Geometry Class 9th Mathematics Term 1 Tamilnadu Board Solution

Class 9th Mathematics Term 1 Tamilnadu Board Solution
Exercise 4.1
  1. Find the complement of each of the following angles. i. 63° ii. 24° iii. 48° iv.…
  2. Find the supplement of each of the following angles. i. 58° ii. 148° iii. 120°…
  3. Find the value of x in the following figures. i. ii.
  4. The angle which is two times its complement. Find the angles in each of the…
  5. The angle which is four times its supplement. Find the angles in each of the…
  6. The angles whose supplement is four times its complement. Find the angles in…
  7. The angle whose complement is one sixth of its supplement. Find the angles in…
  8. Supplementary angles are in the ratio 4:5. Find the angles in each of the…
  9. Two complementary angles are in the ratio 3:2. Find the angles in each of the…
  10. Find the values of x, y in the following figures.
  11. Find the values of x, y in the following figures.
  12. Find the values of x, y in the following figures.
  13. Let l1 || l2 and m1 is a transversal. If ∠F = 65°, find the measure of each of…
  14. For what value of x will l1 and l2 be parallel lines. i. ii.
  15. The angles of a triangle are in the ratio of 1:2:3. Find the measure of each…
  16. In ΔABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the…
  17. In the given figure at right, side BC of ΔABC is produced to D. Find ∠A and ∠C.…
Exercise 4.3
  1. If an angle is equal to one third of its supplement, its measure is equal toA.…
  2. In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to…
  3. The complement of an angle exceeds the angle by 60°. Then the angle is equal…
  4. Find the measure of an angle, if six times of its complement is 12° less than…
  5. In the given figure, ∠ B:∠ C = 2:3, find ∠ B A. 120° B. 52° C. 78° D. 130°…
  6. ABCD is a parallelogram, E is the mid - point of AB and CE bisects ∠ BCD. Then ∠…

Exercise 4.1
Question 1.

Find the complement of each of the following angles.

i. 63° ii. 24°

iii. 48° iv. 35°

v. 20°


Answer:

i. We know that two angles are said to be complementary if their sum is 90°.


So, the complement of 63° = 90° - 63° = 27°


ii. We know that two angles are said to be complementary if their sum is 90°.


So, the complement of 24° = 90° - 24° = 66°


iii. We know that two angles are said to be complementary if their sum is 90°.


So, the complement of 48° = 90° - 48° = 42°


iv. We know that two angles are said to be complementary if their sum is 90°.


So, the complement of 35° = 90° - 35° = 55°


v. We know that two angles are said to be complementary if their sum is 90°.


So, the complement of 20° = 90° - 20° = 70°



Question 2.

Find the supplement of each of the following angles.

i. 58° ii. 148°

iii. 120° iv. 40°

v. 100°


Answer:

i. We know that two angles are said to be supplementary if their sum is 180°.


So, the supplement of 58° = 180° - 58° = 122°


ii. We know that two angles are said to be supplementary if their sum is 180°.


So, the supplement of 148° = 180° - 148° = 32°


iii. We know that two angles are said to be supplementary if their sum is 180°.


So, the supplement of 120° = 180° - 120° = 60°


iv. We know that two angles are said to be supplementary if their sum is 180°.


So, the supplement of 40° = 180° - 40° = 140°


v. We know that two angles are said to be supplementary if their sum is 180°.


So, the supplement of 100° = 180° - 100° = 80°



Question 3.

Find the value of x in the following figures.

i. 

ii. 


Answer:

i. In the given figure,


AB is a straight line, hence all the angles formed on it are supplementary.


⇒ (x – 20)° + x° + 40° = 180°


⇒ 2x + 20° = 180°


⇒ 2x = 180° - 20°


⇒ 2x = 160°


⇒ x = 80°


ii. In the given figure,


AB is a straight line, hence all the angles formed on it are supplementary.


⇒ (x + 30)° + (115 – x)° + x° = 180°


⇒ x + 145° = 180°


⇒ x = 180° - 145°


⇒ x = 35°



Question 4.

Find the angles in each of the following.

The angle which is two times its complement.


Answer:

Let the complement of an angle be x°


According to the question,


Angle = 2x°


We know that two angles are said to be complementary if their sum is 90°.


So, 2x° + x° = 90°


⇒ 3x° = 90°


⇒ x° = 30°


∴ Angle = 2x° = 2× 30° = 60°, Compliment = x° = 30°



Question 5.

Find the angles in each of the following.

The angle which is four times its supplement.


Answer:

Let the supplement of an angle be x°


According to the question,


Angle = 4x°


We know that two angles are said to be supplementary if their sum is 180°.


So, 4x° + x° = 180°


⇒ 5x° = 180°


⇒ x° = 36°


∴ Angle = 4x° = 4× 36° = 144°, Compliment = x° = 36°



Question 6.

Find the angles in each of the following.

The angles whose supplement is four times its complement.


Answer:

Let the compliment of an angle be x°


According to the question,


Supplement of the angle = 4x°


We know that two angles are said to be supple mentary if their sum is 180°.


So, required angle = 180° - 4x° ..(I)


Also, if the two angles are complimentary their sum is 90°.


⇒ required angle = 90° - x° …(II)


Equating I and II,


90° - x° = 180° - 4x°


⇒ 3x° = 90°


⇒ x° = 30°


Hence, required angle = 90° - x° = 90° - 30° = 60°



Question 7.

Find the angles in each of the following.

The angle whose complement is one sixth of its supplement.


Answer:

Let the supplement of an angle be x°


According to the question,


Compliment of the angle 


We know that two angles are said to be supple-mentary if their sum is 180°.


So,  ..(I)


Also, if the two angles are complimentary their sum is 90°.


 …(II)


Equating I and II,





⇒ x° = 108°


Hence, required angle = 180° - x° = 180° - 108° = 72°



Question 8.

Find the angles in each of the following.

Supplementary angles are in the ratio 4:5.


Answer:

Let the supplementary angles be 4x and 5x


We know that two angles are said to be supple mentary if their sum is 180°.


⇒ 4x + 5x = 180°


⇒ 9x = 180°


⇒ x = 20°


So, the supplementary angles will be 4× 20° = 80° and 5× 20° = 100°



Question 9.

Find the angles in each of the following.

Two complementary angles are in the ratio 3:2.


Answer:

Let the complimentary angles be 3x and 2x


We know that two angles are said to be complimentary if their sum is 90°.


⇒ 2x + 3x = 90°


⇒ 5x = 90°


⇒ x = 18°


So, the complimentary angles will be 2×18° = 36° and 3×18° = 54°



Question 10.

Find the values of x, y in the following figures.



Answer:

Given: ∠ADC = 3x and ∠BDC = 2x


Here AB is the straight line.


∴ ∠ADC and ∠BDC are linear pair.


⇒ ∠ADC + ∠BDC = 180°


⇒ 3x + 2x = 180°


⇒ 5x = 180°


⇒ x = 36°



Question 11.

Find the values of x, y in the following figures.



Answer:

Given: ∠ADC = (3x + 5)° and ∠BDC = (2x – 25)°


Here AB is the straight line.


∴ ∠ADC and ∠BDC are linear pair.


⇒ ∠ADC + ∠BDC = 180°


⇒ 3x + 5 + 2x - 25 = 180°


⇒ 5x - 20 = 180°


⇒ 5x = 200°


⇒ x = 40°



Question 12.

Find the values of x, y in the following figures.



Answer:

Given: ∠AOD = y° ,∠DOC = 90°,∠COB = x° ,∠AOE = 3x and ∠BOE = 60°


Here AB is the straight line.


∴ ∠AOE and ∠BDC are linear pair.


⇒ ∠ADC + ∠BOE = 180°


⇒ 3x + 60° = 180°


⇒ 3x = 120°


⇒ x = 40°


Also, ∠AOD, ∠DOC and ∠BOC are linear pair.


⇒ ∠AOD + ∠DOC + ∠BOC = 180°


⇒ y + 90° + x° = 180°


⇒ y = 90° - 40°


⇒ y = 50°



Question 13.

Let l1 || l2 and m1 is a transversal. If ∠F = 65°, find the measure of each of the remaining angles.



Answer:

Given: ∠F = 65°


∠B = ∠F = 65° {Corresponding angle}


∠H = ∠F = 65° {vertically opposite angle}


∠D = ∠B = 65° {vertically opposite angle}


Also, ∠C + ∠F = 180° {co-interior angles are supplementary}


⇒ ∠C = 180° - 65°


⇒ ∠C = 115°


∠E = ∠C = 115° {Alternate interior angle}


∠G = ∠C = 115° {Corresponding angle}


∠A = ∠G = 115° {Alternate exterior angle}



Question 14.

For what value of x will l1 and l2 be parallel lines.

i. 

ii. 


Answer:

(i) Given: l1||l2


∠1 = (2x + 20)° and ∠2 = (3x-10)°


∵ Corresponding angles are equal


∴ ∠1 = ∠2


⇒ (2x + 20)° = (3x-10)°


⇒ x = 30°


ii. Given: l1||l2


∠1 = (2x)° and ∠2 = (3x + 20)°


∵ Co-interior angles are supplementary


∴ ∠1 + ∠2 = 180°


⇒ (2x)° + (3x + 20)° = 180°


⇒ 5x = 160°


⇒ x = 32°



Question 15.

The angles of a triangle are in the ratio of 1:2:3. Find the measure of each angle of the triangle.


Answer:

Let the angles of a triangle be x, 2x and 3x.


We know that sum of the angles of a triangle is 180°.


⇒ x + 2x + 3x = 180°


⇒ 6x = 180°


⇒ x = 30°


So, the angles of the triangle are 30°, 2×30° = 60° and 3× 30° = 90°.



Question 16.

In ΔABC, ∠A + ∠B = 70° and ∠B + ∠C = 135°. Find the measure of each angle of the triangle.


Answer:

Given ∠A + ∠B = 70° and ∠B + ∠C = 135°


We know that sum of the angles of a triangle is 180°.


∠A + ∠B + ∠C = 180° …I


Also, ∠A + ∠B + ∠B + ∠C = 70° + 135°


⇒ (∠A + ∠B + ∠C) + ∠B = 205°


From I,


⇒ 180° + ∠B = 205°


⇒ ∠B = 205° - 180°


⇒ ∠B = 25°


Putting in given equations,


∠A + ∠B = 70° and ∠B + ∠C = 135°


⇒ ∠A + 25° = 70° and 25° + ∠C = 135°


⇒ ∠A = 45° and ∠C = 110°


So, the angles of the triangle are 45°, 25° and 110°.



Question 17.

In the given figure at right, side BC of ΔABC is produced to D. Find ∠A and ∠C.



Answer:

Given ∠B = 40° and ∠ACD = 120°


∵ ∠ACD is an external angle


∴ ∠ACD = ∠A + ∠B


⇒ ∠A + 40° = 120°


⇒ ∠A = 80°


We know that sum of the angles of a triangle is 180°.


∠A + ∠B + ∠C = 180°


⇒ 80° + + 40° + ∠C = 180°


⇒ ∠C = 180° - 120°


⇒ ∠C = 60°




Exercise 4.3
Question 1.

If an angle is equal to one third of its supplement, its measure is equal to
A. 40°

B. 50°

C. 45°

D. 55°


Answer:

Let the required angle be x and the supplement of x is y.


As we know if two angles are supplement of each other, then sum of those two angles is 180°.


⇒ x + y = 180


⇒ y = 180 – x


Now, according to question;


Required angle is equal to one third of its supplement;


i.e. 


⇒ y = 3x


Now putting the value of y in this equation,


⇒ 180 – x = 3x


⇒ 


⇒ 


⇒ 


Hence, required angle is 45°.


Hence, correct option is (c).


Question 2.

In the given figure, OP bisect ∠ BOC and OQ bisect ∠ AOC. Then ∠ POQ is equal to


A. 90°

B. 120°

C. 60°

D. 100°


Answer:

Suppose that


BOC = 2x and ∠ AOC = 2y


Now, by question; OP and OQ is the bisect or of ∠ BOC and ∠ AOC respectively.


⇒ 


Similarly,



Now,


∠ POQ = ∠ POC + ∠ COQ = x + y


So, we have to find ∠ POQ i.e.x + y Now, ∠ BOC + ∠ AOC = 180


⇒ 2x + 2y = 180


⇒ 2(x + y) = 180


⇒ x + y = 180/2


⇒ x + y = 90°


Hence, ∠ POQ = 90°.


Question 3.

The complement of an angle exceeds the angle by 60°. Then the angle is equal to
A. 25°

B. 30°

C. 15°

D. 35°


Answer:

Let the required angle be x and it’s complement is y.


So, 


⇒ y = 90 – x


Now, according to question;


Complement of the angle exceeds the angle by 60° i.e. y exceeds the x by 60°.


⇒ y – 60 = x


Now, putting the value of your in this equation,


⇒ (90 – x) – 60 = x


⇒ 30 – x = x


⇒ 2x = 30


⇒ x = 15°


Hence, required angle is 15°.


Question 4.

Find the measure of an angle, if six times of its complement is 12° less than twice of its supplement.
A. 48°

B. 96°

C. 24°

D. 58°


Answer:

Let the required angle be x and it’s complement is y and it’s supplement is z.


⇒ x + y = 90


i.e. y = 90 – x


And, x + z = 180


i.e. z = 180 – x


Now, according to question,


Six times of complement of x is 12°


less than twice of its supplement.


i.e. 6y = 2z – 12


Putting the value of y and z in this equation,


⇒ 6 (90 – x) = 2 (180 – x) – 12


⇒ 540 – 6x = 360 – 2x – 12


⇒ 6x – 2x = 540 – 360 + 12


⇒ 4x = 192


⇒ 


⇒ x = 48°


Hence required angle is 48°.


Question 5.

In the given figure, ∠ B:∠ C = 2:3, find ∠ B


A. 120°

B. 52°

C. 78°

D. 130°


Answer:

Let ∠ B and ∠ C be 2x and 3x respectively.


Now, As we know sum of two interior angles of a triangle is equal to third exterior angle;


⇒ ∠ B + ∠ C = 130°


⇒ 2x + 3x = 130°


⇒ 5x = 130°


⇒ x = 26°


So, ∠ B = 2x = 2×26° = 52°


Hence, required angle is 52°.


Question 6.

ABCD is a parallelogram, E is the mid – point of AB and CE bisects ∠ BCD. Then ∠ DEC is
A. 60°

B. 90°

C. 100°

D. 120°


Answer:

Given: ABCD is a parallelogram and E is the mid – point of AB such that CE bisects ∠ BCD.


construction: Join DE. The figure is given below:



Let ∠ BCD = 2x


⇒ ∠ BCE = ∠ ECD = ∠ BCD/2


Now, let ∠ ADC = 2y


And as we joined DE it will also bisect ∠ ADC.


Therefore, ∠ EDC = 


Now, as we know sum of adjacent angles of parallelogram is equal to 180°.


⇒ ∠ BCD + ∠ ADC = 180°


⇒ 2x + 2y = 180°


⇒ 2 (x + y) = 180°


⇒ x + y = 90°


Now, In triangle CED,


∠ CED + ∠ EDC + ∠ DCE = 180°


⇒ ∠ CED + x + y = 180°


⇒ ∠ CED + 90° = 180°


⇒ ∠ CED = 180° – 90°


⇒ ∠ CED = 90°


Hence required angle is 90°.