Class 9

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.1

Question 1.
Find the lateral surface area and total surface area of the following right prisms.
i)
L.S.A. = 4l²
= 4 × 4²
= 64cm²
T.S.A = 6l²
= 6 × 4² = 96cm²

ii) L.S.A. =2h(l + b)
= 2 × 5 (8 + 6)
= 10 × 14 = 140 cm²
T.S.A. = 2 (lb + bh + lh)
= 2(8 × 6 + 6 × 5 + 8 × 5)
= 2 (48 + 30 + 40)
= 236 cm²

Question 2.
The total surface area of a cube is 1350 sq.m. Find its volume.
Solution:
Given T.S.A. of a cube 6l² = 1350
l² = \(\frac{1350}{6}\)
l² = 225
∴ l = \(\sqrt{225}\) = 15m
∴ Volume of the cube = l³
= 15 × 15 × 15
= 3375 m³

Question 3.
Find the area of four walls of a room (Assume that there are no doors or windows) if its length 12 m; breadth 10 m and height 7.5 m.
Solution:
Length of the room = 12 m
Breadth of the room = 10 m
Height of the room = 7.5 m
Area of four walls of the room
A = 2h (l + b)
A = 2 × 7.5 (12 + 10)
= 15 × 22
= 330 m²

Question 4.
The volume of a cuboid is 1200 cm³. The length is 15 cm and breadth is 10 cm. Find its height.
Solution:
Length of a cuboid, ‘l’ = 15 cm
Breadth of the cuboid, b = 10 cm
Volume of the cuboid, V = lbh = 1200.cm³
Let the height = h
∴ 15 × 10 × h = 1200
∴ h = \(\frac{1200}{15 \times 10}\)
= 8 cm

Question 5.
How does the total surface area of a box change if
i) Each dimension is doubled ?
Solution:
Let the original dimensions be Length – l units
Breadth – b units
Height – h units
Then T.S.A = 2 (lb + bh + lh)
If the dimensions are doubled then
Length = 2l
Breadth = 2b
Height = 2h
T.S.A. = 2 (2l. 2b + 2b . 2h + 2l . 2h)
= 2 (4lb + 4bh + 4lh)
= 4 × [2 (lb + bh + lh]
= 4 × original T.S.A.
i.e., T.S.A. increases by 4 times.

ii) Each dimension is tripled ?
Solution:
Let the original and changed dimensions are l, b, h and 31, 3b, 3h
Original T.S.A. = 2 (lb + bh + lh)
Changed T.S.A
= 2 (3l . 3b + 3b . 3h + 3l. 3h)
Changed T S.A. = 2 (9lb + 9bh + 9lh)
= 9 × [2 (lb + bh + lh)]
= 9 [original T.S.A.]
Thus original T.S.A. increased by 9 times if each dimension is tripled.

Question 6.
The base of a prism is triangular in shape with sides 3 cm, 4 cm and 5 cm. Find the volume of the prism if its height is 10 cm.
Solution:
Volume of triangular prism = Area of the base × height
Sides of the triangle are 3 cm, 4 cm and 5 cm.
Area = s (s – a) (s – b) (s – c)

∴ Volume of the prism = 6 × 10 = 60 cm³
(OR)
3 cm, 4 cm and 5 cm are the sides of right triangle.
∴ Area of the triangle
= \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) × 3 × 4 = 6 cm²
Volume of prism = base area × height
= 6 × 10 = 60cm³

Question 7.
A regular square pyramid is 3 m height and the perimeter of its base is16 m Find the volume of the pyramid.
Solution:
Perimeter of the base = 16 m
Height of the pyramid 3 m
Volume of the pyramid
= \(\frac { 1 }{ 3 }\) × volume of prism
= \(\frac { 1 }{ 3 }\) × (base area x height)
= \(\frac { 1 }{ 3 }\) × 4 × 4 × 3= 16m [4 × side=16 ∴ side = 4 m Area = s² = 4 × 4]

Question 8.
An Olympic swimming pool is in the shape of a cuboid of dimensions 50 m long and 25 m wide. If it is 3 m deep throughout, how many litres of water does it hold ?
Solution:
Dimensions of the swimming pool are
Length = 50 m
Breadth = 25 m
Deep = 3 m
∴ Volume of the swimming pool
V = lbh
V = 50 × 25 × 3 = 3750 m³
∴ It can hold 37,50,000 litres of water.
[∵ 1 m³ = 1000 lit.]

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 9th Lesson Statistics Exercise 9.2

Question 1.
Weights of parcels in a transport office are given below.

Find the mean weight of the parcels.
Solution:

Weight in kg xi	No. of parcels fi	x1fi
50	25	1250
65	34	2210
75	38	2850
90	40	3600
110	47	5170
120	16	1920

Σf_i = 200
Σf_ix_i = 17000
\(\begin{array}{l}
\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{17000}{200}=\frac{170}{2} \\
\overline{\mathrm{x}}=85
\end{array}\)
Mean = 85

Question 2.
Number of familles In a village in correspondence with the number of children are given below.

Find the mean number of children per family.
Solution:

No. of childrens xi	No. of families fi	x1fi
0	11	0
1	25	25
2	32	64
3	10	30
4	5	20
6	1	5

Σf_i = 84
Σf_ix_i = 144
\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{144}{84}\)
Mean = 1.714285

Question 3.
If the mean of the following frequency distribution is 7.2, find value of ‘k’.

Solution:

Σf_i = 40 + k;
Σf_ix_i = 260 + 10k
Given that \(\overline{\mathrm{x}}\) = 7.2
But \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{1} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
7.2 = \(\frac{260+10 k}{40+k}\)
288.0 + 7.2k = 260 + 10k
10k – 7.2k = 288 – 260
2.8k = 28
k = \(\frac{28}{2.8}\) = 10

Question 4.
Number of villages with respect to their population as per India census 2011 are given below.

Find the average population in each village.
Solution:

Population (in thousands xi)	Villages fi	x1fi
12	20	240
5	15	75
30	32	960
20	35	700
15	36	540
8	7	56

Σf_i = 145 Σf_ix_i = 2571 thousands
\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Mean = \(\frac{2571}{145}\) = 17.731 thousands

Question 5.
A FLATOUN social and financial educational programme initiated savings programme among the high school children in Hyderabad district. Mandal wise savings in a month are given in the following table.

Mandal	No. of schools	Total amount saved (in rupees
Amberpet	6	2154
Thirumalgiri	6	2478
Saidabad	5	975
Khairathabad	4	912
Secunderabad	3	600
Bahadurpura	9	7533

Find arithmetic mean of school wise savings in each mandal. Also find the arithmetic mean of saving of all schools.
Solution:

Σf_i = 33
Σf_ix_i = 14652
Mean = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\bar{x}=\frac{14652}{33}\) = ₹ 444 (Mean savings per school)

Question 6.
The heights of boys and girls of IX class of a school are given below.

Compare the heights of the boys and girls.
[Hint: Fliid median heights of boys and girls]
Solution:

Boys median class =\(\frac{37+1}{2}=\frac{38}{2}\)= 19th observation
∴ Median height of boys = 147 cm
Girls median class = \(\frac{29+1}{2}=\frac{30}{2}\) = 15th observation
∴ Median height of girls = 152 cm

Question 7.
Centuries scored and number of cricketers in the world are given below.

Find the mean, median and mode of the given data.
Solution:

Question 8.
On the occasion of New year’s day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet is given as follows

Find the mean, median and mode of the given data.
Solution:

N = Σf_i = 150
Σf_ix_i = 12000
Mean = \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{12000}{150}=80\)
Median = average of (\(\frac{N}{2}+1\) and \(\frac{N}{2}\) terms = average of 75 and 76 observation = 75
Mode = 50

Question 9.
The mean (average) weight of three students is 40 kg. One of the students Ranga weighs 46 kg. The other two students, Rahim and Reshma have the same weight.
Find Rahim’s weight. cgigB)
Solution:
Weight of Ranga = 46 kg
Weight of Reshma = Weight of Rahim = x kg say
Average = \(\frac{\text { Sum of the weights }}{\text { Number }}\) = 40kg
∴ 40 = \(\frac{46+x+x}{3}\)
3 x 40 = 46 + 2x
2x = 120 – 46 = 74
∴ x = \(\frac{74}{2}\) = 37 .
∴ Rahim’s weight = 37 kg.

Question 10.
The donations given to an orphanage home by the students of different classes of a secondary school are given below.

Class	Donation by each student in (Rs)	No. of students donated
VI	5	15
VII	7	15
VIII	10	20
IX	15	16
X	20	14

Find the mean, median and mode of the data.
Solution:

Σf_i = 80
Σf_ix_i = 900
Mean \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{900}{80}=11.25\)
Median = Average of \(\left(\frac{\mathrm{N}}{2}\right)\) and \(\left(\frac{\mathrm{N}}{2}+1\right)\) terms of \(\frac{80}{2},\left(\frac{80}{2}+1\right)\) terms
= average of 40 and 41 terms = ₹10
Mode = ₹ 10

Question 11.
There are four unknown numbers. The mean of the first two numbers is 4 and the mean of the first three is 9. The mean of all four numbers is 15; if one of the four numbers is 2 find the other numbers.
Solution:
We know that mean = \(\frac{\text { sum }}{\text { number }}\)
Given that, Mean of 4 numbers = 15
⇒ Sum of the 4 numbers = 4 x 15 = 60
Mean of the first 3 numbers = 9
⇒ Sum of the first 3 numbers = 3 x 9 = 27
Mean of the first 2 numbers = 4
⇒ Sum of the first 2 numbers = 2 x 4 = 8
Fourth number = sum of 4 numbers – sum of 3 numbers = 60 – 27 = 33
Third number = sum of 3 numbers – sum of 2 numbers = 27 – 8 = 19
Second number = Sum of 2 numbers – given number = 8-2 = 6
∴ The other three numbers are 6, 19, 33.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at ‘O’. Join A to O. Show that (i) OB = OC (ii) AO bisects ∠A.

Solution:
Given that in ΔABC
AB = AC
Bisectors of ∠B and ∠C meet at ‘O’.
To prove
i) OB = OC
∠B = ∠C (Angles opposite to equal, sides)
\(\frac{1}{2} \angle \mathrm{B}=\frac{1}{2} \angle \mathrm{C}\) (Dividing both sides by 2)
∠OBC = ∠OCB
⇒ OB = OC (∵ Sides opposite to equal angles in ΔOBC)

ii) AO bisects ∠A.
In ΔAOB and ΔAOC
AB = AC (given)
BO = CO (already proved)
∠ABO = ∠ACO (∵ ∠B =∠C)
∴ ΔAOB ≅ ΔAOC
⇒ ∠BAO = ∠CAO [ ∵ CPCT of ΔAOB and ΔAOC]
∴ AO is bisector of ∠A.

Question 2.
In ΔABC, AD is the perpendicular bisector of BC (see given figure). Show that ΔABC is an isosceles triangle in which AB = AC

Solution:
Given that AD ⊥ BC; AD = DC
In ΔABD and ΔACD
AD = AD (common)
BD = DC (given)
∠ADB = ∠ADC (given)
∴ ΔABD ≅ ΔACD (∵ SAS congruence)
⇒ AB = AC (CPCT of ΔABD and ΔACD)

Question 3.
ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Solution:
Given that AC = AB; BD ⊥ AC; CE ⊥ AB
In ΔBCD and ΔCBE
∠BDC = ∠CEB (90° each)
∠BCD = ∠CBE (∵ angles opp. to equal sides of a triangle)
BC = BC
∴ ΔBCD ≅ ΔCBE (∵ AAScongruence)
⇒ BD = CE (CPCT)

Question 4.
ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal (see figure). Show that
i) ΔABD ≅ ΔACE
ii) AB = AC i.e., ABC is an isosceles triangle.

Solution:
Given that BD ⊥ AC; CE ⊥ AC
BD = CE
Now in ΔABD and ΔACE
∠ADB = ∠AEC (∵ given 90°)
∠A = ∠A (commori angle)
BD = CE
∴ ΔABD = ΔACE (∵ AAS congruence)
⇒ AB = AC (∵ C.P.C.T)

Question 5.
ΔABC and ΔDBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Solution:

Given that ΔABC and ΔDBC are isosceles.
To prove ∠ABD = ∠ACD
Join A and D.
Now in ΔABD and ACD
AB = AC (∵ equal sides of isosceles triangles)
BD = CD (∵ equal sides of isosceles triangles)
AD = AD (∵ common side)
∴ ΔABD ≅ ΔACD (∵ SSS congruence)
⇒ ∠ABD = ∠ACD (CPCT)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 9 Statistics Ex 9.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 9th Lesson Statistics Exercise 9.1

Question 1.
Write the mark wise frequencies in the following frequency distribution table.

Solution:

Question 2.
The blood groups of 36 students of IX class are recorded as follows.

Represent the data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these groups ?

Blood group	A	B	AB	O
Frequency	10	9	2	15

From the table, most common group is O and rarest group is AB.

Question 3.
Three coins were tossed 30 times simultaneously. Each time the occurring was noted down as follows :

Prepare a frequency distribution table for the data given above.
Solution:

No. of heads	0	1	2	3
Frequency	3	10	10	7

Question 4.
A T.V. channel organized a SMS (Short Message Service) poll on prohibition on smoking giving options like A – complete prohibitions, B – prohibition in public places only, C – not necessary. SMS results in one hour were

Represent the above data as grouped frequency distribution table. How many appropriate answers were received ? What was the majority of people’s opinion ?
Solution:

Options	A	B	C
Frequency(f)	19	36	10

Total appropriate answers received = 19 + 36 + 10 = 65
Majority of people’s opinion is prohibition in public places only i.e., B.

Question 5.
Represent the data in the given bar graph as frequency distribution table.

Solution:

Question 6.
Identify the scale used on the axes of the given graph. Write the frequency distribu- tion from it.
Solution:
Frequency distribution table :

Class	No. of students
I	40
II	55
III	65
IV	30
V	15

Scale : X – axis : 1 cm = 1 class interval
Y – axis : 1 cm = 10 students

Question 7.
The marks of 30 students of a class, obtained in a test (out of 75), are given below : 42, 21, 50, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29, 59, 61, 33, 17, 17, 39, 44, 42, 39, 14, 7, 27, 19, 54, 51. Form a frequency table with equal class intervals.
(Hint: One of them being 0 – 10)
Solution:

Question 8.
The electricity bill (in rupees) of 25 houses in a locality are given below. Construct a grouped frequency distribution table with a class size of 75.
170, 212, 252, 225, 310, 712, 412, 425, 322, 325, 192, 198, 230, 320, 412, 530, 602, 724, 370, 402, 317, 403, 405, 372, 413.
Solution:
The least value of observations = 170
The height value of observations = 724

Question 9.
A company manufactures car batteries of a particular type. The life (in years) of 40 batteries were recorded as follows.

Construct a grouped frequency distribution table with exclusive classes for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution:

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.4

Question 1.
ABC is a triangle. D is a point on AB such that AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such that AE = \(\frac { 1 }{ 4 }\) AC. If DE = 2 cm find BC.
Solution:

Given that D and E are points on AB and AC.
Such that AD = \(\frac { 1 }{ 4 }\) AB and AE = \(\frac { 1 }{ 4 }\) AC
Let X, Y be midpoints of AB and AC.
Joint D, E and X, Y.
Now in ΔAXY; D, E are the midpoints of sides AX and AY.
∴ DE // XY and DE = \(\frac { 1 }{ 2 }\) XY
⇒ 2 cm = \(\frac { 1 }{2 }\) XY
⇒ XY = 2 x 2 = 4cm
Also in ΔABC; X, Y are the midpoints of AB and AC.
∴ XY//BC and XY = \(\frac { 1 }{2 }\) BC
4 cm = \(\frac { 1 }{2 }\) BC
⇒ BC = 4 x 2 = 8 cm

Question 2.
ABCD is a quadrilateral. E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.

Solution:
Given that E, F, G and H are the midpoints of the sides of quad. ABCD.
In ΔABC; E, F are the midpoints of the sides AB and BC.
∴ EF//AC and EF = \(\frac { 1 }{2 }\) AC
Also in ΔACD; HG // AC
and HG = \(\frac { 1 }{ 2 }\) AC
∴ EF // HG and EF = HG
Now in □EFGH; EF = HG and EF // HG
∴ □EFGH is a parallelogram.

Question 3.
Show that the figure formed by joining the midpoints of sides of a rhom¬bus successively is a rectangle.
Solution:

Let □ABCD be a rhombus.
P, Q, R and S be the midpoints of sides of □ABCD
In ΔABC,
P, Q are the midpoints of AB and BC.
∴ PQ//AC and PQ = \(\frac { 1 }{2 }\)AC …………………..(1)
Also in ΔADC, ,
S, R are the midpoints of AD and CD.
∴ SR//AC and SR = \(\frac { 1 }{2 }\)AC ………………(2)
From (1) and (2);
PQ // SR and PQ = SR
Similarly QR // PS and QR = PS
∴ □PQRS is a parallelogram.
As the diagonals of a rhombus bisect at right angles.
∠AOB – 90°
∴ ∠P = ∠AOB = 90°
[opp. angles of //gm PYOX] Hence □PQRS is a rectangle as both pairs of opp. sides are equal and parallel, one angle being 90°.

Question 4.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.

Solution:
□ABCD is a parallelogram. E and F are the mid points of AB and CD.
∴ AE = \(\frac { 1 }{2 }\)AB and CF = \(\frac { 1 }{2 }\)CD
Thus AE = CF [∵ AB – CD]
Now in □AECF, AE = CF and AE ||CF
Thus □AECF is a parallelogram.
Now in ΔEQB and ΔFDP
EB = FD [Half of equal sides of a //gm]
∠EBQ = ∠FDP[alt. int.angles of EB//FD]
∠QEB = ∠PFD
[∵∠QED = ∠QCF = ∠PFD]
∴ ΔEQB ≅ ΔFPD [A.S.A. congruence]
∴ BQ = DP [CPCT] ……………… (1)
Now in ΔDQC; PF // QC and F is the midpoint of DC.
Hence P must be the midpoint of DQ
Thus DP = PQ …………….. (2)
From (1) and (2), DP = PQ = QB
Hence AF and CE trisect the diagonal BD.

Question 5.
Show that the line segments joining the mid points of the opposite sides of a quadrilateral and bisect each other.

Solution:
Let ABCD be a quadrilateral.
P, Q, R, S are the midpoints of sides of □ABCD.
Join (P, Q), (Q, R), (R, S) and (S, P).
In ΔABC; P, Q are the midpoints of AB and BC.
∴ PQ // AC and PQ = \(\frac { 1 }{2 }\)AC ………….(1)
Also from ΔADC
S, R are the midpoints of AD and CD
SR // AC and SR = \(\frac { 1 }{2 }\) AC …………………(2)
∴ From (1) & (2)
PQ = SR and PQ //SR
∴ □PQRS is a parallelogram.
Now PR and QS are the diagonals of □ PQRS.
∴ PR and QS bisect each other.

Question 6.
ABC is a triangle right angled at’C’. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that
i) D is the midpoint of AC
ii) MD ⊥ AC
iii) CM = MA= \(\frac { 1 }{2 }\)AB

Solution:

Given that in ΔABC; ∠C = 90°
M is the midpoint of AB.
i) If ‘D’ is the midpoints of AC.
The proof is trivial.
Let us suppose D is not the mid point of AC.
Then there exists D’ such that AD’ = D’C
Then D’M is a line parallel to BC through M.
Also DM is a line parallel to BC through M.
There exist two lines parallel to same line through a point M.
This is a contradiction.
There exists only one line parallel to a given line through a point not on the line.
∴ D’ must coincides with D
∴ D is the midpoint of AC

ii) From (i) DM // BC
Thus ∠ADM = ∠ACB = 90°
[corresponding angles]
⇒ MD ⊥ AC

iii) In ∆ADM and ∆CDM
AD = CD [ ∵ D is midpoint from (i)]
∠ADM = ∠MDC (∵ 90° each)
DM = DM (Common side)
∴ ∆ADM = ∆CDM (SAS congruence)
⇒ CM = MA (CPCT)
CM = \(\frac { 1 }{2 }\) AB (∵ M is the midpoint of AB)
∴ CM = MA = \(\frac { 1 }{2 }\)AB

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.3

Question 1.
The opposite angles of a parallelogram are (3x – 2)° and (x + 48)°. Find the measure of each angle of the parallelogram.
Solution:
Given that the opposite angles of a parallelogram are (3x – 2)° and (x + 48)°
Thus 3x – 2 = x + 48
(∵ opp. angles of a //gm are equal)
3x – x = 48 + 2
2x = 50
x = \(\frac{50}{2} \) = 25°
∴ The given angles are (3 x 25 – 2)° and (25 + 48) °
= (75 – 2)° and 73° = 73° and 73°
We know the consecutive angles are supplementary.
∴ The other two angles are (180°-73°) and (180°-73°)
= 107° and 107°
∴ The four angles are 73°, 107°, 73° and 107°.

Question 2.
Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.
Solution:
Let the smallest angle = x
Then its consecutive angle = 180 – x°
By problem (180 – x)° = (2x- 24)°
(∵ opp. angles are equal)
180 + 24 = 2x + x
3x = 204
x = \(\frac{204}{3} \) = 68°
∴ The angles are
68°; (2 x 68 – 24)°; 68°; (2 x 68 – 24)°
= 68°, 112°, 68°, 112°

Question 3.
In the given figure ABCD is a paral-lelogram and E is the mid point of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.

Solution:

Given that □ABCD is a parallelogram.
E is the midpoint of BC.
Let G be the midpoint of AD.
Join G, E.
Now in ΔAFD, GE is the line joining the midpoints G, E of two sides AD and FD.
∴GE // AF and GE = \(\frac{1}{2}\)AF
But GE = AB [ ∵ ABEG is a parallelo¬gram and AB, GE forms a pair of opp. sides]
\(\frac{1}{2}\) = AB ⇒ AF = 2AB
Hence Proved.

Question 4.
In the given figure ABCD is a paral¬lelogram. P, Q are the midpoints of sides AB and DC respectively. Show that
Solution:

□ABCD is a parallelogram.
P, Q are the mid points of AB and CD.
Join Q, P.
Now AB = CD (Opp. sides of a //gm)
\(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD
PB = QC
Also PB // QC.
Now in □PBCQ;
PB = QC; PB//QC
Hence □PBCQ is a parallelogram.

Question 5.
ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD//BA as shown in the figure. Show that i) ∠DAC = ∠BCA
ii) ABCD is a parallelogram.

Solution:
Given that AABC is isosceles; AB = AC
AD is bisector of ∠QAC

i) In ΔABC, AB = AC ⇒ ∠B = ∠ACB
(angles opp. to equal sides)
Also ∠QAC = ∠B + ∠ACB
∠QAC = ∠BCA + ∠BCA
(∵∠BCA = ∠B)
⇒ \(\frac{1}{2}\)∠QAC = \(\frac{1}{2}\) [2 ∠BCA]
⇒ ∠DAC = ∠BCA [ ∵ AD is bisector of ∠QAC]

ii) From (i) ∠DAC = ∠BCA
But these forms a pair of alt. int. angles for the pair of lines AD and BC; AC as a transversal.
∴ AD//BC
In □ABCD ; AB // DC; BC // AD
□ABCD is a parallelogram.

Question 6.
ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure). Show that 1) ΔAPB ≅ ΔCQD ii) AP = CQ.

Solution:
Given that □ABCD is a parallelogram.
BD is a diagonal.
AP ⊥ BD and CQ ⊥ BD
i) In ΔAPB and ΔCQD
AB = CD ( ∵ Opp. sides of //gm ABCD)
∠APB = ∠CQD (each 90°)
∠PBA = ∠QDC (alt. int. angles for the lines AB and DC)
∴ ΔAPB ≅ ΔCQD (AAS congruence)

ii) From (1) ΔAPB ≅ ΔCQD
⇒ AP = CQ (CPCT)

Question 7.
In Δs ABC and Δs DEF, AB = DC and AB//DE; BC = EF and BC//EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that
i) ABED is a parallelogram
ii) BCFE is a parallelogram
iii) AC = DF
iv) ΔABC = ΔDEF

Solution:

Given that in ΔABC and ΔDEF
AB = DE and AB // DE
BC = EF and BC//EF.
i) In □ABED AB//ED and AB = ED
Hence □ABED is a parallelogram.

ii) In □BCFE; BC = EF and BC//EF
Hence □BCFE is a parallelogram.

iii) ACFD is a parallelogram (In a paral-lelogram opposite sides are equal).
So, AC = DF.

iv) Consider ΔABC = ΔDEF
AB = DE (given);
AC = DF (proved)
BC = EF (given)
∴ ΔABC ≅ ΔDEF (SSS congruency rule).

Question 8.
ABCD is a parallelogram. AC and BD are the diagonals intersect at ‘O’. P and Q are the points of trisection of the diagonal BD. Prove that CQ//AP and also AC bisects PQ.

Solution:
Given □ABCD is a parallelogram;
BD is a diagonal.
P, Q are the points of trisection of BD.
In ΔAPB and ΔCQD
AB = CD (•.• Opp. sides of //gm ABCD)
BP = DQ (given)
∠ABP = ∠CDQ (alt. int. angles for the lines AB//DC, BD as a transversal)
ΔAPB = ΔCQD (SAS congruence)
Similarly in ΔAQD and ΔCPB
AD = BC (opp. sides of //gm ABCD)
DQ = BP (given)
∠ADQ = ∠CBP (all int. angles for the lines AD//BC, BD as a transversal)
ΔAQD ≅ ΔCPB
Now in □APCQ
AP = CQ (CPCT of AAPB, ACQD)
AQ = CP (CPCT of AAQD and ACPB)
∴ □APCQ is a parallelogram.
∴ CQ//AP (opp. sides of//gm APCQ)
Also AC bisects PQ. [ ∵ diagonals of //gm APCQ]

Question 9.
ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution:

Given that ABCD is a square.
E, F, G, H are the mid points of AB, BC, CD and DA.
Also AE = BF = CG = DH
In ΔABC; E, F are the mid points of sides AB and BC.
∴ EF//AC and EF = \(\frac{1}{2}\) AC
Similarly GH//AC and GH = AC
GF//BD and GF = \(\frac{1}{2}\) BD
HE//BD and HE = \(\frac{1}{2}\) BD

But AC = BD (∵ diagonals of a square)
∴ EF = FG = GH = HE
Hence EFGH is a rhombus.
Also AC ⊥ BD
(∵ diagonals of a rhombus)
∴ In //gm OIEJ [ ∵ 0I // EJ; IE // OJ]
We have ∠IOJ = ∠E
[ ∵ Opp. angles of a //gm]
∴ ∠E – 90°
Hence in quad. EFGH; all sides are equal and one angle is 90°.
∴ EFGH is a square.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.2

Question 1.
In the given figure ABCD is a parallelogram. ABEF is a rectangle. Show that ΔAFD ≅ ΔBEC

Solution:
Given that □ABCD is a parallelogram.
□ABEF is a rectangle.
In ΔAFD and ΔBEC
AF = BE ( ∵ opp. sides of rectangle □ABEF)
AD = BC (∵ opp. sides of //gm □ABCD)
DF = CE (∵ AB = DC = DE + EC , AB = EF = DE + DF)
∴ ΔAFD ≅ ΔBEC (SSS congruence)

Question 2.
Show that the diagonals of a rhombus divide it into four congruent triangles.
Solution:

□ABCD is a rhombus.
Let AC and BD meet at O’.
In ΔAOB and ΔCOD
∠OAB = ∠OCD (alt.int. angles)
AB = CD (def. of rhombus)
∠OBA = ∠ODC ………………….(1) (alt. int. angles)
∴ ΔAOB ≅ ΔCOD (ASA congruence)
Thus AO = OC (CPCT)
Also ΔAOD ≅ ΔCOD …………..(2)
[ ∵ AO = OC; AD = CD; OD = OD SSS congruence]
Similarly we can prove
ΔAOD ≅ ΔCOB ……………. (3)
From (1), (2) and (3) we have
ΔAOB ≅ ΔBOC ≅ ΔCOD ≅ ΔAOD
∴ Diagonals of a rhombus divide it into four congruent triangles.

Question 3.
In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O. Prove that ∠COD = \(\frac{1}{2}\) (∠A + ∠B) .
(OR)
In a quadrilateral ABCD, the bisectors of ∠A and ∠B are intersects at ‘O’ then prove that ∠AOB = \(\frac{1}{2}\) (∠C + ∠D)
Solution:

In a quadrilateral □ABCD
∠A + ∠B + ∠C + ∠D = 360°
(angle sum property)
∠C + ∠D = 360° – (∠A + ∠B)
\(\frac{1}{2}\) (∠C + ∠D) = 180 – \(\frac{1}{2}\) (∠A + ∠B) ………….. (1)
(∵ dividing both sides by 2) .
But in ΔCOD
\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D + ∠COD = 180°
\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D = 180° – ∠COD
∴\(\frac{1}{2}\)(∠C +∠D) = 180° -∠COD………….(2)
From (1) and (2);
180° – ∠COD = 180° – \(\frac{1}{2}\) (∠A + ∠B)
∴ ∠COD = \(\frac{1}{2}\) (∠A + ∠B)
Hence proved.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.1

Question 1.
State whether the statements are true or false.
i) Every parallelogram is a trapezium.
ii) All parallelograms are quadrilaterals.
iii) All trapeziums are parallelograms.
iv) A square is a rhombus.
v) Every rhombus is a square.
vi) All parallelograms are rectangles.
Solution:
i) True
ii) True
iii) False
iv) True
v) False
vi) False

Question 2.
Complete the following table by writing YES if the property holds for the particular quadrilateral and NO if property does not holds.

Solution:

Question 3.
ABCD is a trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.
Solution:
Given that in □ABCD AB || CD; AD = BC
Mark a point ‘E’ on AB such DC = AE.

Join E, C.
Now in AECD quadrilateral
AE // DC and AE = DC
∴ □AECD is a parallelogram.
∴ AD//EC
∠DAE = ∠CEB (corresponding angles) ……………..(1)
In ΔCEB; CE = CB (∵ CE = AD)
∴ ∠CEB = ∠CBE (angles opp. to equal sides) …………….. (2)
From (1) & (2)
∠DAE = ∠CBE
⇒ ∠A = ∠B
Also ∠D = ∠AEC (∵ Opp. angles of a parallelogram)
= ∠ECB + ∠CBE [ ∵ ∠AEC is ext. angle of ΔBCE] |
= ∠ECB + ∠CEB [ ∵∠CBE = ∠CEB]
= ∠ECB + ∠ECD [∵ ∠ECD = ∠CEB alt. int. angles]
= ∠BCD = ∠C
∴ ∠C = ∠D

Question 4.
The four angles of a quadrilateral are in the ratio of 1 : 2 : 3 : 4. Find the measure of each angle of the quadri-lateral.
Solution:
Given that, the ratio of angles of a quad-rilateral = 1 : 2 : 3 : 4
Sum of the terms of the ratio
= 1 +2 + 3 + 4= 10
Sum of the four interior angles of a quadrilateral = 360°
∴ The measure of first angle
= \(\frac{1}{10}\) × 360° = 36°
The measure of second angle
= \(\frac{2}{10}\) × 360° = 72°
The measure of third angle
= \(\frac{3}{10}\) × 360° = 108°
The measure of fourth angle
= \(\frac{4}{10}\) × 360° = 144°

Question 5.
ABCD is a rectangle, AC is diagonal. Find the angles of ΔACD. Give reasons.
Solution:

Given that □ABCD is a rectangle;
AC is its diagonal.
In ΔACD; ∠D = 90° [ ∵ ∠D is also angle of the rectangle]
∠A + ∠C = 90° [ ∵ ∠D = 90° ⇒ ∠A + ∠C = 180°-90° = 90°]
(i.e,,) ∠D right angle and
∠A, ∠C are complementary angles.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:

Let a ΔABC be right angled at ∠B.
Then ∠A + ∠C = 90°
(i.e.,) ∠A and ∠C are both acute.
Now, ∠A < ∠B ⇒ BC < AC
Also ∠C < ∠B ⇒ AB < AC
∴ AC, the hypotenuse is the longest side.

Question 2.
In the given figure, sides AB and AC of ΔABC are extended “to points P and Q respectively. Also ∠PBC < ∠QCB. Show that AC > AB.

Solution:
From the figure,
∠PBC = ∠A + ∠ACB
∠QCB = ∠A + ∠ABC
Given that ∠PBC < ∠QCB
⇒∠A + ∠ACB < ∠A + ∠ABC
⇒ ∠ACB < ∠ABC
⇒ AB < AC
⇒ AC > AB
Hence proved.

Question 3.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Given that ∠B < ∠A; ∠C < ∠D
∠B < ∠A ⇒ AO < OB [in ΔAOB] ……………… (1)
∠C < ∠D ⇒ OD < OC [in ΔCOD]…… (2)
Adding (1) & (2)
AO + OD < OB + OC
AD < BC
Hence proved.

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A >∠C and ∠B > ∠D.

Solution:

Given that AB and CD are the smallest and longest sides of quadrilateral ABCD.
From the figure,
In ΔBCD
∠1 > ∠2 [∵ DC > BC] ………………(1)
In ΔBDA
∠4 > ∠3 [∵ AD > AB] ………….(2)
Adding (1) & (2)
∠1 + ∠4 > ∠2 + ∠3
∠B > ∠D
Similarly,
In ΔABC, ∠6 < ∠7 [ ∵AB < BC] ……………….(3)
In ΔACD
∠5 < ∠8 …………. (4)
Adding (3) & (4)
∠6 + ∠5 < ∠7 + ∠8
∠C < ∠A ⇒ ∠A > ∠C
Hence proved.

Question 5.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that
∠PSR > ∠PSQ.

Solution:
Given that PR > PQ;
∠QPS =∠RPS
PR> PQ
∠Q > ∠R
Now ∠Q +∠QPS > ∠R + ∠RPS
⇒ 180° – (∠Q + ∠QPS) < 180° – (∠R + ∠RPS)
⇒ ∠PSQ < ∠PSR ⇒ ∠PSR > ∠PSQ
Hence proved.

Question 6.
If two sides of a triangle measure 4 cm and 6 cm find all possible measurements (positive integers) of the third side. How many distinct triangles can be obtained ?
Solution:
Given that two sides of a triangle are 4 cm and 6 cm.
∴ The measure of third side > Differ-ence between other two sides.
third side > 6 – 4
third side > 2
Also the measure of third side < sum of other two sides
third side <6 + 4 < 10
∴ 2 < third side <10
∴ The measure of third side may be 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm
∴ Seven distinct triangles can be obtained.

Question 7.
Try to construct a triangle with 5 cm, 8 cm and 1 cm. Is it possible or not ? Why ? Give your justification.
Solution:
As the sum (6 cm) of two sides 5 cm and 1 cm is less than third side. It is not possible to construct a triangle with the given measures.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.3

Question 1.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that, (i) AD bisects BC (ii) AD bisects ∠A.
Solution:

Given that in ΔABC, AB = AC
and AD ⊥ BC
i) Now in ΔABD and ΔACD
AB = AC (given)
∠ADB = ADC (given AD ⊥ BC)
AD = AD (common)
∴ ΔABD ≅ ΔACD (∵ RHS congruence)
⇒ BD = CD (CPCT)
⇒ AD, bisects BC.

ii) Also ∠BAD = ∠CAD
(CPCT of ΔABD ≅ ΔACD )
∴ AD bisects ∠A.

Question 2.
Two sides AB, BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see figure). Show that:
(i) ΔABM ≅ ΔPQN
ii) ΔABC ≅ ΔPQR

Solution:
Given that
AB = PQ
AM = PN
i) Now in ΔABM and ΔPQN
AB = PQ (given)
AM = PN (given)
BM = QN (∵ BC = QR ⇒ \(\frac { 1 }{ 2 }\)BC = \(\frac { 1 }{ 2 }\)QR ⇒ BM = QN)
∴ ΔABM ≅ ΔPQN
(∵ SSS congruence)

ii) In ΔABC and ΔPQR
AB = PQ (given)
BC = QR (given)
∠ABC = ∠PQN [∵ CPCT of ΔABM and ΔPQN from (i)]
∴ ΔABC ≅ ΔPQR
(∵ SAS congruence)

Question 3.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:

In ΔABC altitude BE and CF are equal.
Now in ΔBCE and ΔCBF
∠BEC = ∠CFB (∵ given 90°)
BC = BC (common; hypotenuse)
CF = BE (given)
∴ ΔBEC ≅ ΔCBF
⇒ ∠EBC = ∠FCB (∵ CPCT)
But these are also the interior angles opposite to sides AC and AB of ΔABC.
⇒ AC = AB
Hence proved.

Question 4.
ΔABC is an isosceles triangle in which AB = AC. Show that ∠B = ∠C.
(Hint : Draw AP ⊥ BQ (Using RHS congruence rule)
Solution:

Given the ΔABC is an isosceles triangle and AB = AC
Let D be the mid point of BC; Join A, D.
Now in ΔABD and ΔACD
AB = AC (given)
BD = DC (construction)
AD = AD (common)
∴ ΔABD ≅ ΔACD (∵ SSS congruence)
⇒ ∠B = ∠C [∵ CPCT]

Question 5.
ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Solution:

Given that in ΔDBC; AB = AC; AD = AB
In ΔABC
∠ABC + ∠ACB = ∠DAC …………… (1)
[∵ exterior angle]
In ΔACD
∠ADC + ∠ACD = ∠BAC ………………(2)
Adding (1) & (2)
∠DAC + ∠BAC = 2 ∠ACB + 2∠ACD
[∵ ∠ABC = ∠ACB
∠ADC = ∠ACD]
180° = 2 [∠ACB + ∠ACD]
180° = 2[∠BCD]
∴ ∠BCD = \(\frac{180^{\circ}}{2}\) = 90°
(or)
From the figure
∠2 = x + x = 2x
∠1 = y + y = 2y
∠1 + ∠2 = 2x + 2y
180° = 2 = (x + y)
∴ x + y  = \(\frac{180^{\circ}}{2}\) = 90°
Hence proved.

Question 6.
ABC is a right angled triangle in which ∠A = 90° and AB = AC, Show that ∠B = ∠C.
Solution:

Given ΔABC; AB – AC
Join the mid point D of BC to A.
Now in ΔADC and ΔADB
AD = AD (common)
AC = AB (giyen)
DC = DB (construction)
⇒ ΔADC ≅ ΔADB
⇒ ∠C = ∠B (CPCT)

Question 7.
Show that the angles of an equilateral triangle are 60° each.
Solution:

Given ΔABC is an equilateral triangle
AB = BC = CA
∠A = ∠B (∵ angles opposite to equal sides)
∠B = ∠C (∵ angles opposite to equal sides)
⇒ ∠A = ∠B = ∠C = x say
Also ∠A+∠B + ∠C =180°
⇒ x + x + x = 180°
3x = 180°
⇒ x = \(\frac{180}{3}\) = 60°
Hence proved.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1

Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD What can you say about BC and BD ?

Solution:
Given that AC = AD
∠BAC = ∠BAD (∵ AB bisects∠A)
Now in ΔABC and ΔABD
AC = AD (∵ given)
∠BAC = ∠BAD (Y given)
AB = AB (common side)
∴ ΔABC ≅ ΔABD
(∵ SAS congruence rule)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA, prove that i) ΔABD ≅ΔBAC ii) BD = AC
iii) ∠ABD = ∠BAC.

Solution :
i) Given that AD = BC and
∠DAB = ∠CBA
Now in ΔABD and ΔBAC
AB = AB (∵ Common side)
AD = BC (∵ given)
∠DAB = ∠CBA (∵ given)
∴ ΔABD ≅ ΔBAC
(∵ SAS congruence)
ii) From (i) AC = BD (∵ CPCT)
iii) ∠ABD = ∠BAC [ ∵ CPCT from (i)]

Question 3.
AD and BC are equal and perpendi-culars to a line segment AB. Show that CD bisects AB.

Solution:
Given that AD = BC; AD ⊥ AB; BC ⊥ AB
In ΔBOC and ΔAOD
∠BOC = ∠AOD (∵ vertically opposite angles)
∴ ΔOBC = ΔOAD (∵ right angle)
BC = AD
ΔOBC ≅ ΔOAD (∵ AAS congruence)
∴ OB = OA (∵ CPCT)
∴ ‘O’ bisects AB
Also OD = OC
∴ ‘O’ bisects CD
⇒ AB bisects CD

Question 4.
l and m are two parallel lines inter-sected by another pair of parallel lines p and q. Show that ΔABC ≅ ΔCDA.

Solution:
Given that l // m; p // q.
In ΔABC and ΔCDA
∠BAC = ∠DCA (∵ alternate interior angles)
∠ACB = ∠CAD
AC = AC
∴ ΔABC ≅ ΔCDA (∵ ASA congruence)

Question 5.
In the figure given below AC = AE; AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
Given that AC = AE, AB = AD and
∠BAD = ∠EAC
In ΔABC and ΔADE
AB = AD
AC = AE
∠BAD = ∠EAC
∴ ΔABC ≅ ΔADE (∵ SAS congruence)
⇒ BC = DE (CPCT)

Question 6.
In right triangle ABC, right angle is at ‘C’ M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that

i) ΔAMC = ΔBMD
ii) ∠DBC is a right angle
iii) ΔDBC = ΔACB
iv) CM = \(\frac{1}{2}\) AB
Solution:
Given that ∠C = 90°
M is mid point of AB;
DM = CM (i.e., M is mid point of DC)

i) In ΔAMC and ΔBMD
AM = BM (∵ M is mid point of AB)
CM = DM ( ∵ M is mid point of CD)
∠AMC = ∠BMD ( ∵ Vertically opposite angles)
∴ ΔAMC ≅ ΔBMD
(∵ SAS congruence)

ii) ∠MDB = ∠MCA
(CPCT of ΔAMC and ΔBMD)
But these are alternate interior angles for the lines DB and AC and DC as transversal.
∴DB || AC
As AC ⊥ BC; DB is also perpendicular to BC.
∴ ∠DBC is a right angle.

iii) In ΔDBC and ΔACB
DB = AC (CPCT of ΔBMD and ΔAMC)
∠DBC = ∠ACB = 90°(already proved)
BC = BC (Common side)
∴ ΔDBC ≅ ΔACB (SAS congruence rule)

iv) DC = AB (CPCT of ΔDBC and ΔACB)
\(\frac { 1 }{ 2 }\) DC = \(\frac { 1 }{ 2 }\) AB (Dividing both sides by 2)
CM = \(\frac { 1 }{ 2 }\)AB

Question 7.
In the given figure ΔBCD is a square and ΔAPB is an equilateral triangle.
Prove that ΔAPD ≅ ΔBPC.
[Hint: In ΔAPD and ΔBPC; \(\overline{\mathbf{A D}}=\overline{\mathbf{B C}}\), \(\overline{\mathbf{AP}}=\overline{\mathbf{BP}}\) and ∠PAD = ∠PBC = 90° – 60° = 30°]

Solution:
Given that □ABCD is a square.
ΔAPB is an equilateral triangle.
Now in ΔAPD and ΔBPC
AP = BP ( ∵ sides of an equilateral triangle)
AD = BC (∵ sides of a square)
∠PAD = ∠PBC [ ∵ 90° – 60°]
∴ ΔAPD ≅ ΔBPC (by SAS congruence)

Question 8.
In the figure given below ΔABC is isosceles as \(\overline{\mathbf{A B}}=\overline{\mathbf{A C}} ; \overline{\mathbf{B A}}\) and \(\overline{\mathbf{CA}}\) are produced to Q and P such that \(\overline{\mathbf{A Q}}=\overline{\mathbf{AP}}\). Show that \(\overline{\mathbf{PB}}=\overline{\mathbf{QC}}\) .
(Hint: Compare ΔAPB and ΔACQ)

Solution:
Given that ΔABC is isosceles and
AP = AQ
Now in ΔAPB and ΔAQC
AP = AQ (given)
AB = AC (given)
∠PAB = ∠QAC (∵ Vertically opposite angles)
∴ ΔAPB ≅ ΔAQC (SAS congruence)
∴ \(\overline{\mathbf{PB}}=\overline{\mathbf{QC}}\) (CPCT of ΔAPB and ΔAQC)

Question 9.
In the figure given below AABC, D is the midpoint of BC. DE ⊥ AB, DF ⊥ AC and DE = DF. Show that ΔBED ≅ AΔCFD.

Solution:
Given that D is the mid point of BC of ΔABC.
DF ⊥ AC; DE = DF
DE ⊥ AB
In ΔBED and ΔCFD
∠BED = ∠CFD (given as 90°)
BD = CD (∵D is mid point of BC)
ED = FD (given)
∴ ΔBED ≅ ΔCFD (RHS congruence)

Question 10.
If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.
Solution:

Let ΔABC be a triangle.
The bisector of ∠A bisects BC
To prove: ΔABC is isosceles
(i.e., AB = AC)
We know that bisector of vertical angle divides the base of the triangle in the ratio of other two sides.
∴ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{BC}}\)
Thus \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = 1( ∵ given)
⇒ AB = AC
Hence the Triangle is isosceless.

Question 11.
In the given figure ΔABC is a right triangle and right angled at B such that ∠BCA = 2 ∠BAC. Show that the hypotenuse AC = 2BC.

[Hint : Produce CB to a point D that BC = BD]
Solution:

Given that ∠B = 90°; ∠BCA = 2∠BAC
To prove : AC = 2BC
Produce CB to a point D such that
BC = BD
Now in ΔABC and ΔABD
AB = AB (common)
BC = BD (construction)
∠ABC =∠ABD (∵ each 90°)
∴ ΔABC ≅ ΔABD
Thus AC = AD and ∠BAC = ∠BAD = 30° [CPCT]
[ ∵ If ∠BAC = x then
∠BCA = 2x
x + 2x = 90°
3x = 90°
⇒ x = 30°
∴ ∠ACB = 60°]
Now in ΔACD,
∠ACD = ∠ADC = ∠CAD = 60°
∴∠ACD is equilateral ⇒ AC = CD = AD
⇒ AC = 2BC (∵ C is mid point)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.4

Question 1.
Give the graphical representation of the following equation
a) on the number line and b) on the Cartesian plane.
l) x = 3
ii) y + 3 = 0
iii) y = 4
iv) 2x – 9 = 0
v) 3x + 5 = 0
Solution:
i) x = 3 is a line parallel to Y-axis at a distance of 3 units on the right side of the origin.
ii) y + 3 = 0 y = – 3 is a line parallel to X-axis, below the origin.
iii) y = 4 is a line parallel to X-axis at a distance of 4 units above the origin.
iv) 2x – 9 = 0
⇒ x = \(\frac{9}{2}\) = 4.5 is a line parallel to Y-axis at a distance of 4.5 units, right side of the zero.
v) 3x + 5 = 0
⇒ 3x = -5 x = \(\frac{-5}{3}\) is a line parallel to Y – axis at a distance of \(\frac{5}{3}\)units on the left side of the origin.

x = 3

x	3	3	3
y	-1	2	4

y + 3 = 0

X	-4	6	8
y	-3	-3	-3

y = 4

x	-2	3	5	8
y	4	4	4	4

2x – 9 = 0

X	4.5	4.5	4.5
y	-2	4	6

3x + 5 = 0

x	\(\frac{-5}{3}\)	\(\frac{-5}{3}\)	\(\frac{-5}{3}\)
y	-1	2	4

Question 2.
Give the graphical representation of 2x – 11 = 0 as an equation in i) one variable ii) two variables
Solution:
2x – 11 = 0

x	5.5	5.5	5.5
y	-3	1	5

Question 3.
Solve the equation 3x + 2 = 8x – 8 and represent the solution on
i) the number line ii) the Cartesian plane.
Solution:
Given that 3x + 2 = 8x – 8
3x – 8x = – 8 – 2
– 5x = -10
x = \(\frac{-10}{-5}\) = 2

X	2	2	2
y	5	6	4

Question 4.
Write the equation of the line parallel to X-axis and passing through the point i) (0, – 3) ii) (0,4) iii) (2, – 5) iv) (3,4)
Solution:
i) The given point is (0, – 3)
Equation of a line parallel to X-axis is y = k
∴ Required equation is y = – 3 or y + 3 = 0

ii) The given point is (0, 4)
Equation of a line parallel to X-axis is y = k
∴ Required equation isy = 4ory-4 = 0

iii) The given point is (2, – 5)
Equation of a line parallel to X-axis is y = k
∴ Required equation isy = -5 or y + 5 = 0

iv) The given point is (3, 4)
Equation of a line parallel to X-axis is y = k
∴ Required equation isy = 4 or y – 4 = 0

Question 5.
Write the equation of the line parallel to Y-axis passing through the point
i) (- 4, 0)
ii) (2,0)
iii) (3, 5)
(iv) (- 4, – 3)
Solution:
Equation of a line parallel to Y-axis is x = k
∴ The required equations are
i) Through the point (- 4, 0) ⇒ the equation is x = – 4 or x + 4 = 0
ii) Through the point (2, 0) ⇒ the equation isx = 2orx-2 = 0
iii) Through the point (3, 5) ⇒ the equation isx = 3orx-3 = 0
iv) Through the point (- 4,-3) ⇒ the equation is x = – 4 or x + 4 = 0

Question 6.
Write the equation of three lines that are
1) Parallel to the X-axis
Solution:
y = 3
y = -4
y = 6

ii) Parallel to the Y-axis
Solution:
x = – 2
x = 3
x = 4