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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.2

10th Class Maths 8th Lesson Similar Triangles Ex 8.2 Textbook Questions and Answers

Question 1.
In the given figure, ∠ADE = ∠B
i) Show that △ABC ~ △ADE
ii) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.

Answer:
i) Given: △ABC and ∠ADE = ∠B
R.T.P: △ABC ~ △ADE.
Proof: In △ABC and △ADE
∠A = ∠A [∵ Common]
∠B = ∠ADE [∵ Given]
∴ ∠C = ∠AED [∵ By Angle Sum property of triangles] △ABC ~ △ADE by AAA similarity condition.]

ii) AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.
To find DE; △ABC ~ △ADE
Hence,
\(\frac{AB}{AD}\) = \(\frac{BC}{DE}\) = \(\frac{AC}{AE}\)
[∵ Ratios of corresponding sides are equal]

Question 2.
The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
Answer:

Given: △ABC ~ △PQR
Perimeter of △ABC = 30 cm.
Perimeter of △PQR = 20 cm.
AB = 12 cm.
To find: \(\overline{\mathrm{PQ}}\)
Ratio of perimeters = 30 : 20 = 3 : 2
Let the length of the side corresponding to the side with length 12 cm be x.
Then 30 : 20 : : 12 : x
30x = 20 x 12
\(x = \frac{20 \times 12}{30}\) = 8 cm

Question 3.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp-post is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Answer:

Given:
A lamp-post \(\overline{\mathrm{AB}}\) of height = 3.6 m
= 360 cm.
Speed of the girl = 1.2 m/sec.
Distance travelled in 4 sec = Speed x Time = 1.2 × 4 = 4.8 m = 480 cm.
\(\overline{\mathrm{CD}}\), height of the girl = 90 cm.
Let the length of the shadow at a distance of 4.8 m from the lamp post = x cm.
From the figure,
△ABE ~ △DCE
[∵ ∠B = ∠C = 90°
∠E = ∠C common
(A.A. similarity)]
Hence,
\(\frac{AB}{DC}\) = \(\frac{BE}{CE}\) = \(\frac{AE}{DE}\)
∴ \(\frac{360}{90}\) = \(\frac{480+x}{x}\)
⇒ 4 = \(\frac{480+x}{x}\)
⇒ 4x = 480 + x
⇒ 4x – x = 480
⇒ 3x = 480
⇒ x = 160 cm = 1.6 m
∴ Length of the shadow = 1.6 m

Question 4.
CM and RN are respectively the medians of similar triangles △ABC and △PQR. Prove that
i) △AMC ~ △PNR
ii) \(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\)
iii) △CMB ~ △RNQ

Answer:
Given : △ABC ~ △PQR
CM is a median through C of △ABC.
RN is a median through R of △PQR.
R.T.P:
i) △AMC ~ △PNR.
Proof: In △AMC and △PNR,
\(\frac{AC}{PR}\) = \(\frac{AM}{PN}\) and ∠A = ∠P [∵ In △ABC, △PQR and M, N are the mid-points of AB and PQ]
∴ △AMC ~ △PNR
[∵ SAS similarity condition]

ii) \(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\)
Proof: From (i) we have
△AMC ~ △PNR
Hence \(\frac{AC}{PQ}\) = \(\frac{AM}{PN}\) = \(\frac{CM}{RN}\)
[∵ Ratio of corresponding sides of two similar triangles are equal]
Thus, \(\frac{CM}{RN}\) = \(\frac{AM×2}{PN×2}\)
[Multiplying both numerator and the denominator by 2]
\(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\) [2AM = AB; 2PN = PQ]

iii) △CMB ~ △RNQ
Proof: In △CMB and △RNQ
∠B = ∠Q [Corresponding angles of △ABC and △PQR]
Also, \(\frac{BC}{RQ}\) = \(\frac{BM}{QN}\)

Thus, △CMB ~ △RNQ by S.A.S similarity condition.

Question 5.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point ‘O’. Using the criterion of similarity for two triangles, show that \(\frac{OA}{OC}\) = \(\frac{OB}{OD}\).
Answer:

Given : □ ABCD, AB || DC
The diagonals AC and BD intersect at ‘O’.
R.T.P: \(\frac{OA}{OC}\) = \(\frac{OB}{OD}\)
Construction: Draw EF || AB, passing through ‘O’.
Proof: In △ACD, OE || CD [∵ Construction]
Hence \(\frac{OA}{OC}\) = \(\frac{EA}{ED}\) …….. (1)
(∵ Line drawn parallel to one side of a triangle divides other two sides in the same ratio – Basic proportionality theorem)
Also in △ABD, EO || AB [Construction] Hence,
\(\frac{EA}{ED}\) = \(\frac{OB}{OD}\) ……… (2)
(∵ Basic proportionality theorem) From (1) and (2), we have
\(\frac{OA}{OC}\) = \(\frac{OB}{OD}\)
∴ Hence proved.

Question 6.
AB, CD, PQ are perpendicular to BD. AB = x, CD = y and PQ = z, prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\).

Answer:
Given ∠B = ∠Q = ∠D = 90°
Thus, AB || PQ || CD.
Now in △BQP, △BDC
∠B = ∠B (Common)
∠Q = ∠D (90°)
∠P = ∠C [∵ Angle Sum property of triangles]
∴ △BQP ~ △BDC
(by A.A.A similarity condition)
Hence \(\frac{BQ}{BD}\) = \(\frac{PQ}{CD}\)
[∵ Ratio of corresponding sides is equal] Also in △DQP and △DBA
∠D = ∠D (Common)
∠Q = ∠B (90°)
∴ △DQP ~ △DBA (by A.A. similarity condition)
\(\frac{QD}{BD}\) = \(\frac{PQ}{AB}\)
[ Ratio of corresponding sides is equal]
Adding (1) and (2), we get

Question 7.
A flag pole 4 m tall casts a 6 m., shadow. At the same time, a nearby building casts a shadow of 24 m. How tall is the building?
Answer:
Given: 4 m length flag pole casts a shadow 6 m.
Let x m length/tall building casts a shadow 24 m.

Let AB be the length of flag pole = 4 m.
Shadow of AB = BC = 6 m.
PQ be the building = x m (say)
QR, the shadow of the building = 24 m
From the figure,
∠A = ∠P
∠B = ∠Q
∴ △ABC ~ △PQR by A.A. similarity condition
Hence \(\frac{AB}{PQ}\) = \(\frac{BC}{QR}\)
[∵ Ratio of corresponding angles is equal]
\(\frac{4}{6}\) = \(\frac{x}{24}\)
x = \(\frac{24 \times 4}{6}\) = 16 m
∴ Height of the building = 16 m.

Question 8.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of △ABC and △FEG respectively. If △ABC ~ △FEG then show that
i) \(\frac{CD}{GH}\) = \(\frac{AC}{FG}\)
ii) △DCB ~ △HGE
iii) △DCA ~ △HGF
Answer:

Given: △ABC ~ △FEG.
CD is the bisector of ∠C and GH is the bisector of ∠G.
R.T.P.:
i) \(\frac{CD}{GH}\) = \(\frac{AC}{FG}\)
In △ACD and △FGH
∠A = ∠F
[∵ Corresponding angles of △ABC and △FEG]
∠ACD = ∠FGH
[∵ ∠C = ∠G ⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠ACD = ∠FGH]
∴ By A.A. similarity condition, △ACD ~ △FGH
\(\frac{AC}{FG}\) = \(\frac{CD}{GH}\) = \(\frac{AD}{FH}\)
[∵ Ratio of the Corresponding angles is equal]
⇒ \(\frac{AC}{FG}\) = \(\frac{CD}{GH}\) [Q.E.D]

ii) △DCB ~ △HGE
In △DCB and △HGE,
∠B = ∠E
[∵ Corresponding angles of △ABC and △FEG]
∠DCB = ∠HGE
[∵ ∠C = ∠G ⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠DCB = ∠HGE]
∴ △DCB ~ △HGE . (by A.A. similarity condition)

iii) △DCA ~ △HGF
In △DCA and △HGF
∠A = ∠F
\(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠DCA = ∠HGF
[∵ Corresponding angles of the similar triangles]
∴ △DCA ~ △HGF
[ A.A. similarity condition]

Question 9.
AX and DY are altitudes of two similar triangles △ABC and △DEF. Prove that AX : DY = AB : DE.
Answer:

Given: △ABC ~ △DEF.
AX ⊥ BC and DY ⊥ EF.
R.T.P.: AX : DY = AB : DE.
Proof: In △ABX and △DEY ∠B = ∠E [∵ Corresponding angles of △ABC and △DEF]
∠AXB = ∠DYE [given]
∴ △ABX ~ △DEY
(by A.A. similarity condition)
Hence \(\frac{AB}{DE}\) = \(\frac{BX}{EY}\) = \(\frac{AX}{DY}\)
[∵ Ratios of corresponding sides of similar triangles are equal]
⇒ AX : DY = AB : DE [Q.E.D.]

Question 10.
Construct a triangle shadow similar to the given △ABC, with its sides equal to \(\frac{5}{3}\) of the corresponding sides of the triangle ABC.
Answer:

Steps of construction :

1. Draw a △ABC with certain measures.
2. Draw a ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
3. Locate 8 points (B₁, B₂, …., B₈) on \(\overrightarrow{\mathrm{BX}}\) such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈.
4. Join B₅, C.
5. Draw a line parallel to B₅C through which it intersects BC extended at C’.
6. Draw a line parallel to AC through ‘C’ which meets \(\overrightarrow{\mathrm{BA}}\) produced at A’.
7. △A’BC’ is the required triangle.

Question 11.
Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it,whose sides are 2/3 of the corresponding sides of the first triangle.
Answer:

Steps of construction:

1. Draw △ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ on \(\overrightarrow{\mathrm{BX}}\) such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃, C.
5. Draw a line parallel to B₃C through B₂ meeting BC at C’.
6. Draw a line parallel to BA through C’ meeting BA at A’.
7. △A’BC’ is the required triangle.

Question 12.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Answer:

Steps of construction:

1. Draw AABC in which BC = 8 cm and altitude AD = 4 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₂C.
5. Draw a line parallel to B₂C through B₃ meeting BC produced C’.
6. Draw a line paral1e1 to AC through C’ meeting BA produced at A’.
7. △A’BC’ is the required triangle.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.4

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.4 Textbook Questions and Answers

Question 1.
Find the slope of the line joining the two given points.
i) (4,-8) and (5,-2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+8}{5-4}\)
= 6

ii) (0, 0) and (√3, 3)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{\sqrt{3}-0}\)
= \(\frac{3}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\)
= √3

iii) (2a, 3b) and (a, -b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-b-3b}{a-2a}\)
= \(\frac{-4b}{-a}\)
= \(\frac{4b}{a}\)

iv) (a, 0) and (0, b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{b-0}{0-a}\)
= \(\frac{-b}{a}\)

v) A (-1.4, -3.7), B (-2.4, 1.3).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\)
= -5

vi) A (3, -2), B (-6, -2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+2}{-6-3}\)
= 0

vii) A (-3\(\frac{1}{2}\), 3), B (-7, 2\(\frac{1}{2}\)).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{2 \frac{1}{2}-3}{-7+3 \frac{1}{2}}\)
= \(\frac{-\frac{1}{2}}{-3 \frac{1}{2}}\)
= \(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1}{7}\)

viii) A(0, 4), B(4, 0)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-4}{4-0}\)
= \(\frac{-4}{4}\)
= -1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.4

10th Class Maths 8th Lesson Similar Triangles Ex 8.4 Textbook Questions and Answers

Question 1.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Answer:

Given : □ ABCD is a rhombus.
Let its diagonals AC and BD bisect each other at ‘O’.
We know that “the diagonals in a rhombus are perpendicular to each other”.
In △AOD; AD² = OA² + OD² ………. (1)
[Pythagoras theorem]
In △COD; CD² = OC² + OD² ………. (2)
[Pythagoras theorem]
In △AOB; AB² = OA² + OB² ………. (3)
[Pythagoras theorem]
In △BOC; BC² = OB² + OC² ………. (4)
[Pythagoras theorem]
Adding the above equations we get AD² + CD² + AB² + BC² = 2 (OA² + OB2 + OC² + OD²)

Question 2.
ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE² + CD² = AC² + DE².

Answer:
Given: In △ABC; ∠B = 90°
D and E are points on AB and BC.
R.T.P.: AE² + CD² = AC² + DE²
Proof: In △BCD, △BCD is a right triangle right angled at B.
∴ BD² + BC² = CD² ……… (1)
[∵ Pythagoras theorem states that hypotenuse² = side² + side²]
In △ABE; ∠B = 90°
Adding (1) and (2), we get
BD² + BC² + AB² + BE² – CD² + AE²
(BD² + BE²) + (AB² + BC²) = CD² + AE²
DE² + AC² – CD² + AE² [Q.E.D.]
[∵ (i) In △DBE, ∠B = 90° and DE² = BD² + BE²
(ii) In △ABC, ∠B = 90° and AB² + BC²]

Question 3.
Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Answer:
Given: △ABC, an equilateral triangle;
AD – altitude and the side is a units, altitude h units.
R.T.P: 3a² = 4h²

Proof: In △ABD, △ACD
∠B = ∠C [∵ 60°]
∠ADB = ∠ADC [∵ 90°]
∴ ∠BAD = ∠DAC [∵ Angle sum property]
Also, BA = CA
∴ △ABD s △ACD (by SAS congruence condition)
Hence, BD = CD = \(\frac{1}{2}\)BC = \(\frac{a}{2}\) [∵ c.p.c.t]
Now in △ABD, AB² = AD² + BD²
[∵ Pythagoras theorem]
a² = h² + \(\left(\frac{a}{2}\right)^{2}\)
a² = h² + \(\frac{a^{2}}{4}\)
h² = \(\frac{4 a^{2}-a^{2}}{4}\)
∴ h² = \(\frac{3 a^{2}}{4}\)
⇒ 4h² = 3a² (Q.E.D)

Question 4.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Answer:

Given: In △PQR, ∠P = 90° and PM ⊥ QR.
R.T.P : PM² = QM . MR
Proof: In △PQR; △MPR
∠P = ∠M [each 90°]
∠R = ∠R [common]
∴ △PQR ~ △MPR ……… (1)
[A.A. similarity]
In △PQR and △MQP,
∠P = ∠M (each 90°)
∠Q = ∠Q (common)
∴ △PQR ~ △MQP ……… (2)
[A.A. similarity]
From (1) and (2),
△PQR ~ △MPR ~ △MQP [transitive property]
∴ △MPR ~ △MQP
\(\frac{MP}{MQ}\) = \(\frac{PR}{QP}\) = \(\frac{MR}{MP}\)
[Ratio of corresponding sides of similar triangles are equal]
\(\frac{PM}{QM}\) = \(\frac{MR}{PM}\)
PM . PM = MR . QM
PM² = QM . MR [Q.E.D]

Question 5.
ABD is a triangle right angled at A and AC ⊥ BD.
Show that (i) AB² = BC BD
(ii) AD² = BD CD
(iii) AC² = BC DC.

Answer:
Given: In △ABD; ∠A = 90° AC ⊥ BD
R.T.P.:
i) AB² = BC . BD
Proof: In △ABD and △CAB,
∠BAD = ∠ACB [each 90°]
∠B = ∠B [common]
∴ △ABD ~ △CBA
[by A.A. similarity condition]
Hence, \(\frac{AB}{BC}\) = \(\frac{BD}{AB}\) = \(\frac{AD}{AC}\)
[∵ Ratios of corresponding sides of similar triangles are equal]
\(\frac{AB}{BD}\) = \(\frac{BC}{AB}\)
⇒ AB . AB = BC . BD
∴ AB² = BC . BD

ii) AD² = BD . CD
Proof: In △ABD and △CAD
∠BAD = ∠ACD [each 90°]
∠D = ∠D (common)
∴ △ABD ~ △CAD [A.A similarity]
Hence, \(\frac{AB}{AC}\) = \(\frac{BD}{AD}\) = \(\frac{AD}{CD}\)
⇒ \(\frac{BD}{AD}\) = \(\frac{AD}{CD}\)
AD . AD = BD . CD
AD² = BD . CD [Q.E.D]

iii) AC² = BC . DC
Proof: From (i) and (ii)
△ACB ~ △DCA
[∵ △BAD ~ △BCA ~ △ACD
Hence, \(\frac{AC}{DC}\) = \(\frac{BC}{AC}\) = \(\frac{AB}{AD}\)
\(\frac{AC}{DC}\) = \(\frac{BC}{AC}\)
AC . AC = BC . DC
AC² = BC . DC [Q.E.D]

Question 6.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Answer:

Given: In △ABC; ∠C = 90°; AC = BC.
R.T.P.: AB² = 2AC²
Proof: In △ACB; ∠C = 90°
Hence, AC² + BC² = AB²
[Square of the hypotenuse is equal to sum of the squares of the other two sides – Pythagoras theorem]
⇒ AC² + AC² = AB² [∵ AC = BC given]
⇒ AB² = 2AC² [Q.E.D.]

Question 7.
‘O’ is any point in the interior of a triangle ABC.
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB, show that
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer:
Given: △ABC; O’ is an interior point of △ABC.
OD ⊥ BC, OE ⊥ AC, OF ⊥ AB.
R.T.P.:
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
Proof: In OAF, OA² = AF² + OF² [Pythagoras theorem]
⇒ OA² – OF² = AF² …….. (1)
In △OBD,
OB² = BD² + OD²
⇒ OB² – OD² = BD² …….. (2)
In △OCE, OC² = CE² + OE²
OC² – OE² = CE² ……… (3)
Adding (1), (2) and (3) we get,
OA² – OF² + OB² – OD² + OC² – OE² = AF² + BD² + CE²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ……… (4)

ii) AF² + BD² + CE² = AE² + CD² + BF²
In △OAE,
OA² = AE² + OF² ……… (1)
⇒ OA² – OE² = AE²
In △OBF, OB² = BF² + OF²
OB² – OF² = BF² ……… (2)
In △OCD, OC² = OD² + CD²
OC² – OD² = CD² ……… (3)
Adding (1), (2) and (3) we get
OA² – OE² + OB² – OF² + OC² – OD² = AE² + BF² + CD²
⇒ OA² + OB² + OC² – OD² – OE² – OF² = AE² + CD² + BF²
⇒ AF² + BD² + CE² = AE² + CD² + BF² [From problem (i)]

Question 8.
A wire attached to vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Answer:

Height of the pole AB = 18 m.
Length of the wire AC = 24 m.
Distance beween the pole and the stake be ‘d’ meters.
By Pythagoras theorem,
Hypotenuse² = side² + side²
24² = 18² + d²
d² = 24² – 18² = 576 – 324 = 252
= \(\sqrt{36 \times 7}\)
∴ d = 6√7 m.

Question 9.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Answer:

Let the height of the first pole AB = 6 m.
Let the height of the second pole CD = 11 m.
Distance between the poles AC = 12 m.
From the figure □ ACEB is a rectangle.
∴ AB = CE = 6 m
ED = CD – CE = 11 – 6 = 5 m
Now in △BED; ∠E = 90°; DE = 5 m; BE = 12 m
BD² = BE² + DE²
[hypotenuse² = side² + side² – Pythagoras theorem]
= 12² + 5²
= 144 + 25
BD² = 169
BD = √l69 = 13m
∴ Distance between the tops of the poles = 13 m.

Question 10.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9AD² = 7AB².
Answer:

In △ABE, ∠E = 90°
⇒ \(\overline{\mathrm{AB}}\) is hypotenuse.
∴ AB² = AE² + BE²
AE² = AB² – BE²
⇒ AE² = AB² – \(\left(\frac{BC}{2}\right)^{2}\)
= AE² = AB² – \(\left(\frac{AB}{2}\right)^{2}\) (∵ AB = BC)
⇒ AE² = \(\frac{3}{4}\)AB² ……… (1)
In △ADE, ∠E = 90°
⇒ \(\overline{\mathrm{AD}}\) is hypotenuse.
⇒ AD² = AE² + DE²
⇒ AE² = AD² + DE²

⇒ 28 AB² = 36 AD²
⇒ 7 AB² = 9 AD²
Hence proved.

Question 11.
In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8 AE² = 3 AC² + 5 AD².

Answer:
In △ABC, ∠B=90°
⇒ \(\overline{\mathrm{AD}}\) is hypotenuse.
AC2 = AB2 + BC2
3AC² = 3AB² + 3BC² …….. (1)
In △ABD, ∠B = 90°
⇒ AD is hypotenuse.
∴ AD² = AB² + BD² = AB² + \(\left(\frac{BC}{3}\right)^{2}\)
⇒ AD² = AB² + \(\frac{\mathrm{BC}^{2}}{9}\)
⇒ 5 AD² = 5 AB² + \(\frac{5 \mathrm{BC}^{2}}{9}\) …….. (2)
(1) + (2)
3 AC² + 5 AD² = 3 AB² + 3 BC² + 5 AB² + \(\frac{5}{9} \mathrm{BC}^{2}\)
= 8AB² + \(\frac{32}{9} \mathrm{BC}^{2}\) ……… (3)
Now in △ABE, ∠B = 90°
⇒ \(\overline{\mathrm{AE}}\) is hypotenuse.
⇒ AE² = AB² + BE² = AB² + \(\left(\frac{2}{3} BC\right)^{2}\)
= AB² + \(\frac{4}{9} \mathrm{BC}^{2}\)
⇒ AE² = 8AB² + \(\frac{32}{9} \mathrm{BC}^{2}\) ……… (4)
∴ RHS of (3) and (4) are equal.
∴ LHS of (3) and (4) are equal.
∴ 8 AE² = 3 AC² + 5 AD².
Hence proved.

Question 12.
ABC is an isosceles triangle right angled at B. Equilateral triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of △ABE and △ACD.
Answer:

Given: △ABC, AB = BC and ∠B = 90°
△ABE on AB; △ACD on AC are equiangular triangles.
Let equal sides of the isosceles right triangle, AB = BC = a (say)
Then, in △ABC, ∠B = 90°
AC² – AB² + BC²
[hypotenuse² = side² + side² – Pythagoras theorem] = a² + a² = 2a²
Since, △ABE ~ △ACD
\(\frac{\Delta \mathrm{ABE}}{\Delta \mathrm{ACD}}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}\)
[∵ Ratio of areas of two similar tri-angles is equal to the ratio of squares of their corresponding sides]
= \(\frac{a^{2}}{2 a^{2}}\) = \(\frac{1}{2}\)
△ABE : △ACD = 1 : 2.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1

10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers

Question 1.
A joker’s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:

Radius of the cap (r) = 7 cm
Height of the cap (h) = 24 cm
Slant height of the cap (l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{7^{2}+24^{2}}\)
= \(\sqrt{49+576}\)
= √625
= 25
∴ l = 25 cm.
Lateral surface area of the cap = Cone = πrl
L.S.A. = \(\frac{22}{7}\) × 7 × 25 = 550 cm².
∴ Area of the sheet required for 10 caps = 10 x 550 = 5500 cm².

Question 2.
A sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.
Answer:

Radius of the cylinder, r = 7 cm
Height of the cylinder, h = 35 cm
T.S.A. of the cylinder with lids at both ends = 2πr(r+h)
= 2 × \(\frac{22}{7}\) × 7 × (7 + 35)
= 2 × \(\frac{22}{7}\) × 7 × 42 = 1848 cm².
Area of thin paper required for 100 cylinders = 100 × 1848
= 184800 cm²
= \(\frac{184800}{100 \times 100}\) m²
= 18.48 m².

Question 3.
Find the volume of right circular cone with radius 6 cm. and height 7 cm.
Answer:

Base radius of the cone (r) = 6 cm.
Height of the cone (h) = 7 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7
= 264 c.c. (Cubic centimeters)
∴ Volume of the right circular cone = 264 c.c.

Question 4.
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.
Answer:
Base of cylinder and cone be the same.

CSA / LSA of cylinder = 2πrh
CSA of cone = πrl
The lateral surface area of a cylinder is equal to the curved surface area of cone.
∴ 2πrh = πrl
⇒ \(\frac{h}{l}=\frac{\pi r}{2 \pi r}\)
⇒ \(\frac{h}{l}\) = \(\frac{1}{2}\)
∴ h : l = 1 : 2

Question 5.
A self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm², then how many caps can be manufactured from that paper sheet?
Answer:

Radius of the cap (conical cap) (r) = 3 cm
Height of the cap (h) = 4 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
(by Pythagoras theorem)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
C.S.A. of the cap = πrl
= \(\frac{22}{7}\) × 3 × 5
≃ 47.14 cm²
Number of caps that can be made out of 1000 cm² = \(\frac{1000}{47.14}\) ≃ 21.27
∴ Number of caps = 21.

Question 6.
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.
Answer:
Given dimensions are:
Cone:
Radius = r
Height = h
Volume (V) = \(\frac{1}{3}\)πr²h

Cylinder:
Radius = r
Height = h
Volume (V) = πr²h

Ratio of volumes of cylinder and cone = πr²h : \(\frac{1}{3}\)πr²h
= 1 : \(\frac{1}{3}\)
= 3 : 1
Hence, their volumes are in the ratio = 3 : 1.

Question 7.
A solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?
Answer:

Diameter of the cylinder (d) = 7 cm
Radius of the base (r) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 11 cm
Volume of the cylinder V = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 11 = 423.5 cm³
∴ Total volume of 50 rods = 50 × 423.5 cm³ = 21175 cm³.

Question 8.
A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use π = 3.14)
Answer:

Diameter of the heap (conical) (d) = 12 cm
∴ Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Height of the cone (h) = 8 m
Volume of the cone, V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 8
= 301.71 m³.

Question 9.
The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height?
Answer:
C.S.A. of a cone = πrl = 4070 cm²
Diameter of the cone (d) = 70 cm
Radius of the cone = r = \(\frac{d}{2}\) = \(\frac{70}{2}\) = 35 cm
Let its slant height be ‘l’.
By problem,
πrl = 4070 cm²
\(\frac{22}{7}\) × 35 × l = 4070
110 l = 4070
l = \(\frac{4070}{110}\) = 37 cm
∴ Its slant height = 37 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.2

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers

Question 1.
Choose the correct answer and give justification for each.
(i) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60°
b) 30°
c) 45°
d) 90°
Answer: [ d ]
If radius is not perpendicular to the tangent, the tangent must be a secant i.e., 90°.

(ii) From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is
a) 7 cm
b) 12 cm
c) 15 cm
d) 24.5 cm
Answer: [ a ]

O – centre of the circle
OP – a circle radius = ?
OQ = 25 cm
PQ = 24 cm
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
25² = OP² + 24²
OP² = 625 – 576
OP² = 49
OP = √49 = 7 cm.

iii) If AP and AQ are the two tangents a circle with centre O, so that ∠POQ = 110°. Then ∠PAQ is equal to

a) 60°
b) 70°
c) 80°
d) 90°
Answer: [ b ]
In □ OPAQ,
∠OPA = ∠OQA = 90°
∠POQ = 110°
∴ ∠O + ∠P + ∠A + ∠Q = 360°
⇒ 90° + 90° + 110° + ∠PAQ – 360°
⇒ ∠PAQ = 360° – 290° = 70°

iv) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
a) 50°
b) 60°
c) 70°
d) 80°
Answer: [None]

If ∠APB = 80°
then ∠AOB = 180° – 80° = 100°
[∴ ∠A + ∠B = 90° + 90° = 180°]

v) In the figure XY and XV are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XV at B then ∠AOB =
a) 80°
b) 100°
c) 90°
d) 60°

Answer: [ c ]

Question 2.
Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Given: Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner/small circle and chord to the larger circle.
Let ‘P’ be the point of contact.
Construction: Join OP and OB.
In △OPB ;
∠OPB = 90°
[radius is perpendicular to the tangent]
OP = 3cm OB = 5 cm
Now, OB² = OP² + PB²
[hypotenuse² = Adj. side² + Opp. side², Pythagoras theorem]
5² = 3² + PB²
PB² = 25 – 9 = 16
∴ PB = √l6 = 4cm.
Now, AB = 2 × PB
[∵ The perpendicular drawn from the centre of the circle to a chord, bisects it]
AB = 2 × 4 = 8 cm.
∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:

Given: A circle with centre ‘O’.
A parallelogram ABCD, circumscribing the given circle.
Let P, Q, R, S be the points of contact.
Required to prove: □ ABCD is a rhombus.
Proof: AP = AS …….. (1)
[∵ tangents drawn from an external point to a circle are equal]
BP = BQ ……. (2)
CR = CQ ……. (3)
DR = DS ……. (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∵ Opposite sides of a parallelogram are equal]
2AB = 2AD
AB = AD
Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus □ ABCD is a rhombus (Q.E.D.)

Question 4.
A triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See below figure). Find the sides AB and AC.

Answer:
The given figure can also be drawn as

Given: Let △ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm.
i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.
It is given that BD = 9 cm
CD = 3 cm

∵ Lengths of two tangents drawn from an external point to a circle are equal.
∴ BF = BD = 9 cm
CD = CE = 3 cm
AF = AE = x cm say
∴ The sides of die triangle are
12 cm, (9 + x) cm, (3 + x) cm
Perimeter = 2S = 12 + 9 + x + 3 + x
⇒ 2S = 24 + 2x
or S = 12 + x
S – a = 12 + x – 12 = x
S – b = 12 + x – 3 – x = 9
S – c = 12 + x – 9 – x = 3
∴ Area of the triangle

Squaring on both sides we get,
27 (x² + 12x) = (36 + 3x)²
27x² + 324x = 1296 + 9x² + 216x
⇒ 18x² + 108x- 1296 = 0
⇒ x² + 6x – 72 = 0
⇒ x² + 12x – 6x – 72 = 0
⇒ x (x + 12) – 6 (x + 12) = 0
⇒ (x – 6) (x + 12) = 0
⇒ x = 6 or – 12
But ‘x’ can’t be negative hence, x = 6
∴ AB = 9 + 6 = 15 cm
AC = 3 + 6 = 9 cm.

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.
Answer:
Steps of construction:

1. Draw a circle with centre ‘O’ and radius 6 cm.
2. Take a point P outside the circle such that OP =10 cm. Join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tan¬gents of lengths 8 cm each.

Proof: In △OAP
OA² + AP² = 6² + 8²
= 36 + 64 = 100
OP² = 10² = 100
∴ OA² + AP² = OP²
Hence AP is a tangent.
Similarly BP is a tangent.

Question 6.
Construct, a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer:
Steps of construction:

1. Draw two concentric circles with centre ‘O’ and radii 4 cm and 6 cm.
2. Take a point ‘P’ on larger circle and join O, P.
3. Draw the perpendicular bisector of OP which intersects it at M.
4. Taking M as centre and PM or MO as radius draw a circle which intersects smaller circle at Q.
5. Join PQ, which is a tangent to the smaller circle.

Question 7.
Draw a circle with the help of a bangle, take a point outside the circle. Con-struct the pair of tangents from this point to the circle measure them. Write conclusion.
Answer:
Steps of construction:

1. Draw a circle with the help of a bangle.
2. Draw two chords AB and AC. Perpendicular bisectors of AB and AC meets at ‘O’ which is the centre of the circle.
3. Taking an outside point P, join OP.
4. Let M be the midpoint of OP. Taking M as centre OM as radius, draw a circle which intersects the given circle at R and S. Join PR, PS which are the required tangents.

Conclusion: Tangents drawn from an external point to a circle are equal.

Question 8.
In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.
Answer:
Let ABC be a right triangle right angled at P.
Consider a circle with diametere AB.
From the figure, the tangent to the circle at B meets BC in Q.
Now QB and QP are two tangents to the circle from the same point P.
QB = QP …….. (1)
Also, ∠QPC = ∠QCP
∴ PQ = QC (2)
From (1) and (2);
QB = QC Hence proved.

Question 9.
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? [Hint: The distance of two points to the point of contact is the same.
Answer:
Only two tangents can be drawn from a given point outside the circle.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.3

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Answer:
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|2(0+4)-1(-4-3)+2(3-0)|\)
= \(\frac{1}{2}|8+7+6|\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\) sq.units

ii) (-5, -1), (3, -5), (5, 2)
Answer:
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|-5(-5-2)+3(2+1)+5(-1+5)|\)
= \(\frac{1}{2}|35+9+20|\)
= \(\frac{64}{2}\)
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Answer:
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(0-2)+3(2-0)+0(0-0)|\)
= \(\frac{1}{2}|6|\)
= 3 sq.units

Or

△AOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Answer:
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
⇒ \(\frac{1}{2} \mid 7(1-\mathrm{K})+5(\mathrm{~K}+2)+3(-2-1)\) = 0
⇒ \(|7-7 K+5 K+10-9|\) = 0
⇒ \(|-2 \mathrm{~K}+8|\) = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = \(\frac{8}{2}\)
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Answer:
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|8(-4+5)+\mathrm{K}(-5-1)+2(1+4)|\) = 0
⇒ \(|8-6 \mathrm{~K}+10|\) = 0
⇒ \(|18-6 \mathrm{~K}|\) = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
Answer:
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|\mathrm{~K}(3+1)+2(-1-\mathrm{K})+4(\mathrm{~K}-3)|\) = 0
⇒ \(|4K-2-2K+4K-12|\) = 0
⇒ \(|6 \mathrm{~K}-14|\) = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\) = \(\frac{7}{3}\)
∴ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer:

Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}\).

Area of a triangle ABC =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|\)
= \(\frac{1}{2}|8|\)
= 4 sq.units
Area of △DEF = \(\frac{1}{2}|1(2-1)+1(1-0)+0(0-2)|\)
= \(\frac{1}{2}|1+1|\)
= \(\frac{2}{2}\)
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3).
Answer:

Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)

Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
Answer:
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
now

now formula for area of triangle using Heron’s formula = △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
∴ s = \(\frac{3+4+5}{2}\) = \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1)

∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
= \(\frac{5+4+3}{2}\)
= \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.
Evaluate the following:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Answer:
Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

ii) (sin θ + cos θ)² + (sin θ – cos θ)²
Answer:
Given (sin θ + cos θ)² + (sin θ – cos θ)²
= (sin² θ + cos² θ + 2 sin θ cos θ) + (sin² θ + cos² θ – 2 sin θ cos θ) [∵ (a + b)² = a² + b² + 2ab
(a – b)² = a² + b² – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin² θ + cos² θ = 1]
= 1 + 1
= 2

iii) (sec² θ – 1) (cosec² θ – 1)
Answer:
Given (sec² θ – 1) (cosec² θ – 1)
= tan² θ × cot² θ [∵ sec² θ – tan² θ = 1; cosec² θ – cot² θ = 1]
= tan² θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

Question 2.
Show that (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:
Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.

[∵ (a + b)(a + b) = (a + b)²]; (a – b)(a + b) = a² — b²]
= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)
= \(\frac{1+\sin A}{\cos A}\)
= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A = R.H.S.

Question 4.
Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan² A
Answer:

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.
Answer:

Question 6.
Simplify sec A (1 – sin A) (sec A + tan A)
Answer:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A . sin A) (sec A + tan A)
= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec² A – tan² A [∵ sec² A – tan² A = 1]
= 1

Question 7.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Answer:
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A . cosec A) + (cos² A – sec² A + 2 cos A . sec A) [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + cosec² A + 2 sin A . \(\frac{1}{\sin A}\) + sec² A + 2 cos A . \(\frac{1}{\cos A}\)
[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]
= 1 +(1 + cot² A) + 2 + (1 + tan² A) + 2
[∵ sin² A + cos² A = 1; cosec² A = 1 + cot² A; sec² A = 1 + tan² A]
= 7 + tan² A + cot² A
= R.H.S.

Question 8.
Simplify (1 – cos θ) (1 + cos θ) (1 + cot² θ)
Answer:
Given that
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
= (1 – cos² θ) (1 + cot² θ)
[∵ (a – b) (a + b) = a² – b²]
= sin² θ. cosec² θ [∵ 1 – cos² θ = sin² θ; 1 + cot² θ = cosec² θ]
= sin² θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ?
Answer:
Given that sec θ + tan θ = p ,
We know that sec² θ – tan² θ = 1
sec² θ – tan² θ = (sec θ + tan θ) (sec θ – tan θ)
= p (sec θ – tan θ)
= 1 (from given)
⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)
Answer:
Method-I:
Given that cosec θ + cot θ = k
R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)

Method – II:
Given that cosec θ + cot θ = k ……..(1)
We know that cosec² θ – cot² θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a² – b² = (a -b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)
By solving (1) and (2)

According to identity cos² θ + sin² θ = 1

Hence proved.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.
In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.
Prove that △PQR is an isosceles triangle.
Answer:
Given : In △PQR,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.
Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
Hence, ST || QR (Converse of Basic proportionality theorem)
∠PST = ∠PQR …….. (1)
(Corresponding angles for the lines ST || QR)
Also, ∠PST = ∠PRQ ……… (2) given
From (1) and (2),
∠PQR = ∠PRQ
i.e., PR = PQ
[∵ In a triangle sides opposite to equal angles are equal]
Hence, APQR is an isosceles triangle.

Question 2.
In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:
Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)
∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.
In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:
In △ABC, DE || AC
Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)
[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
Also in △ABE, DF || AE
Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)
From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
Answer:

Given: In △ABC; D is the mid-point of AB.
A line ‘l’ through D, parallel to BC, meeting AC at E.
R.T.P: E is the midpoint of AC.
Proof:
DE || BC (Given)
then
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)
Also given ‘D’ is mid point of AB.
Then AD = DB.
⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1
⇒ AE = EC
∴ ‘E’ is mid point of AC
∴ The line bisects the third side \(\overline{\mathrm{AC}}\).
Hence proved.

Question 5.
Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
Answer:
Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.
R.T.P : DE || BC.
Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)
also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)
From (1) and (2)
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
∴ DE || BC by Basic proportionality theorem.
Hence proved.

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:
Given: △PQR, DE || OQ; DF || OR
R.T.P: EF || QR
Proof:
In △POQ;
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)
[∵ ED || QO, Basic proportionality theorem]
In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]
From (1) and (2),
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.
Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.
In the given figure A, B and C are points on OP, OQ and OR respec¬tively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:
Given:
In △PQR, AB || PQ; AC || PR
R.T.P : BC || QR
Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)
(∵ Basic Proportional theorem)
and in △OPR, Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)
From (1) and (2), we can write
\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)
Then consider above condition in △OQR then from (3) it is clear.
∴ BC || QR [∵ from converse of Basic Proportionality Theorem]
Hence proved.

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).
Answer:
Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.
R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:
Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In △ACD, EO || CD
∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In △ABD, EO || AB
Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)
[∵ Basic proportionality theorem]
\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]
From (1) and (2),
\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)
⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Answer:

Steps of construction:

1. Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
2. Construct an acute angle ∠BAX at A.
3. Mark off 5 + 3 = 8 equal parts (A₁, A₂, …., A₈) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius.
4. Join A₈ and B.
5. Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A₅ meeting AB at C.
6. Now the point C divides AB in the ratio 5:3.
7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.2

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.2 Textbook Questions and Answers

Question 1.
Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 :3.
Answer:
Given points P (-1, 7) and Q (4, – 3). Let ‘R’ be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 2:3. Then

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Answer:
Given points A (4, – 1) and B (- 2, – 3) Let P and Q be the points of trisection
of \(\overline{\mathrm{AB}}\), then AP = PQ = QB.

∴ P divides \(\overline{\mathrm{AB}}\) internally in the ratio 1 : 2.

Also, Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.

Question 3.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer:
Let the point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in a ratio of m : n

⇒ 6m – 3n = -(m + n) = -m – n
⇒ 6m + m = – n + 3n
⇒ 7m = 2n
⇒ \(\frac{m}{n}=\frac{2}{7}\)
⇒ m : n = 2 : 7
∴ The point (-1, 6) divides the given line segment in a ratio of 2 : 7.

Question 4.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer:

Given: ▱ ABCD is a parallelogram where A (1, 2), B (4, y), C (x, 6) and D (3, 5).
In a parallelogram, diagonals bisect each other.
i.e., the midpoints of the diagonals coincide with each other.
i.e.,midpoint of AC = midpoint of BD

⇒ 1 + x = 7 and 8 = y + 5
⇒ x = 7 – 1 and y = 8 – 5
∴ x = 6 and y = 3.

Question 5.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Answer:

Given:
A circle with centre ‘C’ (2, -3). \(\overline{\mathrm{AB}}\) is a diameter where
B = (1, 4); A = (x, y).
C is the midpoint of AB.
[∵ Centre of a circle is the midpoint of the diameter]

4 = x + 1 and – 6 = y + 4
⇒ x = 4 – 1 = 3 and y = -6 – 4 = -10
A (x, y) = (3, -10)

Question 6.
If A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the segment AB.
Answer:
Given: A (- 2, – 2) and B (2, – 4)
P lies on AB such that AP = latex]\frac{3}{7}[/latex] AB

i.e., P divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 4 By section formula,

Question 7.
Find the coordinates of points which divide the line segment joining A (-4, 0) and B (0, 6) into four equal parts.
Answer:
Given, A (- 4, 0) and B (0, 6).
Let P, Q and R be the points which divide AB into four equal parts.

P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3, Q → 1 : 1 and R → 3 : 1 Use section formula to find P, Q and R.
Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 8.
Find the coordinates of the points which divides the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer:
Given, A (- 2, 2) and B (2, 8).
Let P, Q and R be the points which divide AB into four equal parts.

Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 9.
Find the coordinates of the point which divide the line segment joining the points (a + b, a-b) and (a-b, a + b) in the ratio 3 : 2 internally.
Answer:
Given : A (a + b, a – b) and B (a – b, a + b).
Let P (x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 2 internally.
Section formula

Question 10.
Find the coordinates of centroid of the triangle with following vertices:
i) (-1, 3), (6, -3) and (-3, 6)
Answer:
Given: △ABC in which- A (- 1, 3), B (6, -3) and C (-3, 6)

ii) (6, 2), (0, 0) and (4, -7)
Answer:
Given: The three vertices of a triangle are A (6, 2), B (0, 0) and C (4, – 7).
Centroid (x, y)

iii) (1,-1), (0, 6) and (-3, 0)
Answer:
Given: (1, -1), (0, 6) and (-3, 0) are the vertices of a triangle.
Centroid (x, y)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.1

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers

Question 1.
Fill in the blanks.
i) A tangent to a circle intersects it in ——— point(s). (one)
ii) A line intersecting a circle in two points is called a ———. (secant)
iii) The number of tangents drawn at the end of the diameter is ———. (two)
iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
v) We can draw ——— tangents to a given circle. (infinite)

Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Answer:
Given: A circle with centre O and radius OP = 5 cm
\(\overline{\mathrm{PQ}}\) is a tangent and OQ = 12 cm
We know that ∠OPQ = 90°
Hence in △OPQ
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
12² = 5² + PQ²
∴ PQ² = 144 – 25 .
PQ² = 119
PQ = √119

Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:
Steps:

1. Draw a circle with some radius.
2. Draw a chord of the circle.
3. Draw a line parallel to the chord intersecting the circle at two distinct points.
4. This is secant of the circle (l).
5. Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Answer:
Given: A circle with radius OP = 9 cm
A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm
Now in △POQ, ∠P = 90°
OP² + PQ² – OQ²
9² + PQ² = 15²
PQ² = 15² – 9²
PQ² = 144
∴ PQ = √144 = 12 cm.
Hence the length of the tangent =12 cm.

Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel.
Answer:
A circle with a diameter AB.
PQ is a tangent drawn at A and RS is a tangent drawn at B.
R.T.P: PQ || RS.
Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.
∴ OA ⊥ PQ ……….(1)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
Similarly, OB ⊥ RS ……….(2)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
But, OA and OB are the parts of AB.
i.e., AB ⊥ PQ and AB ⊥ RS.
∴ PQ || RS.
O is the centre, PQ is a tangent drawn at A.
∠OAQ = 90°
Similarly, ∠OBS = 90°
∠OAQ + ∠OBS = 90° + 90° = 180°
∴ PQ || RS.
[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.1

10th Class Maths 13th Lesson Probability Ex 13.1 Textbook Questions and Answers

Question 1.
Complete the following statements:
i) Probability of an event E + Probability of the event ‘not E’ = 1 .
ii) The probability of an event that cannot happen is zero.
Such an event is called an impossible event.
iii) The probability of an event that is certain to happen is  1  such an event is called sure or certain event.
iv) The sum of the probabilities of all the elementary events of an experiment is 1 .
v) The probability of an event is greater than or equal to zero and less than or equal to 1 .

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
Answer:
Equally likely. Since both have the same probability \(\frac{1}{2}\).

ii) A player attempts to shoot a basket-ball. She/he shoots or misses the shot.
Answer:
Equally likely. Since both have the same probability \(\frac{1}{2}\).

iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely. Since both have the same probability \(\frac{1}{2}\).

iv) A baby is born. It is a boy or a girl.
Equally likely. Since both the events have the same probability \(\frac{1}{2}\).

Question 3.
If P(E) = 0.05, what is the probability of not E?
Answer:
Given: P(E) = 0.05
Hence, P(E) + P(\(\overline{\mathrm{E}}\)) = 1, where P(\(\overline{\mathrm{E}}\)) is the probability of ‘not E’
0.05 + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\)) = 1 -0.05 = 0.95.

Question 4.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Answer:
Bag contains only lemon flavoured candies.
i) Taking an orange flavoured candy is an impossible event and hence the probability is zero.
ii) Also taking a lemon flavoured candy is a sure event and hence its probability is 1.

Question 5.
Rahim removes all the hearts from the cards. What is the probability of
i. Picking out an ace from the remaining pack.
ii. Picking out a diamond.
iii. Picking out a card that is not a heart.
iv. Picking out the Ace of hearts.
Answer:
Total number of cards in the deck = 52.
Total number of hearts in the deck of cards =13.
When Hearts are removed, remaining cards = 52 – 13 = 39.
i)Picking out an Ace:
Number of outcomes favourable to Ace = 3 [∵ ♦ A, ♥ A, ♠ A, ♣ A]
Total number of possible outcomes from the remaining cards = 39
– after removing Hearts.
Probability = P(A)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{39}\) = \(\frac{1}{13}\)

ii) Picking out a diamond:
Number of favourable outcomes to diamonds (♦) = 13
Total number of possible outcomes = 39
∴ p(♦) = \(\frac{13}{39}\) = \(\frac{1}{3}\)

iii) Picking out a card that is ‘not a heart’:
As all hearts are removed, the remain-ing cards are all non-heart cards. So the picked card will be definitely a non-heart card. So this is a sure event.
Hence its probability is one
P(E) = \(\frac{39}{39}\) = 1

iv) Picking out the Ace of Hearts:
a) As all the heart cards are removed the left over cards will have three suits (i) spades, (ii), clubs, (iii) dia¬monds of each 13.
Hence total outcomes = 3 × 13 = 39 But among them there is no Ace of heart. So number of favourable outcomes for picking Ace of heart = zero.
∴ Probability P(E) = \(\frac{0}{39}\) = 0
So it is an impossible event.

b) If picking from the rest of the cards, it is an impossible event and hence probability is zero.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.

Question 6.
What is the probability that the 2 students have the same birthday?
Answer:
Let P(E) = The probability that two students not having the same birthday = 0.992
Then P(\(\overline{\mathrm{E}}\)) = The complementary event of E, i.e., two students having the same birthday Also, P(E) + p(\(\overline{\mathrm{E}}\)) = 1
∴ The probability that two students have the same birthday P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – 0.992 = 0.008

Question 7.
A die is thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Answer:
i) When a die is thrown for one time, total number of outcomes = 6
No. of outcomes favourable to a prime number (2, 3, 5) = 3
∴ Probability of getting a prime = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

ii) No. of outcomes favourable to a number lying between 2 and 6 (3, 4, 5) = 3
∴ Probability of getting a number between 2 and 6
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

iii) Number of outcomes favourable to an odd number (1, 3, 5) = 3
∴ Probability of getting an odd number P(odd)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 8.
What is the probability of drawing out a red king from a deck of cards?
Answer: Number of favourable outcomes to red king (♥ K, ♦ K) = 2.
Number of total outcomes = 52
(∵ Number of cards in a deck of cards = 52)
∴ Probability of getting a red king P (Red king)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 9.
Make 5 more problems getting probability using dice, cards or birthdays and discuss with friends and teacher about their solutions.
Answer:
Class-room activity.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.4

10th Class Maths 14th Lesson Statistics Ex 14.4 Textbook Questions and Answers

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Answer:
Since the curve is a less than type graph the data changes to

X – axis – upper limits 1 cm = 50 units.
Y – axis – less than c.f. 1 cm = 5 units.

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Answer:
Given: Upper limits of the classes and less than cumulative frequencies. Therefore required points are (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35)
X – axis – upper limits 1 cm = 2 units.
Y – axis – less than c.f. 1 cm = 4 units.

Number of observations = 35
∴ \(\frac{N}{2}\) = \(\frac{35}{2}\) = 17.5
Locate the point on the ogive whose ordinate is 17.5.
The x – coordinate of this point is the required median.
From the graph, median = 46.5.

Number of observations = n = 35
∴ \(\frac{N}{2}\) = \(\frac{35}{2}\) = 17.5
17.5 belongs to the class 46 – 48
∴ Median class = 46-48
l – lower boundary of class = 46
f – frequency of the median class =14
c.f = 14
Class size = 2
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 46 + \(\frac{17.5-14}{14}\) × 2
= 46 + \(\frac{3.5}{14}\) × 2
= 46 + \(\frac{7}{14}\)
= 46 + \(\frac{1}{2}\)
= 46.5
Here median is 46.5 by either by ways.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution, and draw its ogive.
Answer:
The given data is to be changed to more than frequency distribution type.

A graph is plotted by taking the lower limits on the X – axis and respective of Y – axis.
Scale:
X – axis: 1 cm = 5 units
Y – axis: 1 cm = 5 units