RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

Other Exercises

Question 1.
If ABC and BDE are two equilateral triangles such that D is the mid-ponit of BC, then find ar(∆ABC) : ar(∆BDE).
Solution:
ABC and BDE are two equilateral triangles and D is the mid-point of BC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q1.1
Let each side of AABC = a
Then BD = \(\frac { a }{ 2 }\)
∴ Each side of triangle BDE will be \(\frac { a }{ 2 }\)
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q1.2

Question 2.
In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q2.1
Solution:
In rectangle ABCD,
CD = 6 cm, AD = 8 cm
∴ Area of rectangle ABCD = CD x AD
= 6 x 8 = 48 cm2
∵ DC || AB and AB is produced to F and DC is produced to G
∴ DG || AF
∵ Rectangle ABCD and ||gm CDEF are on the same base CD and between the same parallels
∴ ar(||gm CDEF) = ar(rect. ABCD)
= 48 cm2

Question 3.
In the figure of Q. No. 2, find the area of ∆GEF.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q3.1
Solution:
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q3.2

Question 4.
In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ∆EFG.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q4.1
Solution:
ABCD is a rectangle in which
AB = 10 cm, AD = 5 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q4.2
∵ ABCD is a rectangle
∴DC || AB,
DC is produced to E and AB is produced to G
∴DE || AG
∵ Rectangle ABCD and ||gm ABEF are on the same base AB and between the same parallels
∴ ar(rect. ABCD) = ar(||gm ABEF)
= AB x AD = 10 x 5 = 50 cm2
Now ||gm ABEF and AEFG are on the same
base EF and between the same parallels
∴ area ∆EFG = \(\frac { 1 }{ 2 }\) ar(||gm ABEF)
= \(\frac { 1 }{ 2 }\) x 50 = 25 cm2

Question 5.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find or(∆RAS).
Solution:
In quadrant PLRM, rectangle PQRS is in scribed
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q5.1
Radius of the circle = 13 cm
A is any point on PQ
AR and AS are joined, PS = 5 cm
In right ∆PRS,
PR2 = PS2 + SR2
⇒ (132 = (5)2+ SR2
⇒ 169 = 25 + SR2
⇒ SR2 = 169 – 25 = 144 = (12)2
∴ SR = 12 cm
Area of rect. PQRS = PS x SR = 5x 12 = 60 cm2
∵ Rectangle PQRS and ARAS are on the same
base SR and between the same parallels
∴ Area ARAS = \(\frac { 1 }{ 2 }\) area rect. PQRS 1
= \(\frac { 1 }{ 2 }\) x 60 = 30 cm2

Question 6.
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8 cm and PQ and BD intersect at O, then find area of ∆OPB.
Solution:
In sq. ABCD, P and Q are the mid points of sides AB and CD respectively PQ and BD are joined which intersect each other at O
Side of square AB = 8 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q6.1
∴ Area of square ABCD = (side)2
∵ Diagonal BD bisects the square into two triangle equal in area
∴ Area ∆ABD = \(\frac { 1 }{ 2 }\) x area of square ABCD
= \(\frac { 1 }{ 2 }\) x 64 = 32 cm2
∵ P is mid point of AB of AABD, and PQ || AD
∴ O is the mid point of BD
∴ OP = \(\frac { 1 }{ 2 }\)AD = \(\frac { 1 }{ 2 }\) x 8 = 4 cm
and PB = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 8 = 4 cm
∴ Area ∆OPB = \(\frac { 1 }{ 2 }\)PB x OP
= \(\frac { 1 }{ 2 }\) x4x4 = 8 cm2

Question 7.
ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. If area of ∆ABC is 16 cm2, find the area of ∆DEF.
Solution:
In ∆ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of ∆ABC = 16 cm2
FD is joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q7.1
∵ D is mid point of BC
∴ AD is the median and median divides the triangle into two triangles equal in area
area ∆ADC = \(\frac { 1 }{ 2 }\) ar(∆ABC)
= \(\frac { 1 }{ 2 }\) x 16 = 8 cm2
Similarly, E is mid point of DC
∴ area (∆ADE) = \(\frac { 1 }{ 2 }\) ar(∆ADC)
= \(\frac { 1 }{ 2 }\) x 8 = 4 cm2
∵ F is mid point of AE of ∆ADE
∴ ar(∆DEF) = \(\frac { 1 }{ 2 }\)area (∆ADE)
= \(\frac { 1 }{ 2 }\) x 4 = 2 cm2

Question 8.
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ∆PBQ is 17 cm2, find the area of ∆ASR.
Solution:
In trapezium PQRS,
PS || QR
A and B are points on sides PQ and SR
Such that AB || QR
area of ∆PBQ = 17 cm2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q8.1
∆ABQ and ∆ABR are on the same base AB and between the same parallels
∴ ar(∆ABQ) = ar(∆ABR) …(i)
Similarly, ∆ABP and ∆ABS are on the same base and between the same parallels
∴ ar(ABP) = ar(∆ABS) …(ii)
Adding (i) and (ii)
ar( ∆ABQ) + ar( ∆ABP)
= ar(∆ABR) + ar(∆ABS)
⇒ ar(∆PBQ) = ar(∆ASR)
Put ar(PBQ) = 17 cm2
∴ ar(∆ASR) = 17 cm2

Question 9.
ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3 : 1. If ar(∆PBQ) = 10 cm2, find the area of parallelogram ABCD.
Solution:
In ||gm ABCD, P is mid point on AB,
PC and BD intersect each other at Q
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q9.1
CQ : QP = 3 : 1
ar(∆PBQ) = 10 cm2
In ||gm ABCD,
BD is its diagonal
∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\) ar ||gm ABCD
∴ ar(||gm ABCD) = 2ar(∆ABD) …(i)
In ∆PBC CQ : QP = 3 : 1
∵ ∆PBQ and ∆CQB have same vertice B
∴ 3 x area ∆PBQ = ar(∆CBQ)
⇒ area(∆CBQ) = 3 x 10 = 30 cm2
∴ ar(∆PBC) = 30 + 10 = 40 cm2
Now ∆ABD and ∆PBC are between the
same parallel but base PB = \(\frac { 1 }{ 2 }\) AB
∴ ar(∆ABD) = 2ar(∆PBC)
= 2 x 40 = 80 cm2
But ar(||gm ABCD) = 2ar(∆ABD)
= 2 x 80 = 160 cm2

Question 10.
P is any point on base BC of ∆ABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(∆ABC) = 12 cm2, then find area of ∆EPC.
Solution:
P is any point on base of ∆ABC
D is mid point of BC
DE || PA drawn which meet AC at E
ar(∆ABC) = 12 cm2
AD and PE are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS Q10.1
∵ D is mid point of BC
∴ AD is median
∴ ar(∆ABD) = ar(∆ACD)
= \(\frac { 1 }{ 2 }\) (∆ABC) = \(\frac { 1 }{ 2 }\) x 12 = 6 cm2 …(i)
∵ ∆PED and ∆ADE are on the same base DE and between the same parallels
∴ ar(∆PED) = ar(∆ADE)
Adding ar(∆DCE) to both sides,
ar(∆PED) + ar(∆DCE) = ar(∆ADE) + ar(∆DCE)
ar(∆EPC) = ar(∆ACD)
⇒ ar(∆EPC) = ar(∆ABD) = 6 cm2 [From (i)]
∴ ar(∆EPC) = 6 cm2

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3

Other Exercises

Question 1.
In the figure, compute the area of the quadrilateral.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q1.1
Solution:
In the quadrilateral ABCD,
∠A = 90°, ∠CBD = 90°, AD = 9 cm, BC = 8 cm and CD = 17 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q1.2
In right ∆BCD,
CD = BC2 + BD2 (Pythagoras Theorem)
⇒ (17)2 = (8)2 + BD2
⇒ 289 = 64 + BD2
⇒ BD2 = 289 – 64 = 225 = (15)2
∴ BD = 15 cm
Now in right ∆ABD,
BD2 = AB2 + AD2
⇒ (15)2 = AB2 + (9)2
⇒ 225 = AB2 + 81
⇒ AB2= 225 – 81 = 144 = (12)2
∴ AB = 12 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q1.3

Question 2.
In the figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ∆OTS if PQ = 8 cm.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q2.1
Solution:
In square PQRS, T and U are the mid-points of the sides PS and QR
TU, QS and US are joined
PQ = 8 cm
∴ T and U are mid-points of the opposites sides PS and QR
∴ TU || PQ TO || PQ
In RQS,
T is mid-point of PS and TO || PQ
∴ O is the mid point of SQ 1 1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q2.2

Question 3.
Compute the area of trapezium PQRS in the figure.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q3.1
Solution:
In ∆TQR, ∠RTQ = 90°
∴ QR2 = TQ2 + RT2
⇒ (17)2 = (8)2 + RT2
⇒ 289 = 64 + RT2
⇒ RT2 = 289 – 64 = 225 = (15)2
∴ RT = 15 cm
and PQ = 8 + 8 = 16 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q3.2

Question 4.
In the figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ∆AOB.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q4.1
Solution:
In ∆AOB, ∠AOB = 90°
C is a point on AB such that AC = BC Join OC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q4.2
Since C is the mid-point of hypotenuse of right ∆AOB
∴ AC = CB = OC = 6.5 cm
∴ AB = 6.5 + 6.5 = 13 cm
Now in right ∆AOB
⇒ AB2 = AO2 + OB(Pythagoras Theorem)
⇒ (13)2 = (12)2 + OB2
⇒ 169 = 144 + OB2
⇒ OB2 = 169 – 144 = 25 = (5)2
∴ OB = 5 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q4.3

Question 5.
In the figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q5.1
Solution:
In the trapezium ABCD,
AB = 7 cm
AL = BM = 4 cm
AD = BC = 5 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q5.2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q5.3

Question 6.
In the figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm.If OE = 2 \(\sqrt { 5 } \) , find the area of the rectangle.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q6.1
Solution:
Radius of the quadrant of circle = 2\(\sqrt { 5 } \) units
∴ OD diagonal of rectangle = 10 units (∵ OD = OB = OA = 10 cm)
DE = 2 \(\sqrt { 5 } \) cm
∴ In right ∆OED,
OD2 = OE2 + DEv
(10)2 = OE2 + (2\(\sqrt { 5 } \))2
100 = OE2 + 20
OE2 = 100 – 20 = 80
⇒ OE2 = (4\(\sqrt { 5 } \))2
∴ OE = 4\(\sqrt { 5 } \) cm
∴ Area of rectangle = lxb
= DE x OE
= 2\(\sqrt { 5 } \) x 4\(\sqrt { 5 } \)
= 8 x 5 = 40 cm2

Question 7.
In the figure, ABCD is a trapezium in which AB || DC. Prove that ar( ∆AOD = ar(∆BOC).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q7.1
Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O
∴ ∆ADB and ∆ACB are on the same base AB and between the same parallels
∴ ar(∆ADB = ar(∆ACD)
Subtracting, ar(AAOB) from both sides,
ar(∆ADB) – ar(∆AOB) = ar(∆ACD) – ar(∆AOB)
⇒ ar(∆AOD) = ar(∆BOC)

Question 8.
In the figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(∆ADE) = ar(∆BCF) [NCERT]
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q8.1
Solution:
Given : In the figure, ABCD, ABEF and CDEF are ||gms
To prove : ar(∆ADE) = ar(∆BCF)
Proof: ∴ ABCD is a ||gm
∴ AD = BC
Similarly, in ||gm ABEF
AE = BF
and in ||gm CDEF,
DE = CF
Now, in ∆ADE and ∆BCF
AD = BC (proved)
DE = CF (proved)
AE = BF (proved)
∴ ∆ADE ≅ ∆BCF
∴ ar(∆ADE) = ar(∆BCF) (∵ Congruent triangles are equal in area)

Question 9.
In the figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(∆ABC) = ar(∆ABC)
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q9.1
Solution:
Given : In the figure, ∆ABC and ∆ABD are on the same base AB and line CD is bisected by AB at O i.e. CO = OD
To prove : ar(∆ABC) = ar(∆ABD)
Construction : Draw CL ⊥ AB and DM ⊥ AB
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q9.2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q9.3

Question 10.
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid point of median AD, prove that ar(∆BGC) = 2ar(∆AGC).
Solution:
Given : In ∆ABC, AD is its median. G is mid point of AD. BG and CF are joined
To prove :
(i) ar(∆ADB) = ar(∆ADC)
(ii) ar(∆BGC) = 2ar(∆AGC)
Construction : Draw AL ⊥ BC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q10.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q10.2

Question 11.
A point D is taken on the side BC of a AABC such that BD = 2DC. Prove that ar(∆ABD) = 2ar(∆ADC)
Solution:
Given : In ∆ABC, D is a point on BC such that
BD = 2DC
AD is joined
To prove : ar(∆ABD) = 2ar(∆ADC)
Construction : Draw AL ⊥ BC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q11.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q11.2

Question 12.
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that
(i) ar(∆ADO) = or (∆CDO)
(ii) ar(∆ABP) = ar(∆CBP).
Solution:
Given : In ||gm ABCD, Diagonals AC and BD intersect each other at O
P is any point on BO
AP and CP are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q12.1
To prove :
(i) ar(∆ADO) = ar(∆CDO)
(ii) ar(∆ABP) = ar(∆CBP)
Proof:
(i) In ∆ADC,
O is the mid point of AC
∴ ar(∆ADO) = ar(∆CDO)
(ii) Since O is the mid point of AC
∴ PO is the median of ∆APC
∴ af(∆APO) = or(∆CPO) …(i)
Similarly, BO is the median of ∆ABC
∴ ar(∆ABO) = ar(∆BCO) …(ii)
Subtracting (i) from (ii),
ar(∆ABO) – ar(∆APO) = ar(∆BCO) – ar( ∆CPO)
⇒ ar(∆ABP) = ar(∆CBP)
Hence proved.

Question 13.
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(∆ADF) = ar(∆ECF)
(ii) If the area of ∆DFB = 3 cm2, find the area of ||gm ABCD.
Solution:
Given : In ||gm ABCD, BC is produced to E such that CE = BC
AE intersects CD at F
To prove :
(i) ar(∆ADF) = ar(∆ECF)
(ii) If ar(∆DFB) = 3 cm2, find the area of (||gm ABCD)
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q13.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q13.2

Question 14.
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(∆POA) = ar(∆QOC).
Solution:
Given : In ||gm ABCD, diagonals AC and BD intersect at O
A line through O intersects AB at P and CD at Q
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q14.1
To prove : ar(∆POA) = ar(∆QOC)
Proof : In ∆POA and ∆QOC,
OA = OC (O is mid-point of AC)
∠AOD = ∠COQ (Vertically opposite angles)
∠APO = ∠CQO (Alternate angles)
∴ ar(∆POA) ≅ ar(∆QOC) (AAS criterian)
∴ ar(∆POA) = ar(∆QOC)

Question 15.
In the figure, D and E are two points on BC such that BD = DE = EC. Show that ar(∆ABD) = ar(∆ADE) = ar(∆AEC). [NCERT]
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q15.1
Solution:
Given : D and E are two points on BC such that BD = DE = EC
AD and AE are joined
To prove : ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Construction : From A, draw AL ⊥ BC and XAY || BC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q15.2
Proof: ∵ BD = DE = EC
and ∆ABD, ∆ADE and ∆AEC have equal bases and from the common vertex A
∴ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)

Question 16.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Show that: ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)
Solution:
Given : In quadrilateral ABCD, diagonal AC and BD intersect each other as P
To prove : ar(∆APB) x ar(∆CPD) = ar(APD) x ar(∆BPC)
Construction : Draw AL and CN perpendiculars on BD
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q16.1

Question 17.
If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution:
Given : In ||gm ABCD, P is any point in the ||gm
AP and BP are joined
To prove : ar(∆APB) < \(\frac { 1 }{ 2 }\) ar(||gm ABCD)
Construction : Draw DN ⊥AB and PM ⊥ AM
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q17.1

Question 18.
ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is a point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
Given : In ||gm ABCD, E is a point on AB such that BE = 2EA and F is a point on CD such that DF = 2FC. AE and CE are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q18.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q18.2

Question 19.
In a ∆ABC, P and Q are respectively, the mid-points of AB and BC and R is the mid-point of AP. Prove that
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q19.1
Solution:
Given : In ∆ABC,
P and Q are mid-pionts of AB and BC R is mid-point of AP, PQ, RC, RQ are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q19.2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q19.3
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q19.4
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q19.5
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q19.6

Question 20.
ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.1
(v) Find what portion of the area of parallelogram is the area of AEFG.
Solution:
Given : ABCD is a parallelogram and AG = 2GB, CE = 2DE and BF = 2FC
To prove :
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.2
(v) Find what portion of the area of parallelogram is the area of AFEG.
Construction : Draw EP ⊥ AB and EQ ⊥ BC
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.3
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.4
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.5
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.6
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.7
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.8
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.9
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q20.10

Question 21.
In the figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of ACBX.
(ii) Prove that or(∆ZDE) = ar(∆CZA).
(iii) Prove that ar(∆CZY) = ar(∆EDZ).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q21.1
Solution:
Given : In the figure,
CP || AE and CY || BA
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q21.2
To prove :
(i) Name a triangle equal in area of ∆CBX
(ii) Prove that ar(∆ZDE) = ar(∆CZA)
(iii) ar(BCZY) = ar(∆EDZ)
Proof:
(i) ∆CBX and ∆CYX are on the same base BY and between same parallels.
∴ ar(∆CBX) = ar(∆CYX)
(ii) ∆ADE and ∆ACE are on the same base AE
and between the same parallels (AE || CD)
∴ ar(∆ADE) = ar(∆ACE)
Subtracting ar(∆AZE) from both sides
⇒ ar(∆ADE) – ar(∆AZE) = ar(∆ACE) – ar(∆AZE)
⇒ ar(∆ZDE) = ar(∆ACZ)
⇒ ar∆ZDE = ar∆CZA
(iii) ∵ As ACY and BCY are on the same base CY and between the same parallels
∴ ar(∆ACY) = ar(∆BCY)
Now ar(∆ACZ) = ar(∆ZDE) (Proved)
⇒ ar(∆ACY) + ar(∆CYZ) = ar(∆EDZ)
⇒ ar(∆BCY) + ar(∆CYZ) = ar(∆EDZ)
∴ ar quad. (BCZY) = ar(EDZ)
Hence proved.

Question 22.
In the figure, PSD A is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar(∆PQE) = ar(∆CFD).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q22.1
Solution:
Given : In the figure, PSDA is a ||gm in
which PQ = QR = RS and
AP || BQ || CR || DS
To prove : ar(∆PQE) = ar(∆CFD)
Construction : Join PD
Proof : ∵ PA || BQ || CR || DS
and PQ – QR = RS (Given)
∴ AB = BC = CD
∴ PQ = CD
Now in ABED, F is mid point of ED
∴ EF = FD
Similarly, EF = PE
⇒ PE = FD
In ∆PQE and ∆CFD,
∴ ∠EPQ = ∠FDC (Alternate angles)
PQ = CD
PE = FD (Proved)
∴ APQE ≅ ACFD (SAS cirterion)
∴ ar(∆PQE) = ar(∆CFD)

Question 23.
In the figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = \(\frac { 9 }{ 11 }\) ar(trap. XYBA)
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q23.1
Solution:
Given : In the figure, ABCD is a trapezium in which AB || DC
DC = 40 cm, AB = 60 cm
X and Y are the mid-points of AD and BC respectively
To prove :
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = \(\frac { 9 }{ 11 }\) m(trap. XYBA)
Construction : Join DY and produce it to meet AB produced at P
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q23.2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q23.3
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q23.4

Question 24.
D is the mid-point of side BC of ∆ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(∆BOE) = \(\frac { 1 }{ 8 }\) ar(∆ABC).
Solution:
Given : In ∆ABC, D is mid point of BC, E is mid point BE and O is the mid point of AE. BO, AE, AD are joined.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q24.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q24.2

Question 25.
In the figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q25.1
Solution:
Given : In ∆ABC, X and Y are the mid pionts of AC and AB respectively. Through A, a line parallel to BC is drawn. Join BX and CY and produce them to meet the parallel line through A, at P and Q respectively and intersect each other at O.
To prove : ar(∆ABP) = ar(∆ACQ)
Construction : Join XY and produce it to both sides
Proof : ∵ X and Y are mid points of sides AC and AB
∴ XY || BC
Similarly, XY || PQ
∆BXY and ∆CXY are on the same base XY and between the same parallels
∴ ar(∆BXY) = ar(∆CXY) …(i)
Now, trap. XYAP and trap. XYAQ are on the same base XY and between the same parallels
∴ ar(XYAP) = ar(XYAQ) …(ii)
Adding (i) and (ii),
∴ ar(∆BXY) + ar(∆YAP)
= ar(CXY) + ar(XYAQ)
⇒ ar(∆ABP) = ar(∆ACQ)

Question 26.
In the figure, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q26.1
Solution:
Given : Two ||gm ABCD and ||gm AEFD are on the same base AD. EF is produced to meet CD at Q. Join AF and PD also
To prove :
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD)
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q26.2
Proof:
(i) In ∆AEP and DFQ,
AE = DF (Opposite sides of a ||gm)
∠AEP = ∠DFQ (Corresponding angles)
∠APE = ∠DQF (Corresponding angles)
∴ ∆AEP ≅ ∆DFQ (AAS axiom)
∴ PE = QF (c.p.c.t.)
(ii) and ar(∆AEP) = ar(∆DFQ) …(i)
(iii) ∵ ∆PFA and ∆PFD are on the same base PF and between the same parallels
∴ ar(∆PFA) = ar(∆PFD) …(ii)
From (i) and (ii),
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q26.3

Question 27.
In the figure, ABCD is a ||gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar(||gm DLOP) = ar(||gm BMOQ).
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q27.1
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC. PQ || AB and LM || BC
To prove : ar(||gm DLOP) = ar(||gm BMOQ)
Proof : ∵ Since, a diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(∆ADC) = or(∆ABC)
⇒ ar(∆APO) + or(||gm DLOP) + ar(∆OLC)
= ar(∆AOM) + ar(||gm BMOQ) + ar( ∆OQC) …(i)
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively,
∴ ar(∆APO) = ar(∆AMO) …(ii)
And, ar(∆OLC) = ar(∆OQC) …(Hi)
Subtracting (ii) and (iii) from (i), we get ar(||gm DLOP) = ar(||gm BMOQ)

Question 28.
In a ∆ABC, if L and M are points on AB and AC respectively such that LM || BC.
Prove that:
(i) ar(∆LCM) = ar(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC).
Solution:
Given : In ∆ABC,
L and M are mid points on AB and AC
LM, LC and MB are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q28.1
To prove :
(i) ar(∆LCM) = or(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC)
Proof: ∵ L and M are the mid points of AB and AC
∴ LM || BC
(i) Now ∆LBM and ∆LCM are on the same base LM and between the same parallels
∴ar(∆LBM) = ar(∆LCM) …(i)
⇒ ar(∆LCM) = ar(∆LBM)
(ii) ∵ ∆LBC and ∆MBC are on the same base
BC and between the same parallels
∴ ar(∆LBC) = ar(∆MBC) …(ii)
(iii) a(∆LMB) = ar(∆LMC) [From (i)]
⇒ ar(∆ALM) + ar(∆LMB)
= ar(∆ALM) + ar(∆LMC) [Adding or(∆ALM) to both sides]
⇒ ar(∆ABM) = ar(∆ACL)
(iv) ∵ ar(∆LBC) = ar(∆MBC) [From (ii)]
⇒ ar(∆LBC) – ar(∆BOC) = ar(∆MBC) – ar(∆BOC)
ar(∆LBO) = ar(∆MOC)

Question 29.
In the figure, ABC and BDC are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.2
Solution:
Given : ABC and BDE are two equilateral triangles and D is mid point of BC. AE intersects BC in F
To prove :
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.3
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.4
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.5
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.6
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.7
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q29.8

Question 30.
In the figure, ABC is a right triangle right angled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that [NCERT]
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(∆CFG)
(vii) ar(BCED) = ar(AMBN) + ar(ACFG)
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 Q30.1
Solution:
Given : In ∆ABC, ∠A = 90°
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
AX ⊥ DE meeting DE at X
To prove :
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = 2ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = or(AMBN) + ar(ACFG)
Construction : Join AD, AE, BF and CM
Proof:
(i) In ∆MBC and ∆ABD,
MB=AB (Sides of square)
BC = BD
∠MBC = ∠ABD (Each angle = 90° + ∠ABC)
∴ ∆MBC ≅ ∆ABD (SAS criterian)
∴ ar(∆MBC) = ar(∆ABD) …(i)
(ii) ∵ ∆ABD and rectangle BYXD are on the same base BD and between the same parallels
∴ ar(∆ABD) = \(\frac { 1 }{ 2 }\) ar(rect. BYXD)
⇒ ar(rect. BYXD) = 2ar(∆ABD)
⇒ ar(rect. BYXD) = 2ar(∆MBC) …(ii)
(iii) Similarly, ∆MBC and square MBAN are on the same base MB and between the same parallels
∴ ar(∆MBC) = ar(sq. ABMN) …(iii)
From (ii) and (iii)
ar(sq. ∆BMN) = ar(rect. BYXD)
(iv) In AFCB and ∆ACE,
FC = AC
CB = CE (Sides of squares)
∠FCB = ∠ACE (Each = 90° + ∠ACB)
∴ ∆FCB = ∆ACE (SAS criterian)
(v) ∵ ∆FCB ≅ ∆ACE (Proved)
∴ ar(∆FCB) = ar(∆ACE)
∵∆ACE and rectangle CYXE are on the same base and between the same parallels
∴ 2ar(∆ACE) = ar(CYXC)
⇒ 2ar(∆FCB) = ar(CYXE) …(iv)
(vi) ∵ AFCB and rectangle FCAG are on the base FC and between the same parallels
∴ 2ar(∆FCB) = ar(FCAG) …(v)
From (iv) and (v)
ar(CMXE) = ar(ACFG)
(vii) In ∆ACB.
BC2 = AB2 + AC2 (By Pythagoras Theorem)
⇒ BC x BD = AB x MB + AC x FC
⇒ ar(BCED) = ar(ABMN) + ar(ACFG)
Hence proved.

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 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 12.3

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3

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 12.3

Other Exercises

Question 1.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
Given : In ∆ABC and ∆DEF,
∠B = ∠E = 90°
∠C = ∠F
AB = DE
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q1.1
To prove : ∆ABC = ∆DEF
Proof : In ∆ABC and ∆DEF,
∠B = ∠E (Each = 90°)
∠C = ∠F (Given)
AB = DE (Given)
∆ABC = ∆DEF (AAS axiom)

Question 2.
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Solution:
Given : In ∆ABC, AE is the bisector of vertical exterior ∠A and AE \(\parallel\) BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q2.1
To prove : ∆ABC is an isosceles
Proof: ∵ AE \(\parallel\) BC
∴ ∠1 = ∠B (Corresponding angles)
∠2 = ∠C (Alternate angle)
But ∠1 = ∠2 (∵ AE is the bisector of ∠CAD)
∴ ∠B = ∠C
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Given : In ∆ABC, AB = AC
∠A = 2(∠B + ∠C)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q3.1
To calculate: Base angles,
Let ∠B = ∠C = x
Then ∠A = 2(∠B + ∠C)
= 2(x + x) = 2 x 2x = 4x
∵ Sum of angles of a triangle = 180°
∴ 4x + x + x – 180° ⇒ 6x = 180°
⇒ x= \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30° o
∴ ∠B = ∠C = 30 and ∠A = 4 x 30° = 120

Question 4.
Prove that each angle of an equilateral triangle is 60°. [NCERT]
Solution:
Given : ∆ABC is an equilateral triangle
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q4.1
Proof: In ∆ABC,
AB = AC (Sides of an equilateral triangle)
∴ ∠C = ∠B …(i)
(Angles opposite to equal angles)
Similarly, AB = BC
∴ ∠C = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠A = ∠B = ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\)= 60°

Question 5.
Angles A, B, C of a triangle ABC are equal to each other. Prove that ∆ABC is equilateral.
Solution:
Given : In ∆ABC, ∠A = ∠B = ∠C
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q5.1
To prove : ∆ABC is an equilateral
Proof: In ∆ABC,
∴ ∠B = ∠C (Given)
∴ AC = AB …(i) (Sides opposite to equal angles)
Similarly, ∠C = ∠A
∴ BC =AB …(ii)
From (i) and (ii)
AB = BC = CA
Hence ∆ABC is an equilateral triangle

Question 6.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 90°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q6.1
AB =AC (Given)
∴ ∠C = ∠B (Angles opposite to equal sides)
But ∠B + ∠C = 90° (∵ ∠B = 90°)
∴ ∠B = ∠C = \(\frac { { 90 }^{ \circ } }{ 2 }\) = 45°
Hence ∠B = ∠C = 45°

Question 7.
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
Given : In ∆PQR, PQ = PR
S is a point on PQ and PT || QR
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q7.1
To prove : PS = PT
Proof : ∵ST || QR
∴ ∠S = ∠Q and ∠T = ∠R (Corresponding angles)
But ∠Q = ∠R (∵ PQ = PR)
∴ PS = PT (Sides opposite to equal angles)

Question 8.
In a ∆ABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Solution:
Given : In ∆ABC, AB = AC the bisectors of ∠B and ∠C intersect at O. M is any point on BO produced.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q8.1
To prove : ∠MOC = ∠ABC
Proof: In ∆ABC, AB = BC
∴ ∠C = ∠B
∵ OB and OC are the bisectors of ∠B and ∠C
∴ ∠1 =∠2 = \(\frac { 1 }{ 2 }\)∠B
Now in ∠OBC,
Ext. ∠MOC = Interior opposite angles ∠1 + ∠2
= ∠1 + ∠1 = 2∠1 = ∠B
Hence ∠MOC = ∠ABC

Question 9.
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Solution:
Given : In ∆ABC, P is a point on the bisector of ∠B and from P, RPQ || AB is draw which meets BC in Q
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q9.1
To prove : ∆BPQ is an isosceles
Proof : ∵ BD is the bisectors of CB
∴ ∠1 = ∠2
∵ RPQ || AB
∴ ∠1 = ∠3 (Alternate angles)
But ∠1 == ∠2 (Proved)
∴ ∠2 = ∠3
∴ PQ = BQ (sides opposite to equal angles)
∴ ∆BPQ is an isosceles

Question 10.
ABC is a triangle in which ∠B = 2∠C, D is a point on BC such that AD bisects ∠BAC = 72°.
Solution:
Given: In ∆ABC,
∠B = 2∠C, AD is the bisector of ∠BAC AB = CD
To prove : ∠BAC = 72°
Construction : Draw bisector of ∠B which meets AD at O
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q10.1
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q10.2
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 Q10.3

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5

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 12.5

Other Exercises

Question 1.
ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.
Solution:
Given : In ∆ABC, D is mid-point of BC and DE ⊥ AB, DF ⊥ AC and DE = DF
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 Q1.1
To Prove : ∆ABC is an isosceles triangle
Proof : In right ∆BDE and ∆CDF,
Side DE = DF
Hyp. BD = CD
∴ ∆BDE ≅ ∆CDF (RHS axiom)
∴ ∠B = ∠C (c.p.c.t.)
Now in ∆ABC,
∠B = ∠C (Prove)
∴ AC = AB (Sides opposite to equal angles)
∴ AABC is an isosceles triangle

Question 2.
ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that ∆ABC is an isosceles.
Solution:
Given : In ∆ABC,
BE ⊥ AC and CF ⊥ AB
BE = CF
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 Q2.1
To prove : AABC is an isosceles triangle
Proof : In right ABCE and ABCF Side
BE = CF (Given)
Hyp. BC = BC (Common)
∴ ∆BCE ≅ ∆BCF (RHS axiom)
∴ ∠BCE = ∠CBF (c.p.c.t.)
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
Solution:
Given : A point P lies in the angle ABC and PL ⊥ BA and PM ⊥ BC and PL = PM. PB is joined
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 Q3.1
To prove : PB is the bisector ∠ABC,
Proof : In right ∆PLB and ∆PMB
Side PL = PM (Given)
Hyp. PB = PB (Common)
∴ ∆PLB ≅ ∆PMB (RHS axiom)
∴ ∆PBL = ∆PBM (c.p.c.t.)
∴ PB is the bisector of ∠ABC

Question 4.
In the figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 Q4.1
Solution:
Given : In the figure,
AD ⊥ CD and CB ⊥ CD, AQ = BP and DP = CQ
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 Q4.2
To prove : ∠DAQ = ∠CBP
Proof : ∵ DP = CQ
∴ DP + PQ = PQ + QC
⇒ DQ = PC
Now in right ∆ADQ and ∆BCP
Side DQ = PC (Proved)
Hyp. AQ = BP
∴ ∆ADQ ≅ ∆BCP (RHS axiom)
∴ ∠DAQ = ∠CBP (c.p.c.t.)

Question 5.
Which of the following statements are true (T) and which are false (F):
(i) Sides opposite to equal angles of a triangle may be unequal.
(ii) Angles opposite to equal sides of a triangle are equal.
(iii) The measure of each angle of an equilateral triangle is 60°.
(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.
(v) The bisectors of two equal angles of a triangle are equal.
(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.
(viii)If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.
Solution:
(i) False : Sides opposite to equal angles of a triangle are equal.
(ii) True.
(iii) True.
(iv) False : The triangle is an isosceles triangle.
(v) True.
(vi) False : The triangle is an isosceles.
(vii) False : The altitude an equal.
(viii) False : If one side and hypotenuse of one right triangle on one side and hypotenuse of the other right triangle are equal, then triangles are congruent.
(ix) True.

Question 6.
Fill in the blanks in the following so that each of the following statements is true.
(i) Sides opposite to equal angles of a triangle are …….
(ii) Angle opposite to equal sides of a triangle are …….
(iii) In an equilateral triangle all angles are …….
(iv) In a ∆ABC if ∠A = ∠C, then AB = …….
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ……..
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ……… CE.
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆……
Solution:
(i) Sides opposite to equal angles of a triangle are equal.
(ii) Angle opposite to equal sides of a triangle are equal.
(iii) In an equilateral triangle all angles are equal.
(iv) In a ∆ABC, if ∠A = ∠C, then AB = BC.
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC.
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE.
(vii) In right triangles ABC and DEF, it hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆EFD.

Question 7.
ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Solution:
Given : In square ABCD, X and Y are points on side AD and BC respectively and AY = BX
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 Q7.1
To prove : BY = AX
∠BAY = ∠ABX
Proof: In right ∆BAX and ∆ABY
AB =AB (Common)
Hyp. BX = AY (Given)
∴ ∆BAX ≅ ∆ABY (RHS axiom)
∴ AX = BY (c.p.c.t.)
∠ABX = ∠BAY (c.p.c.t.)
Hence, BY = AX and ∠BAY = ∠ABX.

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 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 13 Linear Equations in Two Variables Ex 13.2

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2

Other Exercises

Question 1.
Two opposite angles of a parallelogram are (3x- 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
∵ Opposite angles of a parallelogram are equal
∴ 3x – 2 = 50 – x
⇒ 3x + x – 50 + 2
⇒ 4x = 52
⇒ x = \(\frac { 52 }{ 4 }\) = 13
∴ ∠A = 3x – 2 = 3 x 13 – 2 = 39° – 2 = 37°
∠C = 50° -x = 50° – 13 = 37°
But∠A + B = 180°
∴ 37° + ∠B = 180°
⇒ ∠B = 180° – 37° = 143°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q1.1
and ∠D = ∠B (Opposite angles of a ||gm)
∴ ∠D = 143°
Hence angles and 37°, 143°, 37°, 143°

Question 2.
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let in ||gm ABCD,
∠A =x
Then ∠B = \(\frac { 2 }{ 3 }\) x
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q2.1
But, ∠A + ∠B = 180° (Sum of two adjacent angles of a ||gm)
⇒ x + \(\frac { 2 }{ 3 }\)x = 180°
⇒ \(\frac { 5 }{ 3 }\)x = 180°
⇒ x = 180° x \(\frac { 3 }{ 5 }\) = 108°
∴ ∠A = 108°
and ∠B = 108° x \(\frac { 2 }{ 3 }\) = 72°
But, ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 108°, ∠D = 72°
Hence angles are 108°, 72°, 108°, 72°

Question 3.
Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.
Solution:
Let smallest angle of a ||gm = x
Then second angle = 2x – 24°
But these are consecutive angles
∴ x + (2x- 24°) = 180°
⇒ x + 2x – 24° = 180°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q3.1
⇒ 3x = 180° + 24° = 204°
⇒ x =\(\frac { { 204 }^{ \circ } }{ 3 }\)  = 68°
∴ Smallest angle = 68°
and second angle = 2x 68° – 24°
= 136°-24° = 112°
∵ The opposite angles of a ||gm are equal Other two angles will be 68° and 112°
∴ Hence angles are 68°, 112°, 68°, 112°

Question 4.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?
Solution:
In a ||gm ABC,
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q4.1
Perimeter = 22cm
and longest side = 6.5 cm
Let shorter side = x
∴ 2x (6.5 + x) = 22
⇒ 13 + 2x = 22
⇒ 2x = 22 – 13 = 9
⇒ x = \(\frac { 9 }{ 2 }\) = 4.5
∴ shorter side = 4.5cm

Question 5.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In ||gm ABCD,
∠D = 135°
But, ∠A + ∠D = 180° (Sum of consecutive angles)
⇒∠A+ 135° = 180°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q5.1
⇒ ∠A = 180° – 135° = 45°
∵ ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠B = 135°

Question 6.
ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.
Solution:
In ||gm ABCD,
∠A = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q6.1

Question 7.
In the figure, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q7.1
Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles) But, ∠A = 75°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q7.2
∴ ∠B = 180° – ∠A = 180° – 75° = 105°
∴ DBA = 105° -60° = 45°
But ∠CDB = ∠DBA (alternate angles)
= 45°
and ∠ADB = ∠DBC = 60°

Question 8.
Which of the following statements are true (T) and which are false (F)?
(i) In a parallelogram, the diagonals are equal.
(ii) In a parallelogram, the diagonals bisect each other.
(iii) In a parallelogram, the diagonals intersect each other at right angles.
(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.
(v) If all the angles of a quadrilateral are equal, it is a parallelogram.
(vi) If three sides of a quadrilateral are equal, it is a parallelogram.
(vii) If three angles of a quadrilateral are equal, it is a parallelogram.
(viii)If all the sides of a quadrilateral are equal it is a parallelogram.
Solution:
(i) False, Diagonals of a parallelogram are not equal.
(ii) True.
(iii) False, Diagonals bisect each other at right angles is a rhombus or a square only.
(iv) False, In a quadrilateral, if opposite sides are equal and parallel, then it is a ||gm.
(v) False, If all angles are equal, then it is a square or a rectangle.
(vi) False, If opposite sides are equal and parallel then it is a ||gm
(vii) False, If opposite angles are equal, then it is a parallelogram.
(viii)False, If all the sides are equal then it is a square or a rhombus but not parallelogram.

Question 9.
In the figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC= BC and DC = 2AD.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q9.1
Solution:
Given : In ||gm ABCD,
∠A = 60°
Bisector of ∠A and ∠B meet at P.
To prove :
(i) AD = DP
(ii) PC = BC
(iii) DC = 2AD
Construction : Join PD and PC
Proof : In ||gm ABCD,
∠A = 60°
But ∠A + ∠B = 180° (Sum of excutive angles)
⇒ 60° + ∠B = 180°
∴ ∠B = 1809 – 60° = 120°
∵ DC || AB
∠PAB = ∠DPA (alternate angles)
⇒ ∠PAD = ∠DPA (∵ ∠PAB = ∠PAD)
∴ AB = DP
(PA is its angle bisector, sides opposite to equal angles)
(ii) Similarly, we can prove that ∠PBC = ∠PCB (∵ ∠PAB = ∠BCA alternate angles)
∴ PC = BC
(iii) DC = DP + PC
= AD + BC [From (i) and (ii)]
= AD + AB = 2AB (∵BC = AD opposite sides of the ||gm)
Hence DC = 2AD

Question 10.
In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove thatAF = 2AB.
Solution:
Given : In ||gm ABCD,
E a mid point of BC
DE is joined and produced to meet AB produced at F
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 Q10.1
To prove : AF = 2AB
Proof : In ∆CDE and ∆EBF
∠DEC = ∠BEF (vertically opposite angles)
CE = EB (E is mid point of BC)
∠DCE = ∠EBF (alternate angles)
∴ ∆CDE ≅ ∆EBF (SAS Axiom)
∴ DC = BF (c.p.c.t.)
But AB = DC (opposite sides of a ||gm)
∴ AB = BF
Now, AF = AB + BF = AB + AB = 2AB
Hence AF = 2AB

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 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 MCQS

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

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

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
In ∆ABC ≅ ∆LKM, then side of ∆LKM equal to side AC of ∆ABC is
(a) LX
(b) KM
(c) LM
(d) None of these
Solution:
Side AC of ∆ABC = LM of ∆LKM (c)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q1.1

Question 2.
In ∆ABC ≅ ∆ACB, then ∆ABC is isosceles with
(a) AB=AC
(b) AB = BC
(c) AC = BC
(d) None of these
Solution:
∵ ∆ABC ≅ ∆ACB
∴ AB = AC (a)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q2.1

Question 3.
In ∆ABC ≅ ∆PQR, then ∆ABC is congruent to ∆RPQ, then which of the following is not true:
(a) BC = PQ
(b) AC = PR
(c) AB = PQ
(d) QR = BC
Solution:
∵ ∆ABC = ∆PQR
∴ AB = PQ, BC = QR and AC = PR
∴ BC = PQ is not true (a)

Question 4.
In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, ∠B = ∠P and BC = PR State which of the congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In two triangles ∆ABC and ∆PQR,
AB = QP, ∠B = ∠P and BC = PR
The condition apply : SAS (a)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q4.1

Question 5.
In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In ∆ABC and ∆PQR,
∠A = ∠R
∠B = ∠P
AB = RP
∴ ∆ABC ≅ ∆PQR (ASA axiom) (b)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q5.1

Question 6.
If ∆PQR ≅ ∆EFD, then ED =
(a) PQ
(b) QR
(c) PR
(d) None of these
Solution:
∵ ∆PQR = ∆EFD
∴ ED = PR (c)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q6.1

Question 7.
If ∆PQR ≅ ∆EFD, then ∠E =
(a) ∠P
(b) ∠Q
(c) ∠R
(d) None of these
Solution:
∵ ∆PQR ≅ ∆EFD
∴ ∠E = ∠P (a)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q7.1

Question 8.
In a ∆ABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C
But Ext. ∠ACD = ∠A + ∠B
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q8.1
∠ACB + ∠ACD = 180° (Linear pair)
∴ ∠ACB + 100° = 180°
⇒ ∠ACB = 180°-100° = 80°
∴ ∠B = ∠ACD = 80°
But ∠A + ∠B 4- ∠C = 180°
∴ ∠A + 80° + 80° = 180°
⇒∠A+ 160°= 180°
∴ ∠A= 180°- 160° = 20° (a)

Question 9.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is
(a) 100°
(b) 120°
(c) 110°
(d) 130°
Solution:
In ∆ABC,
∠A = 2(∠B + ∠C)
= 2∠B + 2∠C
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q9.1
Adding 2∠A to both sides,
∠A + 2∠A = 2∠A + 2∠B + 2∠C
⇒ 3∠A = 2(∠A + ∠B + ∠C)
⇒ 3∠A = 2 x 180° (∵∠A + ∠B + ∠C = 180° )
⇒ 3∠A = 360°
⇒∠A = \(\frac { { 360 }^{ \circ } }{ 3 }\)  = 120°
∴ ∠A = 120° (b)

Question 10.
Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) SSA
(c) ASA
(d) SSS
Solution:
SSA is not the criterion of congruence of triangles. (b)

Question 11.
In the figure, the measure of ∠B’A’C’ is
(a) 50°
(b) 60°
(c) 70°
(d) 80°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q11.1
Solution:
In the figure,
∆ABC ≅ ∆A’B’C’
∴ ∠A = ∠A
⇒3x = 2x- + 20
⇒ 3x – 2x = 20
⇒ x = 20
∠B’A’C’ = 2x + 20 = 2 x 20 + 20
= 40 + 20 = 60° (b)

Question 12.
If ABC and DEF are two triangles such that ∆ABC ≅ ∆FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then, which of the following is true?
(a) DF = 5 cm, ∠F = 60°
(b) DE = 5 cm, ∠E = 60°
(c) DF = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°
Solution:
∵ ∆ABC ≅ ∆FDE,
AB = 5 cm, ∠A = 80°, ∠B = 40°
∴ DF = 5 cm, ∠F = 80°, ∠D = 40°
∴ ∠C =180°- (80° + 40°) = 180° – 120° = 60°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q12.1
∴ ∠E = ∠C = 60°
∴ DF = 5 cm, ∠E = 60° (c)

Question 13.
In the figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ∆ABD is congruent to
(a) ∆EFC
(b) ∆ECF
(c) ∆CEF
(d) ∆FEC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q13.1
Solution:
In the figure, AB ⊥ BE, FE ⊥ BE
BC = DE, AB = EF,
then CD + BC = CD + DE BD = CE
In ∆ABD and ∆CEF,
BD = CE (Prove)
AB = FE (Given)
∠B = ∠E (Each 90°)
∴ ∆ABD ≅ ∆FCE (b)

Question 14.
In the figure, if AE || DC and AB = AC, the value of ∠ABD is
(a) 70°
(b) 110°
(c) 120°
(d) 130°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q14.1
Solution:
In the figure, AE || DC
∴ ∠1 = 70° (Vertically opposite angles)
∴ ∠1 = ∠2 (Alternate angles)
∠2 = ∠ABC (Base angles of isosceles triangle)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q14.2
∴ ABC = 90°
But ∠ABC + ∠ABD = 180° (Linear pair)
⇒ 70° +∠ABD = 180°
⇒∠ABD = 180°-70°= 110°
∴ ∠ABD =110° (b)

Question 15.
In the figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
(a) 52°
(b) 76°
(c) 156°
(d) 104°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q15.1
Solution:
In ∆ABC, AB = AC
AC is produced to E
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q15.2
CD || BA is drawn
∠ABC = 52°
∴ ∠ACB = 52° (∵ AB = AC)
∴ ∠BAC = 180°-(52° +52°)
= 180°-104° = 76°
∵ AB || CD
∴ ∠ACD = ∠BAC (Alternate angles)
= 76°
and ∠BCE + ∠DCB = 180° (Linear pair)
∠BCE + 52° = 180°
⇒∠BCE = 180°-52°= 128°
∠x + ∠ACD = 380°
⇒ x + 76° = 180°
∴ x= 180°-76°= 104° (d)

Question 16.
In the figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q16.1
Solution:
In the figure, AC is the bisector of ∠BAD, AB = 3 cm, AC = 5 cm
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q16.2
In ∆ABC and ∆ADC,
AC = AC (Common)
∠B = ∠D (Each 90°)
∠BAC = ∠DAC (∵ AC is the bisector of ∠A)
∴ ∆ABC ≅ ∆ADC (AAS axiom)
∴ BC = CD and AB = AD (c.p.c.t.)
Now in right ∆ABC,
AC2 = AB2 + BC2
⇒ (5)2 = (3)2 + BC2
⇒25 = 9 + BC2
⇒ BC2 = 25 – 9 = 16 = (4)2
∴ BC = 4 cm
But CD = BC
∴ CD = 4 cm (c)

Question 17.
D, E, F are the mid-point of the sides BC, CA and AB respectively of ∆ABC. Then ∆DEF is congruent to triangle
(a) ABC
(b) AEF
(c) BFD, CDE
(d) AFE, BFD, CDE
Solution:
In ∆ABC, D, E, F are the mid-points of the sides BC, CA, AB respectively
DE, EF and FD are joined
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q17.1
∵ E and F are the mid-points
AC and AB,
∴ EF = \(\frac { 1 }{ 2 }\) BC and EF || BC
Similarly,
DE = \(\frac { 1 }{ 2 }\) AB and DE || AB
DF = \(\frac { 1 }{ 2 }\) AC and DF || AC
∴ ∆DEF is congruent to each of the triangles so formed
∴ ∆DEF is congruent to triangle AFE, BFD, CDE (d)

Question 18.
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =
(a) 55°
(b) 70°
(c) 35°
(d) 110°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q18.1
Solution:
In ∆ABC, AB = AC
AD is median to BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q18.2
∴ BD = DC
In ∆ADB, ∠D = 90°, ∠B = 35°
But ∠B + BAD + ∠D = 180° (Sum of angles of a triangle)
⇒ 35° + ∠BAD + 90° = 180°
⇒∠BAD + 125°= 180°
⇒ ∠BAD = 180°- 125°
⇒∠BAD = 55° (a)

Question 19.
In the figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q19.1
Solution:
In the figure, ABCD and AXYZ are squares
DY = 3 cm, AZ = 2 cm
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q19.2
DZ = DY + YZ
= DY + Z = 3 + 2 = 5 cm
In ∆ADZ, ∠2 = 90°
AD2 + AZ2 + DZ2 = 22 + 52 cm
= 4 + 25 = 29
In ∠ABX, ∠X = 90°
AB2 = AX2 + BX2
AD2 = AZ2 + BX2
(∵ AB = AD, AX = AZ sides of square)
29 = 22 + BX2
⇒ 29 = 4 + BX2
⇒ BX2 = 29 – 4 = 25 = (5)2
∴ BX = 5 cm (a)

Question 20.
In the figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is
(a) 72°
(b) 73°
(c) 74°
(d) 95°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q20.1
Solution:
In the figure, ∠B = 2∠C, AD and BE are the bisectors of ∠A and ∠B respectively,
AB = CD
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q20.2
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS Q20.3

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula 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 VSAQS

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS

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

Other Exercises

Question 1.
Solution:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.

In AABC and ADEF,
∆ABC ≅ ∆DEF
and AB = DE, BC = EF
∴ ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q1.1

Question 2.
Solution:
In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Are the two triangles necessarily congruent?

No, as the triangles are equiangular, so similar.

Question 3.
Solution:
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5 cm and ∠D = 75°. Are two triangles congruent?
Yes, triangles are congruent (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q3.1

Question 4.
Solution:
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
Yes, these are congruent
In two triangles ABC are ADC,
AB = AD (Given)
BC = CD (Given)
and AC = AC (Common)
∴ ∆sABC ≅ AADC (SSS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q4.1

Question 5.
Solution:
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C – 30° and ∠D = 90°. Are two triangles congruent?
Yes, triangles are congruent because,
In ∆ABC, and ∆CDE,
AC = CE
BC = CD ∠C = 30°
∴ ∆ABC ≅ ∆CDE (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q5.1

Question 6.
Solution:
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.

Given : In ∆ABC, AB = AC
BE and CF are two medians
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q6.1
To prove : BE = CF
Proof: In ∆ABE and ∆ACF.
AB = AC (Given)
∠A = ∠A (Common)
AE = AF (Half of equal sides)
∴ ∆ABE ≅ ∆ACF (SAS axiom)
∴ BE = CF (c.p.c.t.)

Question 7.
Solution:
Find the measure of each angle of an equilateral triangle.

In ∆ABC,
AB = AC = BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q7.1
∵ AB = AC
∴ ∠C = ∠B …(i)
(Angles opposite to equal sides)
Similarly,
AC = BC
∴ ∠B = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
∴ ∠A + ∠B + ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\)  = 60°

Question 8.
Solution:
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE.

Given : An equilateral ACDE is formed on the side of square ABCD. AE and BE are joined
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q8.1
To prove : ∆ADE ≅ ∆BCE
Proof : In ∆ADE and ∆BCE,
AD = BC (Sides of a square)
DE = CE (Sides of equilateral triangle)
∠ADE = ∠BCE(Each = 90° + 60° = 150°)
∴ AADE ≅ ABCE (SAS axiom)

Question 9.
Solution:
Prove that the sum of three altitude of a triangle is less than the sum of its sides.

Given : In ∆ABC, AD, BE and CF are the altitude of ∆ABC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q9.1
To prove : AD + BE + CF < AB + BC + CA
Proof : In right ∆ABD, ∠D = 90°
Then other two angles are acute
∵ ∠B < ∠D
∴ AD < AB …(i)
Similarly, in ∆BEC and ∆ABE we can prove thatBE and CF < CA …(iii)
Adding (i), (ii), (iii)
AD + BE -t CF < AB + BC + CA

Question 10.
Solution:
In the figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS Q10.1

Given : In the figure, AB = AC, ∠B = ∠C
To prove : BQ = CP
Proof : In ∆ABQ and ∆ACP
AB = AC (Given)
∠A = ∠A (Common)
∠B = ∠C (Given)
∴ ∆ABQ ≅ ∆ACP (ASA axiom)
∴ BQ = CP (c.p.c.t.)

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS 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 12.6

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6

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 12.6

Other Exercises

Question 1.
In ∆ABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In ∆ABC, ∠A = 40°, ∠B = 60°
But ∠A + ∠B + ∠C = 180°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q1.1
⇒ 40° + 60° + ∠C = 180°
⇒ ∠C = 180° = (40° + 60°)
= 180° – 100° = 80°
∵ ∠C = 80°, which is the greatest angle and
∠A = 40° is the smallest angle
∴ Side AB which is opposite to the greatest angle is the longest and side BC which is opposite to the smallest angle is the shortest.

Question 2.
In a ∆ABC, if ∠B = ∠C = 45°. which is the longest side?
Solution:
In ∆ABC, ∠B = ∠C = 45°
But ∠A + ∠B + ∠C = 180°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q2.1
⇒ ∠A + 45° + 45° = 180°
⇒ ∠A + 90° = 180°
∴ ∠A = 180°-90° = 90°
∴∠A is the greatest
∴ Side BC opposite to it is the longest

Question 3.
In ∆ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that :
(i) AD > CD
(ii) AD > AC
Solution:
Given : In AABC, side BC is produced to D such that BD = BC
∠A = 70° and ∠B = 60°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q3.1
To prove :
(i) AD > CD (ii) AD > AC
Proof: In ∆ABC,
∠A = 70°, ∠B = 60°
But Ext. ∠CBD + ∠CBA = 180° (Linear pair)
∠CBD + 60° = 180° 3
⇒ ∠CBD = 180° – 60° = 120°
But in ∆BCD,
BD = BC
∴ ∠D = ∠BCD
But ∠D + ∠BCD = 180° – 120° = 60°
∴∠D = ∠BCD = \(\frac { { 60 }^{ \circ } }{ 2 }\)  = 30°
and in ∆ABC,
∠A + ∠B + ∠C = 180°
⇒ 70° + 60° + ∠C = 180°
⇒ 130° + ∠C = 180°
∴ ∠C =180°- 130° = 50°
Now ∠ACD = ∠ACB + ∠BCD = 50° + 30° = 80°
(i) Now in ∆ACB,
∠ACD = 80° and ∠A = 70°
∴ Side AD > CD
(Greater angle has greatest side opposite to it)
(ii) ∵ ∠ACD = 80° and ∠D = 30°
∴ AD > AC

Question 4.
Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
We know that in a triangle, sum of any two sides is greater than the third side and 2 cm + 3 cm = 5 cm and 5 cm < 7 cm
∴ This triangle is not possible to draw

Question 5.
In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
Solution:
In ∆ABC, ∠B = 35°, ∠C = 65° and AP is the bisector of ∠BAC which meets BC in P.
Arrange PA, PB and PC in descending order In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q5.1
⇒ ∠A + 35° + 65° = 180°
∠A + 100°= 180°
∴ ∠A =180°- 100° = 80°
∵ PA is a bisector of ∠BAC
∴ ∠1 = ∠2 = \(\frac { { 80 }^{ \circ } }{ 2 }\)  = 40°
Now in ∆ACP, ∠ACP > ∠CAP
⇒ ∠C > ∠2
∴ AP > CP …(i)
Similarly, in ∆ABP,
∠BAP > ∠ABP ⇒ ∠1 > ∠B
∴ BP > AP …(ii)
From (i) and (ii)
BP > AP > CP

Question 6.
Prove that the perimeter of a triangle is greater than the sum of its altitudes
Solution:
Given : In ∆ABC,
AD, BE and CF are altitudes
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q6.1
To prove : AB + BC + CA > AD + BC + CF
Proof : We know that side opposite to greater angle is greater.
In ∆ABD, ∠D = 90°
∴ ∠D > ∠B
∴ AB >AD …(i)
Similarly, we can prove that
BC > BE and
CA > CF
Adding we get,
AB + BC + CA > AD + BE + CF

Question 7.
In the figure, prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q7.1
Solution:
Given : In the figure, ABCD is a quadrilateral and AC is joined
To prove :
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Proof:
(i) In ∆ABC,
AB + BC > AC …(i)
(Sum of two sides of a triangle is greater than its third side)
Similarly in ∆ADC,
CD + DA > AC …(ii)
Adding (i) and (ii)
CD + DA + AB + BC > AC + AC
⇒ CD + DA + AB + BC > 2AC
(ii) In ∆ACD,
CD + DA > CA
(Sum of two sides of a triangle is greater than its third side)
Adding AB to both sides,
CD + DA + AB > CA + AB
But CA + AB > BC (in ∆ABC)
∴ CD + DA + AD > BC

Question 8.
Which of the following statements are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.
Solution:
(i) False. Sum of three sides of a triangle is greater than the sum of its altitudes.
(ii) True.
(iii) True.
(iv) False. Difference of any two sides is less than the third side.
(v) True.
(vi) True.

Question 9.
Fill in the blanks to make the following statements true.
(i) In a right triangle, the hypotenuse is the ……. side.
(ii) The sum of three altitudes of a triangle is ……. than its perimeter.
(iii) The sum of any two sides of a triangle is …….. than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the ….. side opposite to it.
(v) Difference of any two sides of a triangle is……. than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has ……… angle opposite to it.
Solution:
(i) In a right triangle, the hypotenuse is the longest side.
(ii) The sum of three altitudes of a triangle is less than its perimeter.
(iii) The sum of any two sides of a triangle is greater than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.
(v) Difference of any two sides of a triangle is less than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

Question 10.
O is any point in the interior of ∆ABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Solution:
Given : In ∆ABC, O is any point in the interior of the ∆ABC, OA, OB and OC are joined
To prove :
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Construction : Produce BO to meet AC in D.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q10.1
Proof: In ∆ABD,
(i) AB + AD > BD (Sum of any two sides of a triangle is greater than third)
⇒ AB + AD > BO + OD …(i)
Similarly, in ∆ODC,
OD + DC > OC …(ii)
Adding (i) and (ii)
AB + AD + OD + DC > OB + OD + OC
⇒ AB + AD + DC > OB + OC
⇒ AB + AC > OB + OC
(ii) Similarly, we can prove that
BC + AB > OA + OC
and CA + BC > OA + OB
(iii) In ∆OAB, AOBC and ∆OCA,
OA + OB > AB
OB + OC > BC
and OC + OA > CA
Adding, we get
2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OO > \(\frac { 1 }{ 2 }\) (AB + BC + CA)

Question 11.
Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.
Solution:
Given : In quadrilateral ABCD, AC and BD are its diagonals,
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 Q11.1
To prove : AB + BC + CD + DA > AC + BD
Proof: In ∆ABC,
AB + BC > AC …(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly, in ∆ADC,
DA + CD > AC …(ii)
In ∆ABD,
AB + DA > BD …(iii)
In ∆BCD,
BC + CD > BD …(iv)
Adding (i), (ii), (iii) and (iv)
2(AB + BC + CD + DA) > 2AC + 2BD
⇒ 2(AB + BC + CD + DA) > 2(AC + BD)
∴ AB + BC + CD + DA > AC + BD

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 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 12.4

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

Question 1.
In the figure, it is given that AB = CD and AD = BC. Prove that ∆ADC ≅ ∆CBA.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 Q1.1
Solution:
Given : In the figure, AB = CD, AD = BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 Q1.2
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
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 Q2.1
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

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

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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q4.1
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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q5.1
Ext. ∠ACB = ∠A + ∠B
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q5.2

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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q6.1
∴ ∠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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q7.1
∵ 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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q8.1
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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q9.1
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q10.1
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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q10.2

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

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q11.1

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,
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q12.1
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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q13.1
(a) 20°
(b) 50°
(c) 60°
(d) 70°
Solution:
In the figure, EC || AB
∠ECD = 70°, ∠BDO = 20°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q13.2
∵ 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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q14.1
Solution:
In the figure
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q14.2
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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q16.1
Solution:
In the figure, AB ⊥ BC
∠AGF = 32°
∴ ∠CGB = ∠AGF (Vertically opposite angles)
= 32°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q16.2
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q17.1
Solution:
In the figure, BC is produced both sides CA and BA are also produced
In ∆ABC,
∠B = 180° -y
and ∠C 180° – x
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q17.2
∴ 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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q18.1
Solution:
In the figure, l1 || l2
∴ ∠EBA = ∠BAH (Alternate angles)
∴ ∠BAH = 78°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q18.2
⇒ ∠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?
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q19.1
Solution:
In ∆ABC,
∠ACB = 180° – (x + 2x)
= 180° – 3x …(i)
and in ∆BDG,
∠BED = 180° – (2x + y) …(ii)
∠EGC = ∠AGD (Vertically opposite angles)
= 3y
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q19.2
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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q20.1
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q21.1
Solution:
In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q21.2
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q22.1
Solution:
In the figure, AC = BC, BP || CQ
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q22.2
∵ 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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q23.1
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)
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q23.2
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q24.1
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q25.1
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=
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q26.1
Solution:
In ∆ABC, ∠A = x°
and bisectors of ∠B and ∠C meet at O.
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q26.2

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
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q27.1
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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q28.1
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q28.2

Question 29.
In the figure , if l1 || l2, the value of x is
(a) 22 \(\frac { 1 }{ 2 }\)
(b) 30
(c) 45
(d) 60
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q29.1
Solution:
In the figure, l1 || l2
EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q29.2
∵ 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°
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q30.1
Solution:
RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS Q30.2

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

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

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

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

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.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q1.1
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
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q2.1
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
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q3.1
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)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q4.1
⇒ 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)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q5.1
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)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q6.1
∴ ∠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.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q7.1 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°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q8.1
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
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q9.1
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.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q10.1
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

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q11.1
∵ ∠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°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q12.1
Solution:
Given : PQRS is a square and SRT is an equilateral triangle. PT and QT are joined.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q12.2
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.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q13.1
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
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 Q13.2
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.

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