RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

In this chapter, we provide RD Sharma Solutions for Chapter 10 Congruent Triangles Ex 10.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 10 Congruent Triangles Ex 10.1 Maths pdf, free RD Sharma Solutions for Chapter 10 Congruent Triangles Ex 10.1 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 9 Chapter 10 Congruent Triangles Ex 10.1 Download PDF

Question 1.
Write the complement of each of the following angles:
(i) 20°
(ii) 35°
(iii) 90°
(iv) 77°
(v) 30°
Solution:
We know that two angles are complement to each other if their sum is 90°. Therefore,
(i) Complement of 20° is (90° – 20°) = 70°
(ii) Complement of 35° is (90° – 35°) = 55°
(iii) Complement of 90° is (90° – 90°) = 0°
(iv) Complement of 77° is (90° – 77°) = 13°
(v) Complement of 30° is (90° – 30°) = 60°

Question 2.
Write the supplement of each of the following angles:
(i) 54°
(ii) 132°
(iii) 138°
Solution:
We know that two angles are supplement to each other if their sum if 180°. Therefore,
(i) Supplement of 54° is (180° – 54°) = 126°
(ii) Supplement of 132° is (180° – 132°) = 48°
(iii) Supplement of 138° is (180° – 138°) = 42°

Question 3.
If an angle is 28° less than its complement, find its measure.
Solution:
Let required angle = x, then
Its complement = x + 28°
∴  x + x + 28° = 90° ⇒  2x = 90° – 28° = 62°
∴ x = 62∘2  = 31°
∴ Required angle = 31°

Question 4.
If an angle is 30° more than one half of its complement, find the measure of the angle.
Solution:
Let the measure of the required angle = x
∴  Its complement =  90° – x
∴  x = 12 (90° – x) + 30°
2x = 90° – x + 60°
⇒ 2x + x = 90° + 60°
⇒  3x = 150°
⇒ x =  150∘3  = 50°
∴ Required angle = 50°

Question 5.
Two supplementary angles are in the ratio 4 : 5. Find the angles.
Solution:
Ratio in two supplementary angles = 4 : 5
Let first angle = 4x
Then second angle = 5x
∴  4x + 5x = 180
⇒  9x = 180°
∴ x  = 180∘9 = 20°
∴  First angle = 4x = 4 x 20° = 80°
and second angle = 5x
= 5 x 20° = 100°

Question 6.
Two supplementary angles differ by 48°. Find the angles.
Solution:
Let first angle = x                        ”
Then second angle = x + 48°
∴  x + x + 48° = 180°⇒  2x + 48° = 180°
⇒  2x = 180° – 48° = 132°
x= 132∘2 =66°
∴  First angle = 66°
and second angle = x + 48° = 66° + 48° = 114°
∴ Angles are 66°, 114°

Question 7.
An angle is equal to 8 times its complement. Determine its measure.
Solution:
Let the required angle = x
Then its complement angle = 90° – x
∴ x = 8(90° – x)
⇒ x = 720° – 8x ⇒  x + 8x = 720°
⇒ 9x = 720° ⇒ x =  720∘9 = 80°
∴  Required angle = 80°

Question 8.
If the angles (2x – 10)° and (x – 5)° are complementary angles, find x.
Solution:
First complementary angle = (2x – 10°) and second = (x – 5)°
∴ 2x – 10° + x – 5° = 90°
⇒ 3x – 15° = 90° ⇒  3x = 90° + 15° = 105°
∴ x = 105∘3 = 35°
∴  First angle = 2x – 10° = 2 x 35° – 10°
= 70° – 10° = 60°
and second angle = x – 5 = 35° – 5 = 30°

Question 9.
If an angle differ from its complement by 10°, find the angle.
Solution:
Let required angle = x°
Then its complement angle = 90° – x°
∴ x – (90° – x) = 10
⇒  x – 90° + x = 10°⇒  2x = 10° + 90° = 100° 100°
⇒ x =  100∘2 = 50°
∴ Required angle = 50°

Question 10.
If the supplement of an angle is two-third of itself Determine the angle and its supplement.
Solution:
Let required angle = x
Then its supplement angle = 180° – x
∴  (180°-x)= 23x
540° – 3x = 2x ⇒ 2x + 3x = 540°
⇒ 5x = 540°⇒  x = 540∘5 = 108°
-. Supplement angle = 180° – 108° = 72°

Question 11.
An angle is 14° more than its complementary angle. What is its measure?
Solution:
Let required angle = x
Then its complementary angle = 90° – x
∴  x + 14° = 90° – x
x + x = 90° – 14° ⇒  2x = 76°
⇒ x =  76∘2 = 38°
∴  Required angle = 38°

Question 12.
The measure of an angle is twice the measure of its supplementary angle. Find its measure.
Solution:
Let the required angle = x
∴  Its supplementary angle = 180° – x
∴  x = 2(180°-x) = 360°-2x
⇒  x + 2x = 360°
⇒ 3x = 360°
⇒  x = 360∘3 = 120°
∴  Required angle = 120°

Question 13.
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Solution:
Let required angle = x
Then its complement angle = 90° – x
and supplement angle = 180° – x
∴  3(90° – x) = 180° – x
⇒ 270° – 3x = 180° – x
⇒270° – 180° = -x + 3x => 2x = 90°
⇒ x = 45°
∴  Required angle = 45°

Question 14.
If the supplement of an angle is three times its complement, find the angle.
Solution:
Let required angle = x
Then its complement = 90°-  x
and supplement = 180° – x
∴  180°-x = 3(90°-x)
⇒  180° – x = 270° – 3x
⇒  -x + 3x = 270° – 180°
⇒ 2x = 90° ⇒ x = 90∘2 =45°
∴ Required angle = 45°

All Chapter RD Sharma Solutions For Class 9 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 9

All Subject NCERT Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.