ∠C = ∠B = 45°
Hence ∠B = ∠C = 45°

Question 12.
In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°

Solution:
Given : PQRS is a square and SRT is an equilateral triangle. PT and QT are joined.

To prove : (i) PT = QT; (ii) ∠TQR = 15°
Proof : In ∆TSP and ∆TQR
ST = RT (Sides of equilateral triangle)
SP = PQ (Sides of square)
and ∠TSP = ∠TRQ (Each = 60° + 90°)
∴ ∆TSP ≅ ∆TQR (SAS axiom)
∴ PT = QT (c.p.c.t.)
In ∆TQR,
∵ RT = RQ (Square sides)
∠RTQ = ∠RQT
But ∠TRQ = 60° + 90° = 150°
∴ ∠RTQ + ∠RQT = 180° – 150° = 30°
∵ ∠PTQ = ∠RQT (Proved)
∠RQT = \(\frac { { 30 }^{ \circ } }{ 2 }\)  = 15°
⇒ ∠TQR = 15°

Question 13.
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the ponits A and B (see figure). Show that the line PQ is perpendicular bisector of AB.

Solution:
Given : AB is a line segment.
P and Q are points such that they are equidistant from A and B
i.e. PA = PB and QA = QB AP, PB, QA, QB, PQ are joined
To prove : PQ is perpendicular bisector of AB

Proof : In ∆PAQ and ∆PBQ,
PA = PB (Given)
QA = QB (Given)
PQ = PQ (Common)
∴ ∆PAQ ≅ ∆PBQ (SSS axiom)
∴ ∠APQ = ∠BPQ (c.p.c.t.)
Now in ∆APC = ∆BPC
PA = PB (Given)
∆APC ≅ ∆BPC (Proved)
PC = PC (Common)
∴ ∆APC = ∆BPC (SAS axiom)
∴ AC = BC (c.p.c.t.)
and ∠PCA = ∠PCB (c.p.c.t.)
But ∠PCA + ∠PCB = 180° (Linear pair)
∴ ∠PCA = ∠PCB = 90°
∴ PC or PQ is perpendicular bisector of AB

 

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 are helpful to complete your math homework.

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