Srinivas

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 7.2 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 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

y + 3 = 0

y = 4

2x – 9 = 0

3x + 5 = 0

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

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

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

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.3

Question 1.
Draw the graph of each of the following linear equations.
i) 2y = – x + 1
ii) – x + y = 6
iii) 3x + 5y = 15
iv) \(\frac{x}{2}-\frac{y}{3}=3\)
Solution:
i) 2y = – x + 1
⇒ x + 2y = 1
ii) – x + y = 6
iii) 3x + 5y = 15
iv) \(\frac{x}{2}-\frac{y}{3}=3\)
⇒ \(\frac{3 x-2 y}{6}=3\)
⇒ 3x – 2y = 18

Question 2.
Draw the graph of each of the following linear equations and answer the following questions.
i) y – x
ii) y = 2x
iii) y = – 2x
iv) y = 3x
v) y = – 3x
Solution:
i) y = x

ii) y = 2x

iii) y = – 2x

iv) y = 3x

v) y = – 3x

i) Are all these equations of the form y = mx; where m is a real number ?
Solution:
Yes. All the equations are of the form y = mx where m e R.

ii) Are all these graphs passing through the origin ?
Solution:
Yes. All these lines pass through the origin.

iii) What can you conclude about these graphs ?
Solution:
All lines of the form y = mx, pass through the origin.

Question 3.
Draw the graph of the equation 2x + 3y = 11. Find from the graph value of y when x = 1.
Solution:

From the graph; when x = 1 then y = 3.

Question 4.
Draw the graph of the equation y – x = 2. Find from the graph
i) the value of y when x = 4
ii) the value of x when y = – 3
Solution:
The given equation is y – x = 2 or – x + y = 2

i) If x = 4 then y = 6 (∵ from the graph)
ii) When y = – 3 then x = – 5 (∵ from the graph)

Question 5.
Draw the graph of the equation 2x + 3y = 12. Find the solutions from the graph,
(i) Whose y-coordinate is 3 (OR) Whose y-coordinate is 2
(ii) Whose x-coordinate is – 3
Solution:
The given equation is 2x + 3y = 12

i) From the graph when y = 3 then 2x + 3(3) = 12 ⇒ 2x + 9 = 12 ⇒ 2x = 3 ⇒ x = \(\frac{3}{2}\) ; solution is (\(\frac{3}{2}\) , 3) (OR) When y = 2 then 2x + 3(2) = 12 ⇒ 2x + 6 = 12 ⇒ 2x = 6 ⇒ x = 3 solution is (3, 2),
ii) From the graph when x = – 3 then y = 6; solution is (- 3, 6)

Question 6.
Draw the graph of each of the equations given below and also find the coordinates of the points where the graph cuts the coordinate axes.
i) 6x – 3y = 12
Solution:
6x – 3y = 12

From the graph the line cuts the X-axis at (2, 0) and Y-axis at (0, – 4).

ii) -x + 4y = 8
Solution:

From the graph the line cuts the X-axis at (- 8, 0) and Y -axis at (0, 2).

iii) 3x + 2y + 6 = 0
Solution:
3x + 2y + 6 = 0

From the graph the line cuts the X-axis at (- 2, 0) and Y -axis at (0, -3).

Question 7.
Rajiya and Preethi two students of class IX together collected ₹1000 for the Prime Minister Relief Fund for victims of natural calamities. Write a linear equation and
draw a graph to depict the statement. Clfp)
Solution:
Let Rajiya’s contribution to P.M.R.F be = ₹ x
Preethi’s contribution to P.M.R.F be = ₹ y
Then by problem x + y = 1000

Question 8.
Gopaiah sowed w heat and paddy in two fields of total area 5000 sq. meters. Write a linear equation and draw a graph to represent the same.
Solution: Let the wheat be sowed in a land equal to x sq.m,
and the paddy be sowed in a land equal to y sq.m.
∴ By problem x + y = 5000

Question 9.
The force applied on a body of mass 6 kg. is directly proportional to the acceleration produced in the body. Write an equation to express this observation and draw the graph of the equation.
Solution:
Let the lorce = f; mass = 6 kg; acceleration = a
By problem f ∝ a or f = m . a ⇒ f = 6a

Question 10.
A stone is falling from a mountain. The velocity of the stone is given by v = 9.8t.
Draw its graph and find the velocity of the stone 4 seconds after start.
Solution:
Given that, the velocity of the stone v = 9.8 t

The velocity after 4 seconds = v = 9.8 × 4 = 39.2 m/sec².

Question 11.
In an election 60 % of voters cast their votes. Form an equation and draw the graph for this data. Find the following from the graph.
i) The total number of voters, if 1200 voters cast their votes,
ii) The number of votes cast, if the total number of voters are 800.
[Hint: If the number of voters who cast their votes be ‘x’ and the total number of voters be ‘y’ then x = 60 % of y.]
Solution:
Let the total number of votes be = y
Then the number of voters who cast their votes = x
By problem x = 60 % of y

i) From the graph when x = 1200, then y = 2000
ii) From the graph when y = 800 then x = 480

Question 12.
When Rupa was born, her father was 25 years old. Form an equation and draw a graph for this data. From the graph find
i) The age of the father when Rupa is 25 years old.
ii) Rupa’s age when her father is 40 years old.
Solution:
Let her father’s age be = x years.
and Rupa’s age be = y years
By problem x – y = 25 years

i) From the graph age of the father when Rupa is 25 years is 50 years.
ii) Rupa’s age when her father is 40 years is 15 years.

Question 13.
An auto charges ₹15 for first kilometre and ₹ 8 each for subsequent kilometre. For a distance of x km. an amount of ₹y is paid. Write the linear equation representing this information and draw the graph. With the help of graph find the distance travelled if the fare paid is ₹55. How much would have to be paid for 7 kilometres?
Solution:
Charge for the first kilometre = ₹15
Charge for the subsequent kilometres = ₹ 8 per km.
Amount paid = ₹ y when the distance travelled is x km
∴ By problem y = 15 + 8x
∴ 8x – y + 15 = 0

i) When y = 55 then x = 5
ii) From the graph when x = 7 then y = 71

Question 14.
A lending library has fixed charge for the first three days and an additional charges for each day thereafter. John paid ₹ 27 for a book kept for seven days. If the fixed charges be ₹ x and subsequent per day charges be ₹ y; then write the linear equation representing the above information and draw the graph of the same. From the graph if the fixed charge is ₹ 7, find the subsequent per day charge. And if the per day charge is ₹ 4, find the fixed charge, (charge is ₹7)
Solution:
John kept a book for 7 days. He paid ₹ 27
For first three days = ₹ x (fixed)
For the last four days = ₹ 4y (? y for a day)
By problem x + 4y = 27

When x = 7 then y = 5
When y = 4 then x = 11

Question 15.
The parking charges of a car in Hyderabad Railway station for first two hours is ₹ 50 and ₹10 for each subsequent hour. Write down an equation and draw the graph. Find the following charges from the graph.
i) For three hours ii) For six hours iii) How many hours did Rekha park her car if she paid ₹ 80 as parking charges ?
Solution: Let the total money paid be = ₹ y
Parking charges for the first two hours = ₹ 50
Parking charges for total x hours @ ₹10 per hour = 50 + (x – 2) 10
= 50 + 10x – 20
= 10x + 30
∴ By problem y = 10x + 30

i) For three hours = 50 + 10 × 1 = ₹ 60
[ ∵ From the graph we see the same]
ii) For six hours = 50 + 10 × 4 = 50 + 40 = ₹ 90
iii) Rekha parked her car for 5 hours.

Question 16.
Sameera was driving a car with uniform speed of 60 kmph. Draw distance – time graph. From the graph find the distance travelled by Sameera in
i) \(\frac { 1 }{ 2 }\) hours
ii) 2 hours
iii) 3\(\frac { 1 }{ 2 }\) hours
Solution:
Speed of the car = 60 kmph
Let the time taken be = x hours
Then the total distance travelled be = y hours
By problem 60x = y or 60x – y = 0

ii) Distance travelled in 1\(\frac { 1 }{ 2 }\) hours = 120 km
ii) Distance travelled in 2 hours = 120 km
iii) Distance travelled in 3\(\frac { 1 }{ 2 }\) hours = 210 km

Question 17.
The ratio of molecular weight of Hydrogen and Oxygen in water is 1 : 8. Set up an equation between Hydrogen and Oxygen and draw its graph. From the graph find the quantity of Hydrogen if Oxygen is 12 grams. And quantity of Oxygen if
Hydrogen is \(\frac { 3 }{ 2 }\) grams.
[Hint : If the quantities of hydrogen and oxygen ‘x’ and ‘y’ respectively, then
x : y = 1 : 8 ⇒ 8x = y]
Solution:
Let the quantity of Hydrogen = x grams
And the quantity of Oxygen = y grams
By problem 8x = y ⇒ 8x – y = 0

From the graph, the quantity of Hydrogen if Oxygen is 12 gm = \(\frac { 3 }{ 2 }\) g.
From the graph, the quantity of Oxygen if Hydrogen is \(\frac { 3 }{ 2 }\) g = 12 g.

Question 18.
In a mixture of 28 litres, the ratio of milk and water is 5 : 2. Set up the equation between the mixture and milk. Draw its graph. By observing the graph find the quantity of milk in the mixture.
[Hint: Ratio between mixture and milk = 5 + 2:5 = 7:5]
Solution:
Let the quantity of milk in the mixture be = x lit.
quantity of the mixture = y lit.
Ratio of the milk and water = 5:2
Sum of the terms of the ratio = 5 + 2 = 7
Quantity of milk x = \(\frac { 5 }{ 7 }\) y lit.
7x = 5y ⇒ 7x – 5y = 0

From the graph quantity of milk in the mixture = 20 lit.

Question 19.
In countries like U.S.A. and Canada temperature is measured in Fahrenheit whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius F = \(\frac { 9 }{ 5 }\) C + 32
i) Draw the graph of the above linear equation having Celsius on X-axis and Fahrenheit on Y-axis.
ii) If the temperature is 30°C, what is the temperature in Fahrenheit ?
iii) If the temperature is 95°F, what is the temperature in Celsius ?
iv) Is there a temperature that has numerically the same value in both Fahrenheit and Celsius ? If yes, find it.
Solution:
i) Given that F = \(\frac { 9 }{ 5 }\) C + 32

If C = 20, F = \(\frac { 9 }{ 5 }\) × 20 + 32 = 68
If c = 30, F = \(\frac { 9 }{ 5 }\) × 30 + 32 = 86
If C = 35, F = \(\frac { 9 }{ 5 }\) × 35 + 32 = 95
If C = -40, F = \(\frac { 9 }{ 5 }\) × (-40) + 32 = -40

ii) From the graph 30°C = 86°F
iii) 95°F = 35°C
iv) When C = – 40 then F = – 40