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Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.1 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Congruence of Triangles Class 8 Maths Chapter 13 Practice Set 13.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 13.1 Chapter 13 Solutions Answers

Congruence of Triangles Practice Set 13.1 Question 1.
In each pair of triangles in the following figures, parts bearing identical marks are congruent. State the test and correspondence of vertices by which triangles in each pair are congruent.
i.

ii.

iii.

iv.

v.

Solution:
i.

The two triangles are congruent by SAS test in the correspondence XWZ ↔ YWZ.

ii.

The two triangles are congruent by hypotenuse-side test in the correspondence KJI ↔ LJI.

iii.

The two triangles are congruent by SSS test in the correspondence HEG ↔ FGE.

iv.
The two triangles are congruent by ASA test is the correspondence SMA ↔ OPT.

v.

The two triangles are congruent by ASA test or SAS test or SAA test in the correspondence MTN ↔ STN.

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.1 Intext Questions and Activities

Practice Set 13.1 Question 1.
Write answers to the following questions referring to the given figure.

1. Which is the angle opposite to the side DE?
2. Which is the side opposite to ∠E?
3. Which angle is included by side DE and side DF?
4. Which side is included by ∠E and ∠F?
5. State the angles adjacent to side DE. (Textbook pg, no. 81)

Solution:

1. ∠DFE i.e. ∠F is the angle opposite to side DE.
2. Side DF is the side opposite to ∠E.
3. ∠EDF i.e. ∠D is included by side DE and side DF.
4. Side EF is included by ∠E and ∠F.
5. ∠DEF and ∠EDF i.e. ∠E and ∠D are adjacent to side DE.

Congruence of Triangles Class 8th Practice Set 13.1 Question 2.
In the given figure, parts of triangles indicated by identical marks are congruent.
a. Identify the one-to-one correspondence of vertices in which the two triangles are congruent and write the congruence.
b. State with reason, whether the statement, ∆XYZ ≅ ∆STU is right or wrong. (Textbook pg. no. 82)

Solution:
a. From the figure,
S ↔ X, T ↔ Z, U ↔ Y i.e.,
STU ↔ XZY, or SUT ↔ XYZ, or
TUS ↔ ZYX, or TSU ↔ ZXY, or
UTS ↔ YZX, or UST ↔ YXZ

∴ ∆STU ≅ ∆XZY, or ∆SUT ≅ ∆XYZ, or
∆TUS ≅ ∆ZYX, or ∆TSU ≅ ∆ZXY, or
∆UTS ≅ ∆YZX, or ∆UST ≅ ∆YXZ

b. If ∆XYZ ≅ ∆STU, then
∠Y ≅ ∠T, ∠Z ≅ ∠U,
seg XY ≅ seg ST, seg XZ ≅ seg SU
∴ But, all the above statements are wrong. The statement AXYZ ≅ ASTU is wrong.

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.5 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.5 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
If ∆XYZ ~ ∆LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.
Solution:
∆XYZ ~ ∆LMN [Given]
∴ ∠X ≅ ∠L
∠Y ≅ ∠M >
∠Z ≅ ∠N [Corresponding angles of similar triangles]
\( \frac{\mathrm{XY}}{\mathrm{LM}}=\frac{\mathrm{YZ}}{\mathrm{MN}}=\frac{\mathrm{XZ}}{\mathrm{LN}}\) [Corresponding sides of similar triangles]

Question 2.
In ∆XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm. If ∆XYZ ~ ∆PQR and PQ = 8 cm, then find the lengths of remaining sides of ∆PQR.
Solution:
∆XYZ ~ ∆PQR [Given]
∴ \( \frac{\mathrm{XY}}{\mathrm{PQ}}=\frac{\mathrm{YZ}}{\mathrm{QR}}=\frac{\mathrm{XZ}}{\mathrm{PR}}\) [Corresponding sides of similar triangles]

∴ PR = 10 cm
∴ QR = 12 cm, PR = 10cm

Question 3.
Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same signs. Show the lengths of corresponding sides by numbers in proportion.
Solution:

∆GHI ~ ∆STU

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.5 Intext Questions and Activities

Question 1.
We have learnt that if two triangles are equiangular then their sides are in proportion. What do you think if two quadrilaterals are equiangular? Are their sides in proportion? Draw different figures and verify. Verify the same for other polygons. (Textbook pg no 50)
Answer:
If two quadrilaterals are equiangular then their sides will not necessarily be in proportion.
Case 1: The two quadrilaterals are of the same type.

Consider squares ABCD and PQRS.
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CD}}{\mathrm{RS}}=\frac{\mathrm{AD}}{\mathrm{PS}}\)

Case 2: The two quadrilaterals are of different types.

Consider square ABCD and rectangle STUV.
∠A = ∠S, ∠B = ∠T, ∠C = ∠U, ∠D = ∠V

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.4 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.4 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.4 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
In the adjoining figure, point A is on the bisector of ∠XYZ. If AX = 2 cm, then find AZ.

Solution:
AX = 2 cm [Given]
Point A lies on the bisector of ∠XYZ. [Given]
Point A is equidistant from the sides of ∠XYZ. [Every point on the bisector of an angle is equidistant from the sides of the angle]
∴ A Z = AX
∴ AZ = 2 cm

Question 2.
In the adjoining figure, ∠RST = 56°, seg PT ⊥ ray ST, seg PR ⊥ ray SR and seg PR ≅ seg PT. Find the measure of ∠RSP.
State the reason for your answer.

Solution:
seg PT ⊥ ray ST, seg PR ⊥ ray SR [Given]
seg PR ≅ seg PT
∴ Point P lies on the bisector of ∠TSR [Any point equidistant from the sides of an angle is on the bisector of the angle]
∴ Ray SP is the bisector of ∠RST.
∠RSP = 56° [Given]
∴ ∠RSP = \(\frac { 1 }{ 2 }\)∠RST
= \(\frac { 1 }{ 2 }\) x 56°
∴ ∠RSP = 28°

Question 3.
In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.

Solution:
In ∆PQR,
PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given]
Since, 12 > 10 > 8
∴ QR > PQ > PR
∴ ∠QPR > ∠PRQ > PQR [Angle opposite to greater side is greater]
∴ In ∆PQR, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

Question 4.
In ∆FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.

Solution:
In ∆FAN,
∠F + ∠A + ∠N = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 80° + 40° + ∠N = 180°
∴ ∠N = 180° – 80° – 40°
∴∠N = 60°
Since, 80° > 60° > 40°
∴ ∠F > ∠N > ∠A
∴  AN > FA > FN [Side opposite to greater angle is greater]
∴  In ∆FAN, AN is the greatest side and FN is the smallest side.

Question 5.
Prove that an equilateral triangle is equiangular.

Given: ∆ABC is an equilateral triangle.
To prove: ∆ABC is equiangular
i.e. ∠A ≅ ∠B ≅ ∠C …(i) [Sides of an equilateral triangle]
In ∆ABC,
seg AB ≅ seg BC [From (i)]
∴ ∠C = ∠A (ii) [Isosceles triangle theorem]
In ∆ABC,
seg BC ≅ seg AC [From (i)]
∴ ∠A ≅ ∠B (iii) [Isosceles triangle theorem]
∴ ∠A ≅ ∠B ≅ ∠C [From (ii) and (iii)]
∴ ∆ABC is equiangular.

Question 6.
Prove that, if the bisector of ∠BAC of ∆ABC is perpendicular to side BC, then AABC is an isosceles triangle.

Given: Seg AD is the bisector of ∠BAC.
seg AD ⊥ seg BC
To prove: AABC is an isosceles triangle.
Proof.
In ∆ABD and ∆ACD,
∠BAD ≅ ∠CAD [seg AD is the bisector of ∠BAC]
seg AD ≅ seg AD [Common side]
∠ADB ≅ ∠ADC [Each angle is of measure 90°]
∴ ∆ABD ≅ ∆ACD [ASA test]
∴ seg AB ≅ seg AC [c. s. c. t.]
∴ ∆ABC is an isosceles triangle.

Question 7.
In the adjoining figure, if seg PR ≅ seg PQ, show that seg PS > seg PQ.

Solution:
Proof.
In ∆PQR,
seg PR ≅ seg PQ [Given]
∴ ∠PQR ≅ ∠PRQ ….(i) [Isosceles triangle theorem]
∠PRQ is the exterior angle of ∆PRS.
∴ ∠PRQ > ∠PSR ….(ii) [Property of exterior angle]
∴ ∠PQR > ∠PSR [From (i) and (ii)]
i.e. ∠Q > ∠S ….(iii)
In APQS,
∠Q > ∠S [From (iii)]
∴ PS > PQ [Side opposite to greater angle is greater]
∴ seg PS > seg PQ

Question 8.
In the adjoining figure, in AABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD = seg BE.

Solution:
Proof:
In ∆ADB and ∆BEA,
seg BD ≅ seg AE [Given]
∠ADB ≅ ∠BEA = 90° [Given]
seg AB ≅ seg BA [Common side]
∴ ∆ADB ≅ ∆BEA [Hypotenuse-side test]
∴ seg AD ≅ seg BE [c. s. c. t.]

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.4 Intext Questions and Activities

Question 1.
As shown in the given figure, draw ∆XYZ such that side XZ > side XY. Find which of ∠Z and ∠Y is greater. (Textbook pg. no. 41)

Answer:
From the given figure, ∠Z = 25° and ∠Y = 51°
∴ ∠Y is greater.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.3 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.3 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
Find the values of x and y using the information shown in the given figure. Find the measures of ∠ABD and ∠ACD.

Solution:
i. ∠ACB = 50° [Given]
In ∆ABC, seg AC ≅ seg AB [Given]
∴ ∠ABC ≅ ∠ACB [Isosceles triangle theorem]
∴ x = 50°

ii. ∠DBC = 60° [Given]
In ABDC, seg BD ≅ seg DC [Given]
∴ ∠DCB ≅ ∠DBC [Isosceles triangle theorem]
∴ y = 60°

iii. ∠ABD = ∠ABC + ∠DBC [Angle addition property]
= 50° + 60°
∴ ∠ABD = 110°

iv. ∠ACD = ∠ACB + ∠DCB [Angle addition property]
= 50° + 60°
∴ ∠ACD = 110°
∴ x = 50°, y = 60°,
∠ABD = 110°, ∠ACD = 110°

Question 2.
The length of hypotenuse of a right angled triangle is 15. Find the length of median on its hypotenuse.

Solution:
Length of hypotenuse = 15 [Given]
Length of median on the hypotenuse = \(\frac { 1 }{ 2 }\) x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= \(\frac { 1 }{ 2 }\) x 15 = 7.5
∴ The length of the median on the hypotenuse is 7.5 units.

Question 3.
In ∆PQR, ∠Q = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS).

Solution:
i. PQ = 12, QR = 5 [Given]
In APQR, ∠Q = 90° [Given]
∴ PR² = QR² + PQ² [Pythagoras theorem]
= 25 + 144
∴ PR² =169
∴ PR = 13 units [Taking square root of both sides]

ii. In right angled APQR, seg QS is the median on hypotenuse PR.
∴ QS = \(\frac { 1 }{ 2 }\)PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= \(\frac { 1 }{ 2 }\) x 13
∴ l(QS) = 6.5 units

Question 4.
In the given figure, point G is the point of concurrence of the medians of ∆PQR. If GT = 2.5, find the lengths of PG and PT.

Solution:
i. In ∆PQR, G is the point of concurrence of the medians. [Given]
The centroid divides each median in the ratio 2 : 1.
PG : GT = 2 : 1

∴ PG = 2 x 2.5
∴ PG = 5 units

ii. Now, PT = PG + GT [P – G – T]
= 5 + 2.5
∴ l(PG) = 5 units, l(PT) = 7.5 units

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.3 Intext Questions and Activities

Question 1.
Can the theorem of isosceles triangle be proved by doing a different construction? (Textbook pg. no.34)

Solution:
Yes
Construction: Draw seg AD ⊥ seg BC.
Proof:
In ∆ABD and ∆ACD,
seg AB≅ seg AC [Given]
∠ADB ≅ ∠ADC [Each angle is of measure 90°]
seg AD ≅ seg AD [Common side]
∴ ∆ABD ≅ ∆ACD [Hypotenuse side test]
∴ ∠ABD ≅ ∠ACD [c.a.c.t.]
∴ ∠ABC ≅ ∠ACB [B-D-C]

Question 2.
Can the theorem of isosceles triangle be proved without doing any construction? (Textbook pg, no.34)

Solution:
Yes
Proof:
In ∆ABC and ∆ACB,
seg AB ≅ seg AC [Given]
∠BAC ≅ ∠CAB [Common angle]
seg AC ≅ seg AB [Given]
∴ ∆ABC ≅ ∆ACB [SAS test]
∴ ∠ABC ≅ ∠ACB [c. a. c. t.]

Question 3.
In the given figure, ∆ABC is a right angled triangle, seg BD is the median on hypotenuse. Measure the lengths of the following segments.
i. AD
ii. DC
iii. BD
From the measurements verify that BD = \(\frac { 1 }{ 2 }\)AC. (Textbook pg. no. 37)
Solution:
AD = DC = BD= 1.9 cm
AC = AD + DC [A – D – C]
= 1.9 + 1.9
= 2 x 1.9 cm
∴ AC = 2 x BD
∴ BD = \(\frac { 1 }{ 2 }\) AC

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 12.2 8th Std Maths Answers Solutions Chapter 12 Equations in One Variable.

Equations in One Variable Class 8 Maths Chapter 12 Practice Set 12.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 12.2 Chapter 12 Solutions Answers

Equation In One Variable Practice Set 12.2 Question 1.
Mother is 25 years older than her son. Find son’s age, if after 8 years ratio of son’s age to mother’s age will be \(\frac { 4 }{ 9 }\).
Solution:
Let the son’s present age be x years.
∴ Mother’s present age = (x + 25) years
After 8 years,
Son’s age = (x + 8) years
Mother’s age = (x + 25 + 8) = (x + 33) years
Since, the ratio of the son’s age to mother’s age after 8 years is \(\frac { 4 }{ 9 }\).
∴ \(\frac{x+8}{x+33}=\frac{4}{9}\)
∴ 9 (x + 8) = 4 (x + 33)
∴ 9x + 72 = 4x + 132
∴ 9x – 4x = 132 – 72
∴ 5x = 60
∴ x = \(\frac { 60 }{ 5 }\)
∴ x = 12
∴ Son’s present age is 12 years.

8th Std Maths Practice Set 12.2 Question 2.
The denominator of a fraction is greater than its numerator by 12. If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent to \(\frac { 1 }{ 2 }\) . Find the fraction.
Solution:
Let the numerator of the fraction be x.
The denominator of a fraction is greater than its numerator by 12.
∴ Denominator of the fraction = (x + 12)
∴ The required fraction = \(\frac { x }{ x+12 }\)
For the new fraction,
numerator is decreased by 2.
∴ The new numerator = (x – 2)
Also, denominator is increased by 7.
∴ The new denominator = (x + 12) + 7
= (x + 19)
Since, the new fraction is equivalent to \(\frac { 1 }{ 2 }\).
∴ \(\frac{x-2}{x+19}=\frac{1}{2}\)
∴ 2(x – 2) = 1(x + 19)
∴ 2x – 4 = x + 19
∴ 2x – x = 19 + 4
∴ x = 23
∴ The required fraction = \(\frac{x}{x+12}=\frac{23}{23+12}=\frac{23}{35}\)
∴ The required fraction is \(\frac { 23 }{ 35 }\)

Practice Set 12.2 Class 8 Question 3.
The ratio of the weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.
Solution:
Let the weight of zinc in the brass utensil be x gm.
Since, the ratio of the weights of copper to zinc in brass is 13:7.

∴ Weight of copper in the brass utensil = \(\left(\frac{13}{7} x\right)\) gm
The weight of the brass utensil = 700 gm
∴ \(\frac { 13 }{ 7 }x+x=700\)
∴ \(\frac { 13 }{ 7 }x\) x × 7 + x × 7 = 700 × 7
∴ 13x + 7x = 4900
∴ 20x = 4900
∴ \(x=\frac { 4900 }{ 20 }\)
∴ x = 245
∴ The weight of zinc in the brass utensil is 245 gm.

Practice Set 12.2 8th Class Question 4.
Find three consecutive whole numbers whose sum is more than 45 but less than 54.
Solution:
Let the three consecutive whole numbers be (x – 1), x and (x + 1).
∴ Sum of the three numbers
= (x – 1) + x + (x + 1)
= 3x
Given that, the sum of the three numbers is greater than 45 and less than 54.
When the sum of the three numbers is 45,
3x = 45
∴ x = \(\frac { 45 }{ 3 }\)
∴ x = 15
When the sum of the three numbers is 54,
∴ 3x = 54
∴ x = \(\frac { 54 }{ 3 }\)
∴ x = 18
∴ the value of x is greater than 15 and less than 18.
∴ the value of x is either 16 or 17

Case I:
If the value of x is 16, then the three consecutive whole numbers are
(16 – 1), 16,(16 + 1)i.e., 15, 16, 17

Case II:
If the value of x is 17, then the three consecutive whole numbers are (17 – 1), 17, (17 + 1) i.e., 16, 17, 18.
∴ The three consecutive whole numbers are 15, 16, 17 or 16, 17, 18.

Practice Set 12.2 8th Standard Question 5.
In a two-digit number, digit at the ten’s place is twice the digit at unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.
Solution:
Let the digit at unit’s place be x.
The digit at the ten’s place is twice the digit at unit’s place.
∴ The digit at ten’s place = 2x

	Digit in units place	Digit in tens place	Number
Original Number	x	2x	(2x × 10) + x = 20x + x = 21x
New Number	2x	x	(x × 10) + 2x = 10x + 2x = 12x

Since, the sum of the original number and the new number is 66.
∴ 21x + 12x = 66
∴ 33x = 66
∴ x = \(\frac { 66 }{ 33 }\)
∴ x = 2
∴ Original number = 21x = 21 × 2 = 42
∴ the original number is 42.

8th Standard Maths Practice Set 12.2 Question 6.
Some tickets of Rs 200 and some of Rs 100, of a drama in a theatre were sold. The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100. The total amount received by the theatre by sale of tickets was Rs 37000. Find the number of Rs 100 tickets sold.
Solution:
Let the number of tickets sold of Rs 100 be x.
The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100.
∴ Number of tickets sold of Rs 200 = (x + 20)
∴ Total amount received by the theatre through the sale of tickets = 100 × x + 200 × (x + 20)
= 100x + 200x + 4000
= 300x + 4000
Since, the total amount received by the theatre through the sale of tickets = Rs 37000
∴ 300x + 4000 = 37000
∴ 300x = 37000 – 4000
∴ 300x = 33000
∴ \(x=\frac { 33000 }{ 300 }\)
∴ x = 110
∴ 110 tickets of Rs 100 were sold.

8th Maths Practice Set 12.2 Question 7.
Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.
Solution:
Let the three consecutive natural numbers be (x – 1), x and (x + 1).
Here, the smallest number is (x – 1) and the greatest number is (x + 1).
Since, five times the smallest number is 9 more than four times the greatest number.
∴ 5 × (x – 1) = [4 × (x + 1)] + 9
∴ 5x – 5 = 4x + 4 + 9
∴ 5x – 5 = 4x + 13
∴ 5x – 4x = 13 + 5
∴ x = 18 .
∴ the three numbers are (18 – 1), 18, (18 + 1)
i. e., 17, 18, 19
∴ The three consecutive natural numbers are 17,18 and 19.

Raju Sold A Bicycle to Amit at 8 Question 8.
Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs 54. Then he sold the bicycle to Nikhil for Rs 1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.
Solution:
Let the cost price at which Raju purchased the bicycle be Rs x.
Since, Raju sold the bicycle at 8% profit to Amit.
∴ Selling price of bicycle for Raju = x + 8% of x

Since, Amit spent Rs 54 on repairing the bicycle and then sold it to Nikhil for Rs 1134, at no loss and no profit.
∴ Selling price of bicycle + repairing cost = Rs 1134

∴ The cost price of the bicycle at which Raju purchased it is Rs 1000.

Class 8 Maths Practice Set 12.2 Question 9.
A cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.
Solution:
Let the number of runs required by the cricket player to score in the third match be x.
Number of runs scored by the player in first match = 180
Number of runs scored in second match = 257
∴ Total runs scored by the player = 180 + 257 + x = 437 + x
Average of runs in the three matches = \(\frac { 437+x }{ 3 }\)
Since, the average of runs should be 230.
\(\frac { 437+x }{ 3 }=230\)
∴ 437 + x = 230 × 3
∴ 437 + x = 690
∴ x = 690 – 437
∴ x = 253
∴ The cricket player should score 253 runs in the third match.

8th Class Math Practice Set 12.2 Question 10.
Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6, then find Viru’s age.
Solution:
Let Viru’s present age be x years.
Sudhir’s present age is 5 more than three times the age of Viru.
∴ Sudhir’s present age = (3x + 5) years
Anil’s age is half the age of Sudhir.
∴ Anil’s present age = \(\left(\frac{3 x+5}{2}\right)\) years
Since, the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6.

∴ 2 × (24x + 30) = 45x + 75
∴ 48x + 60 = 45x + 75
∴ 48x – 45x = 75 – 60
∴ 3x = 15
∴ x = \(\frac { 15 }{ 3 }\)
∴ x = 5
∴ Viru’s present age is 5 years.

Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.2 Intext Questions and Activities

8th Math Practice Set 12.2 Question 1.
Write correct numbers in the boxes given. (Textbook pg. no. 78)
length is 3 times the breadth

Perimeter of the rectangle = 40
2(__x + __x) = 40
2 × __ x = 40
__ x = 40
x = __
∴ Breadth of rectangle = __ cm and Length of rectangle = __ cm
Solution:
length is 3 times the breadth

Perimeter of the rectangle = 40
∴ 2(3x + 1x) = 40
∴ 2 × 4x = 40
∴ 8x = 40
∴ x = 5
∴ Breadth of rectangle = 5 cm and Length of rectangle = 15 cm

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.1 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.1 Chapter 3 Triangles Questions With Answers Maharashtra Board

Practice Set 3.1 Geometry 9th Standard Question 1.
In the adjoining figure, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.

Solution:
∠A = 70° , ∠B = 40° [Given]
∠ACD is an exterior angle of ∆ABC. [Given]
∴ ∠ACD = ∠A + ∠B
= 70° + 40°
∴ ∠ACD = 110°

Question 2.
In ∆PQR, ∠P = 70°, ∠Q = 65°, then find ∠R.
Solution:
∠P = 70°, ∠Q = 65° [Given]
In ∆PQR,
∠P + ∠Q + ∠R = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + 65° + ∠R = 180°
∴ ∠R = 180° – 70° – 65°
∴ ∠R = 45°

Practice Set 3.1 Geometry 9th Question 3.
The measures of angles of a triangle are x°, (x – 20)°, (x – 40)°. Find the measure of each angle.
Solution:
The measures of the angles of a triangle are x°, (x – 20)°, (x – 40)°. [Given]
∴ x°+ (x – 20)° + (x – 40)° = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 3x – 60 = 180
∴ 3x = 180 + 60
∴ 3x = 240
∴ x = 240
∴ x = \(\frac { 240 }{ 3 }\)
∴ x = 80°
∴ The measures of the remaining angles are
x – 20° = 80° – 20° = 60°,
x – 40° = 80° – 40° = 40°
∴ The measures of the angles of the triangle are 80°, 60° and 40°.

9th Class Geometry Practice Set 3.1 Question 4.
The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.
Solution:
Let the measure of the smallest angle be x°.
One of the angles is twice the measure of the smallest angle.
∴ Measure of that angle = 2x°
Another angle is thrice the measure of the smallest angle.
∴ Measure of that angle = 3x°
∴ The measures of the remaining two angles are 2x° and 3x°.
Now, x° + 2x° + 3x° = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 6x = 180
∴ x = 180
∴ x = \(\frac { 180 }{ 6 }\)
∴ x° = 30°
The measures of the remaining angles are 2x° = 2 x 30° = 60°
3x° = 3 x 30° = 90°
The measures of the three angles of the triangle are 30°, 60° and 90°.

Question 5.
In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.

Solution:
i. ∠NET = 100° and ∠EMR = 140°
∠EMN + ∠EMR = 180°
∴ z +140° =180°
∴ z = 180° -140°
∴ z = 40°

ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair]
∴ 100° + y = 180°
∴ y = 180° – 100°
∴ y = 80°

iii. In ∆ENM,
∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x +80°+ 40°= 180°
∴ x = 180° – 80° – 40°
∴ x = 60°
∴ x = 60°, = 80°, z = 40°

Question 6.
In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

Solution:
i. ∠B AD = 70°, ∠DER = 40° [Given]
line AB || line DE and seg AD is their transversal.
∴ ∠EDA = ∠BAD [Alternate Angles]
∴ ∠EDA = 70° ….(i)
In ∆DRE,
∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A]
∴ ∠DRE = 180° -70° -40°
∴ ∠DRE = 70°

ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair]
∴ 70° + ∠ARE = 180°
∴ ∠ARE = 180°-70°
∴ ∠ARE =110°
∴ ∠DRE = 70°, ∠ARE = 110°

Triangles Class 9 Practice Set 3.1 Question 7.
In ∆ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°, find the measure of ∠AOB.
Solution:
∠OAB = ∠OAC = – ∠BAC ….(i) [Seg AO bisects ∠BAC]
∠OBA = ∠OBC = – ∠ABC …..(ii) [Seg RO bisects ∠ABC]
In AABC,
∠BAC + ∠ABC + ∠ACB = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ ∠BAC + ∠ABC + 70° = 180°
∴ ∠BAC + ∠ABC = 180°- 70°
∴ ∠BAC + ∠ABC = 110°
∴ \(\frac { 1 }{ 2 }\)(∠BAC) + \(\frac { 1 }{ 2 }\) (∠ABC) = \(\frac { 1 }{ 2 }\) x 110° [MuItiplying both sides by \(\frac { 1 }{ 2 }\)]
∴ ∠OAB + ∠OBA = 55° ….(iii) [From (i) and (ii)]
In AOAB,
∠OAB + ∠OBA + ∠AOB = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 55° + ∠AOB = 180° [From (iii)]
∴ ∠AOB = 180°- 55°
∴ ∠AOB = 125°

Question 8.
In the adjoining figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.

Given: line AB || line CD and line PQ is the transversal.
ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively.
To prove: m∠PTQ = 90°
Solution:
Proof:
∠TPB = ∠TPQ = \(\frac { 1 }{ 2 }\)∠BPQ …(i) [Ray PT bisects ∠BPQ]
∠TQD = ∠TQP = \(\frac { 1 }{ 2 }\)∠PQD ….(ii) [Ray QT bisects ∠PQD]
line AB || line CD and line PQ is their transversal. [Given]
∴∠BPQ + ∠PQD = 180° [Interior angles]
∴ \(\frac { 1 }{ 2 }\) (∠BPQ) + \(\frac { 1 }{ 2 }\) (∠PQD) = \(\frac { 1 }{ 2 }\) x 180° [Multiplying both sides by \(\frac { 1 }{ 2 }\)]
∠TPQ + ∠TQP = 90°
In ∆PTQ,
∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 90° + ∠PTQ = 180° [From (iii)]
∴ ∠PTQ = 180° – 90°
= 90°
∴ m∠PTQ = 90°

Triangle Practice Set 3.1 Question 9.
Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.

Solution:
i. ∠c + 100° = 180° [Angles in a linear pair]
∴ ∠c = 180° – 100°
∴ ∠c = 80°

ii. ∠b = 70° [Vertically opposite angles]
iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°]
∠a + 70° + 80° = 1800
∴ ∠a = 180° – 70° – 80°
∴ ∠a = 30°
∴ ∠a = 30°, ∠b = 70°,∠ c = 80°

Practice Set 3.1 Geometry Question 10.
In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,

i. ∠DEG = \(\frac { 1 }{ 2 }\) ∠EDF
ii. EF = FG
Solution:
i. ∠DEG = ∠FEG = x° ….(i) [Ray EG bisects ∠DEF]
∠GFD = ∠GFM = y° …..(ii) [Ray FG bisects ∠DFM]
line DE || line GF and DF is their transversal. [Given]

∴ ∠EDF = ∠GFD [Alternate angles]
∴ ∠EDF = y° ….(iii) [From (ii)]
line DE || line GF and EM is their transversal. [Given]

∴ ∠DEF = ∠GFM [Corresponding angles]
∴ ∠DEG + ∠FEG = ∠GFM [Angle addition property]
∴ x°+ x° = y° [From (i) and (ii)]
∴ 2x° = y°
∴ x° = \(\frac { 1 }{ 2 }\)y°
∴ ∠DEG = \(\frac { 1 }{ 2 }\)∠EDF [From (i) and (iii)]

ii. line DE || line GF and GE is their transversal. [Given]

∴ ∠DEG = ∠FGE …(iv) [Alternate angles]
∴ ∠FEG = ∠FGE ….(v) [From (i) and (iv)]
∴ In ∆FEG,
∠FEG = ∠FGE [From (v)]
∴ EF = FG [Converse of isosceles triangle theorem]

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.1 Intext Questions and Activities

Class 9 Geometry Practice Set 3.1 Question 1. Can you give an alternative proof of the above theorem by drawing a line through point R and parallel to seg PQ in the above figure? (Textbook pg. no. 25)

Solution:
Yes.
Construction: Draw line RM parallel to seg PQ through a point R.
Proof:
seg PQ || line RM and seg PR is their transversal. [Construction]
∴ ∠PRM = ∠QPR ……..(i) [Alternate angles]
seg PQ || line RM and seg QR is their transversal. [Construction]
∴ ∠SRM = ∠PQR ……..(ii) [Corresponding angles]
∴ ∠PRM + ∠SRM = ∠QPR + ∠PQR [Adding (i) and (ii)]
∴ ∠PRS = ∠PQR + ∠QPR [Angle addition property]

3 Triangles Question 2. Observe the given figure and find the measures of ∠PRS and ∠RTS. (Textbook pg. no.25)

Solution:
∠PRS is an exterior angle of ∆PQR.
So from the theorem of remote interior angles,
∠PRS = ∠PQR + ∠QPR
= 40° + 30°
∴ ∠PRS = 70°
∴ ∠TRS=70° …[P – T – R]
In ∆RTS,
∠TRS + ∠RTS + ∠TSR = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ 70° + ∠RTS + 20° = 180°
∴ ∠RTS + 90° = 180°
∴ ∠RTS = 180°
∴ ∠RTS = 90°

9th Class Geometry Triangles Question 3. In the given figure, bisectors of ∠B and ∠C of ∆ABC intersect at point P. Prove that ∠BPC = 90° + \(\frac { 1 }{ 2 }\)∠BAC.
Complete the proof by filling in the blanks. (Textbook pg. no.27)

Solution:
Proof:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ ∠BAC + – ∠ABC + ∠ACB = 180 … [Multiplying each term by \(\frac { 1 }{ 2 }\)]
∴ ∠BAC + ∠PBC + ∠PCB = 90°
∴ ∠PBC + ∠PCB = 90° – 1 ∠BAC ………(i)
In∆BPC,
∠BPC + ∠PBC + ∠PCB = 180° …….[Sum of measures of angles of a triangle]
∴ ∠BPC + 90° – \(\frac { 1 }{ 2 }\)∠BAC = 180° ……[From (i)]
∴ ∠BPC = 180° – 90°\(\frac { 1 }{ 2 }\)∠BAC
= 180°- 90°+ \(\frac { 1 }{ 2 }\)∠BAC
= 90°+ \(\frac { 1 }{ 2 }\)∠BAC

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 2 Parallel Lines.

9th Standard Maths 2 Practice Set 2 Chapter 2 Parallel Lines Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 2 Chapter 2 Parallel Lines Questions With Answers Maharashtra Board

Question 1.
Select the correct alternative and fill in the blanks in the following statements.

i. If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is ____.
(A) 0°
(B) 90°
(C) 180°
(D) 360°
Answer:
(C) 180°

ii. The number of angles formed by a transversal of two lines is _____.
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(C) 8

iii. A transversal intersects two parallel lines. If the measure of one of the angles is 40°, then the measure of its corresponding angle is ______.
(A) 40°
(B) 140°
(C) 50°
(D) 180°
Answer:
(A) 40°

iv. In ∆ABC, ∠A = 76°, ∠B = 48°, then ∠C = _____.
(A) 66°
(B) 56°
(C) 124°
(D) 28°
Answer:
In ∆ABC, ∠A + ∠B + ∠C = 180°
∴ ∠C = 180° – 76° – 48° = 56°
(B) 56°

v. Two parallel lines are intersected by a transversal. If measure of one of the alternate interior angles is 75° then the measure of the other angle is _____.
(A) 105°
(B) 15°
(C) 75°
(D) 45°
Answer:
(C) 75°

Question 2.
Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ∠QPR respectively. Ray PB and ray PA are perpendicular to each other.
Draw a figure showing all these rays and write –
i. A pair of complementary angles
ii. A pair of supplementary angles
iii. A pair of congruent angles.
Solution:

i. Complementary angles:
∠RPQ = 90° [Ray PQ ⊥ ray PR]
∴ ∠RPB + ∠BPQ = 90° [Angle addition property]
∠RPB and ∠BPQ are pair of complementary angles
∠APB = 90° [Ray PA ⊥ ray PB]
∴ ∠APR + ∠RPB = 90°
∠APR and ∠RPB are pair of complementary angles.

ii. Supplementary angles:
∠APB + ∠RPQ = 90° + 90° = 180°
∴ ∠APB and ∠RPQ are a pair of supplementary angles.

iii. Congruent angles:
a. ∠APB = ∠RPQ [Each is of 90°]
b. ∠APB = ∠RPQ
∴ ∠APR + ∠RPB = ∠RPB + ∠BPQ [Angle addition property]
∴ ∠APR = ∠BPQ
∴ ∠APR ≅ ∠BPQ

Question 3.
Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.

Given: line AB || line CD and line EF intersects them at P and Q respectively.
line EF ⊥ line AB
To prove: line EF ⊥ line CD
Solution:
Proof:
line EF ⊥ line AB [Given]
∴ ∠APR = 90° ….(i)
line AB || line CD and line EF is their transversal.
∴ ∠EPB ≅ ∠PQD …..(ii) [Corresponding angles]
∴ ∠PQD = 90° [From (i) and (ii)]
∴ line EF ⊥ line CD

Question 4.
In the given figure, measures of some angles are shown. Using the measures find the measures of ∠x and ∠y and hence show that line l || line m.

Solution:
Proof:
∠x = 130°
∠y = 50° [Vertically opposite angles]
Here, m∠PQT + m∠QTS = 130° + 50° = 180°
But, ∠ PQT and ∠ QTS are a pair of interior angles on lines l and m when line n is the transversal,

∴ line l || line m [Interior angles test]

Question 5.
In the given figure, Line AB || line CD || line EF and line QP is their transversal. If y : z = 3 : 7 then find the measure of ∠x.

Solution:
y : z = 3 : 7 [Given]
Let the common multiple be m
∴ ∠j = 3m and ∠z = 7m ….(i)
line AB || line EF and line PQ is their transversal [Given]
∠x = ∠z
∴ ∠x = 7m …..(ii) [From (i)]
line AB || line CD and line PQ is their transversal [Given]
∠x + ∠y = 180°
∴ 7m + 3m = 180°
∴ 10m = 180°
∴ m = 18°
∴ ∠x = 7m = 7(18°) [From (ii)]
∴ ∠x = 126°

Question 6.
In the given figure, if line q || line r, line p is their transversal and if a = 80°, find the values of f and g.

Solution:
i. ∠a = 80° [Given]
∠g = ∠a [Alternate exterior angles]
∴ ∠g = 80° …..(i)

ii. Now, line q || line r and line p is their transversal.
∴ ∠f + ∠g = 180°
∴ ∠f + 80° = 180° [Interior angles]
∴ ∠f = 180° – 80° [From (i)]
∴ ∠f=100°

Question 7.
In the given figure, if line AB || line CF and line BC || line ED then prove that ∠ABC = ∠FDE.

Given: line AB || line CF and line BC || line ED
To prove: ∠ABC = ∠FDE
Solution:
Proof:
line AB || line PF and line BC is their transversal.
∴ ∠ABC = ∠BCD ….(i) [Alternate angles]
line BC || line ED and line CD is their transversal.
∴ ∠BCD = ∠FDE ….(ii) [Corresponding angles]
∴ ∠ABC = ∠FDE [From (i) and (ii)]

Question 8.
In the given figure, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that □ QXRY is a rectangle.

Given: line AB || line CD
Rays QX, RX, QY, RY are the bisectors of ∠AQR, ∠QRC, ∠BQR and ∠QRD respectively.
To prove: □QXRY is a rectangle.
Proof:
∠XQA = ∠XQR = x° ……(i) [Ray QX bisects ∠AQR]
∠YQR = ∠YQB =y° …….(ii) [Ray QY bisects ∠BQR]
∠XRQ = ∠XRC = u° …….. (iii) [Ray RX bisects ∠CRQ]
∠YRQ = ∠YRD = v° ……(iv) [Ray RY bisects ∠DRQ]
line AB || line CD and line PS is their transversal.

∠AQR+ ∠CRQ= 180° [Interior angles]
(∠XQA + ∠XQR) + (∠XRQ + ∠XRC) = 180° [Angle addition property]
∴ (x + x) + (u + u) = 180° [From (i) and (ii)]
∴ 2x + 2u = 180°
∴ 2(x + u) = 180°
∴ x + u = 90° ……..(v)
In ∆XQR
∠XQR + ∠XRQ + ∠QXR = 180° [Sum of the measures of the angles of triangle is 180°]
∴ x + u + ∠QXR = 180° [From (i) and (iii)]
∴ 90 + ∠QXR = 180° [From (v)]
∴ ∠QXR = 180°- 90°
∴ ∠QXR = 90° …..(vi)
Similarily we can prove that,

∴ y + v = 90°
Hence ∠QYR = 90° ……(vii)
Now, ∠AQR + ∠BQR = 180° [Angles is linear pair]
(∠XQA + ∠XQR) + (∠YQR + ∠YQB) = 180° [Angle addition property]
∴ (x + x) + (y + y) = 180° [From (i) and (ii)]
∴ 2x + 2y = 180°
∴ 2(x+y) = 180°
∴ x +y = 90°
i.e. ∠XQR + ∠YQR = 90° [From (i) and (ii)]
∴ ∠XQY = 90° ….(viii) [Angle addition property]
Similarly we can prove that,
∠XRY = 90° …(ix)
In □QXRY
∠QXR = ∠QYR = ∠XQY = ∠XRY = 90° [From (vi), (vii), (viii) and (ix)]
∴□ QXRY is a rectangle.

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Problem Set 2 Intext Questions and Activities

Question 1.
To verify the properties of angles formed by a transversal of two parallel lines. (Textbook pg. no. 14)
Take a piece of thick coloured paper. Draw a pair of parallel lines and a transversal on it. Paste straight sticks on the lines. Eight angles will be formed. Cut pieces of coloured paper, as shown in the figure, which will just fit at the comers of ∠1 and ∠2. Place the pieces near different pairs of corresponding angles, alternate angles and interior angles and verify their properties.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.2 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.2 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

By SSS test
∆ABC ≅ ∆PQR

By SAS test
∆ XYZ ≅ ∆LMN

By ASA test
∆PRQ ≅ ∆STU

By hypotenuse side test
∆LMN ≅ ∆PTR

Question 2.
Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.
Solution:

From the information shown in the figure,
In ∆ABC and ∆PQR,
∠ABC ≅ ∠PQR
seg BC ≅ seg QR
∠ACB ≅ ∠PRQ
∴ ∆ABC ≅ ∆PQR [ASA test]
∴ ∠BAC ≅ ∠QPR [Corresponding angles of congruent triangles]
seg AB ≅ segPQ and segAC ≅ seg PR [Corresponding sides of congruent triangles]

From the information shown in the figure,
In ∆PTQ and ∆STR,
seg PT ≅ seg ST
∠PTQ ≅ ∠STR [Vertically opposite angles]
seg TQ ≅ seg TR
∴ ∆PTQ ≅ ∆STR [SAS test]
∴ ∠TPQ ≅ ∠TSR and ∠TQP ≅ ∠TRS [Corresponding angles of congruent triangles]
seg PQ ≅ seg SR [Corresponding sides of congruent triangles]

Question 3.
From the information shown in the figure, state the test assuring the congruence of ∆ABC and ∆PQR. Write the remaining congruent parts of the triangles.

Solution:
In ∆BAC and ∆PQR,
seg BA ≅ seg PQ
seg BC ≅ seg PR
∠BAC ≅ ∠PQR = 90° [Given]
∴ ∆BAC ≅ ∆PQR [Hypotenuse side test]
∴ seg AC ≅ seg QR [c.s.c.t.]
∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP [c.a.c.t.]

Question 4.
As shown in the adjoining figure, in ∆LMN and ∆PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

Solution:
In ∆LMN and ∆PNM,
seg LM ≅ seg PN
seg LN ≅ seg PM [Given]
seg MN ≅ seg NM [Common side]
∴ ∆LMN ≅ ∆PNM [SSS test]
∴ ∠LMN ≅ ∠PNM,
∴ ∠MLN ≅ ∠NPM, and ∠LNM ≅ ∠PMN [c.a.c.t.]

Question 5.
In the adjoining figure, seg AB ≅ seg CB and seg AD ≅ seg CD. Prove that ∆ABD ≅ ∆CBD.

Solution:
proof:
In ∆ABD and ∆CBD,
seg AB ≅ seg CB
seg AD ≅ seg CD [Given]
seg BD ≅ seg BD [Common side]
∴ ∆ABD ≅ ∆CBD [SSS test]

Question 6.
In the adjoining figure, ZP ≅ ZR, seg PQ ≅ seg RQ. Prove that APQT ≅ ARQS.

Proof:
In ∆PQT and ∆RQS,
∠P ≅ ∠R
seg PQ ≅ seg RQ [Given]
∠Q ≅ ∠Q [Common angle]
∴ ∆PQT ≅ ∆RQS [ASA test]

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 12.1 8th Std Maths Answers Solutions Chapter 12 Equations in One Variable.

Equations in One Variable Class 8 Maths Chapter 12 Practice Set 12.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 12.1 Chapter 12 Solutions Answers

Equation in One Variable Practice Set 12.1 Question 1. Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.
i. x – 4 = 3, x = – 1, 7, – 7
ii. 9m = 81, m = 3, 9, -3
iii. 2a + 4 = 0, a = 2, – 2, 1
iv. 3 – y = 4, y = – 1, 1, 2
Solution:
i. x – 4 = 3 ….(i)
Substituting x = – 1 in L.H.S. of equation (i),
L.H.S. = (-1) – 4
= – 5
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i),
L.H.S. = (7) – 4
= 3
R.H.S. = 3
∴ L.H.S. = R.H.S.
∴ x = 7 is the solution of the given equation.

Substituting x = – 7 in L.H.S. of equation (i),
L.H.S. = (- 7) – 4
= -11
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 7 is not the solution of the given equation.

ii. 9m = 81 …(i)
Substituting m = 3 in L.H.S. of equation (i),
L.H.S. = 9 × (3)
= 27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i),
L.H.S. = 9 × (9)
= 81
R.H.S. = 81
∴L.H.S. = R.H.S.
∴m = 9 is the solution of the given equation.

Substituting m = – 3 in L.H.S. of equation (i),
L.H.S. = 9 × (- 3)
= -27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = – 3 is not the solution of the given equation.

iii. 2a + 4 = 0 …..(i)
Substituting a = 2 in L.H.S. of equation (i),
L.H.S. = 2 (2) + 4
= 4 + 4
= 8
R.H.S. = 0
∴L.H.S. ≠ R.H.S.
∴a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i),
L.H.S. = 2 (-2)+ 4
= -4 + 4
= 0
R.H.S. = 0
∴L.H.S. = R.H.S.
∴a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),
L.H.S. = 2(1)+ 4
= 2 + 4
= 6
R.H.S. = 0
∴ L.H.S. ≠ R.H.S.
∴a = 1 is not the solution of the given equation.

iv. 3 – y = 4 …(i)
Substituting y = -1 in L.H.S. of equation (i),
L.H.S. = 3 – (- 1)
= 3 + 1
= 4
R.H.S. = 4
∴L.H.S. = R.H.S.
∴y = – 1 is the solution of the given equation.

Substituting y = 1 in L.H.S. of equation (i),
L.H.S. = 3-(1)
= 2
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 1 is not the solution of the given equation.

Substituting y = 2 in L.H.S. of equation (i),
L.H.S. = 3-(2)
= 1
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 2 is not the solution of the given equation.

Practice Set 12.1 Question 2.
Solve the following equations:
i. 17p – 2 = 49
ii. 2m + 7 = 9
iii. 3x + 12 = 2x – 4
iv. 5 (x – 3) = 3 (x + 2)
v. \(\frac { 9x }{ 8 }+1=10\)
vi. \(\frac{y}{7}+\frac{y-4}{3}=2\)
vii. 13x – 5 = \(\frac { 3 }{ 2 }\)
viii. 3 (y + 8) = 10 (y – 4) + 8
ix. \(\frac{x-9}{x-5}=\frac{5}{7}\)
x. \(\frac{y-4}{3}+3 y=4\)
xi. \(\frac{b+(b+1)+(b+2)}{4}=21\)
Solution:
i. 17p – 2 = 49
∴ 17p – 2 + 2 = 49 + 2
…[Adding 2 on both the sides]
∴ 17p = 51
∴ \(\frac{17 p}{17}=\frac{51}{17}\) …[Dividing both the sides by 17]
p = 3

ii. 2m + 7 = 9
∴ 2m + 7 – 7 = 9 – 7
…[Subtracting 7 from both the sides]
∴ 2m = 2
∴ \(\frac{2 m}{2}=\frac{2}{2}\) [Dividing both the sides by 2]
∴ m = 1

iii. 3x + 12 = 2x – 4
∴ 3x + 12 – 12 = 2x – 4 – 12
…[Subtracting 12 from both the sides]
∴ 3x = 2x – 16
∴ 3x – 2x = 2x – 16 – 2x
…[Subtracting 2x from both the sides]
∴ x = – 16

iv. 5 (x – 3) = 3 (x + 2)
∴ 5x – 15 = 3x + 6
∴ 5x – 15 + 15 = 3x + 6 + 15
…[Adding 15 on both the sides]
∴ 5x = 3x + 21
∴ 5x – 3x = 3x + 21 – 3x
…[Subtracting 3x from both the sides]
∴ 2x = 21
∴ \(\frac{2 x}{2}=\frac{21}{2}\) …[Dividing both the sides by 2]
∴ \(x=\frac{21}{2}\)

v. \(\frac { 9x }{ 8 }+1=10\)

vi. \(\frac{y}{7}+\frac{y-4}{3}=2\)

vii. 13x – 5 = \(\frac { 3 }{ 2 }\)

viii. 3 (y + 8) = 10 (y – 4) + 8
∴ 3y + 24 = 10y – 40 + 8
∴ 3y + 24 = 10y – 32
∴ 3y + 24 – 24 = 10y – 32 – 24
…[Subtracting 24 from both the sides]
∴ 3y = 10y – 56
∴ 3y – 10y = 10y – 56
…[Subtracting 10y from both the sides]
∴ – 7y = – 56
∴ \(\frac{-7 y}{-7}=\frac{-56}{-7}\)…[Dividing both the sides by – 7]
∴ y = 8

ix. \(\frac{x-9}{x-5}=\frac{5}{7}\)
∴\(\frac{x-9}{x-5} \times 7(x-5)=\frac{5}{7} \times 7(x-5)\)
…[Multiplying both the sides by 7 (x – 5)]
∴7 (x – 9) = 5 (x – 5)
∴7x – 63 = 5x – 25
∴7x – 63 + 63 = 5x – 25 + 63
…[Adding 63 on both the sides]
∴7x = 5x + 38
∴7x – 5x = 5x + 38 – 5x
…[Subtracting 5x from both the sides]
∴ 2x = 38
∴\(\frac{2 x}{2}=\frac{38}{2}\) …[Dividing both the sides by 2]
∴x = 19

x. \(\frac{y-4}{3}+3 y=4\)
∴\(\frac{y-4}{3} \times 3+3 y \times 3=4 \times 3\)
…[Multiplying both the sides by 3]
∴y – 4 + 9y = 12
∴10y – 4 = 12
∴10y – 4 + 4=12 + 4
…[Adding 4 on both the sides]
∴10y = 16
∴\(\frac{10 y}{10}=\frac{16}{10}\)…[Dividing both the sides by 10]
∴y = \(\frac { 8 }{ 5 }\)

xi. \(\frac{b+(b+1)+(b+2)}{4}=21\)
∴\(\frac{b+(b+1)+(b+2)}{4} \times 4=21 \times 4\)
…[Multiplying both the sides by 4]
∴b + b + 1 + b + 2 = 84
∴3b + 3 = 84
∴3b + 3 – 3 = 84 – 3
…[ Subtracting 3 from both the sides]
∴3b = 81
∴\(\frac{3 b}{3}=\frac{81}{3}[/latex …[Dividing both the sides by 3]
∴b = 27

Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.1 Intext Questions and Activities

Std 8 Maths Practice Set 12.1 Question 1.
Fill in the boxes to solve the following equations. (Textbook pg. no. 75)
i. x + 4 = 9
∴x + 4 – __ = 9 – __
… [Subtracting 4 from both the sides]
∴ x = __

ii. x – 2 = 7
∴x – 2 + __ = 7 + __
… [Adding 2 on both the sides]
∴x = __

iii. [latex]\frac { x }{ 3 }=4\)
∴\(\frac { x }{ 3 }\) × __ = 4 ×__
∴x = __

iv. 4x = 24
∴ __ = __
∴x = __
Solution:
i. x + 4 = 9
∴x + 4 – 4 = 9 – 4
… [Subtracting 4 from both the sides]
∴ x = 5

ii. x – 2 = 7
∴x – 2 + 2 = 7 + 2
… [Adding 2 on both the sides]
∴x = 9

iii. \(\frac { x }{ 3 }=4\)
∴\(\frac { x }{ 3 }\) × 3 = 4 × 3
… [Multiplying both the sides by 3]
∴x = 12

iv. 4x = 24
∴ \(\frac{4 x}{[4]}=\frac{24}{[4]}\)
… [Dividing both the sides by 4]
∴x = 6

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 2 Parallel Lines.

9th Standard Maths 2 Practice Set 2.1 Chapter 2 Parallel Lines Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 2.1 Chapter 2 Parallel Lines Questions With Answers Maharashtra Board

Question :
In the given figure, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.

i. ∠RHD
ii. ∠PHG
iii. ∠HGS
iv. ∠MGK
Solution:
i. ∠DHP = 85° …..(i)
∠DHP + ∠RHD = 180° [Angles in a linear pair]
85° + ∠RHD = 180°
∴ ∠RHD = 180°- 85°
∴ ∠RHD = 95° …..(ii)

ii. ∠PHG = ∠RHD [Vertically opposite angles]
∴ ∠PHG = 95° [From (ii)]

iii. line RP || line MS and line DK is their transversal. [Corresponding angles]
∴ ∠HGS = ∠DHP …..(iii) [From (i)]

iv. ∠HGS = 85° [Vertically opposite angles]
∴ ∠MGK = ∠HGS ∠MGK = 85° [From (iii)]

Question 2.
In the given figure line p line q and line l and line m are tranversals.
Measures of some angles are shown. Hence find the measures of ∠a, ∠b, ∠c, ∠d.

Solution:

i. 110 + ∠a = 180° [Angles in a linear pair]
∴ ∠a = 180° – 110°
∴ ∠a = 70°

ii. consider ∠e as shown in the figure line p || line q, and line lis their transversal.
∠e + 110° = 180° [Interior angles]
∴ ∠e = 180° – 110°
∴ ∠e = 70°
But, ∠b = ∠e [Vertically opposite angles]
∴ ∠b = 70°

iii. line p || line q, and line m is their transversal.
∴ ∠c = 115° [Corresponding angles]

iv. 115° + ∠d = 180° [Angles in a linear pair]
∴ ∠d = 180° – 115°
∴ ∠d = 65°

Question 3.
In the given figure, line 11| line m and line n || line p. Find ∠a, ∠b, ∠c from the given measure of an angle.

Solution:

i. consider ∠d as shown in the figure
line l || line m, and line p is their transversal.
∴ ∠d = 45° [Corresponding angles]
Now, ∠d + ∠b = 180° [Angles in a linear pair]
∴ 45° +∠b = 180°
∴ ∠b = 180° – 45°
∴ ∠b = 135° …..(i)

ii. ∠a = ∠b [Vertically opposite angles]
∴ ∠a = 135° [From (i)]

iii. line n || line p, and line m is their transversal.
∴ ∠c = ∠b [Corresponding angles]
∴ ∠c = 135° [From (i)]

Question 4.
In the given figure, sides of ∠PQR and ∠XYZ are parallel to each other. Prove that, ∠PQR ≅ ∠XYZ.

Given: Ray YZ || ray QRandray YX || ray QP
To prove: ∠PQR ≅ ∠XYZ
Construction: Extend ray YZ in the opposite direction. It intersects ray QP at point S.
Solution:
Proof:
Ray YX || ray QP [Given]
Ray YX || ray SP and seg SY is their transversal [P-S-Q]
∴ ∠XYZ ≅ ∠PSY ……(i) [Corresponding angles]
ray YZ || ray QR [Given]
ray SZ || ray QR and seg PQ is their transversal. [S-Y-Z]
∴ ∠PSY ≅ ∠SQR [Corresponding angles]
∴ ∠PSY ≅ ∠PQR …….. (ii) [P-S-Q]
∴ ∠PQR ≅ ∠XYZ [From (i) and (ii)]

Question 5.
In the given figure, line AB || line CD and line PQ is transversal. Measure of one of the angles is given. Hence find the measures of the following angles.

i. ∠ART
ii. ∠CTQ
iii. ∠DTQ
iv. ∠PRB
Solution:
i. ∠BRT = 105° ….(i)
∠ART + ∠BRT = 180° [Angles in a linear pair]
∴ ∠ART + 105° = 180°
∴ ∠ART = 180° – 105°
∴ ∠ART = 75° …(ii)

ii. line AB || line CD and line PQ is their transversal.
∴ ∠CTQ = ∠ART [Corresponding angles]
∴ ∠CTQ = 75° [From (ii)]

iii. line AB || line CD and line PQ is their transversal.
∴ ∠DTQ = ∠BRT [Corresponding angles]
∴ ∠DTQ = 105° [From (i)]

iv. ∠PRB = ∠ART [Vertically opposite angles]
∴ ∠PRB = 75° [From (ii)]

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.1 Intext Questions and Activities

Question 1.
Angles formed by two lines and their transversal. (Textbook pg, no. 13)
When a transversal (line n) intersects two lines (line l and m) in two distinct points, 8 angles are formed as shown in the figure. Pairs of angles formed out of these angles are as follows:

Pairs of corresponding angles
i. ∠d, ∠h
ii. ∠a, ∠e
iii. ∠c, ∠g
iv. ∠b, ∠f

Pairs of alternate interior angles
i. ∠c, ∠e
ii. ∠b, ∠h

Pairs of alternate exterior angles
i. ∠d, ∠f
ii. ∠a, ∠g

Pairs of interior angles on the same side of the transversal
i. ∠c, ∠h
ii. ∠b, ∠e

Some important properties:
1. When two lines intersect, the pairs of vertically opposite angles formed are congruent.
Example:
In the given diagram,
line l and m intersect at point P.
The pairs of vertically opposite angles that are congruent are:
i. ∠a ≅ ∠b
ii. ∠c ≅ ∠d

2. The angles in a linear pair are supplementary.
Example:
For the given diagram,
∠a and ∠c are in linear pair
∴ ∠a + ∠c = 180°
Also, ∠d and ∠b are in linear pair
∴ ∠d + ∠b = 180°

3. When one pair of corresponding angles is congruent, then all the remaining pairs of corresponding angles are congruent.
Example:
In the given diagram,
If ∠a ≅ ∠b
then ∠e ≅ ∠f, ∠c ≅ ∠d and ∠g ≅ ∠h

4. When one pair of alternate angles is congruent, then all the remaining pairs of alternate angles are congruent.
Example:
For the given diagram,
If ∠e ≅ ∠d, then ∠g ≅ ∠b
Also, ∠a ≅ ∠h and ∠c ≅ ∠f

5. When one pair of interior angles on one side of the transversal is supplementary, then the other pair of interior angles is also supplementary.
Example:
For the given diagram,
If ∠e + ∠b = 180°, then ∠g + ∠d = 180°.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Problem Set 2 Class 9 Answers