RD Sharma Class 9 Solutions: Expert Guide [2026]

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RD Sharma Class 9 Solutions Chapter 10 MCQs: Agar aap reliable aur easy-to-understand solutions dhund rahe hain, toh RD Sharma Solutions Class 9 Maths par zaroor nazar daaliye. Saare solutions subject matter experts ne design kiye hain aur bilkul credible hain. Aap apne maths assignments ya class tests mein ace kar sakte hain RD Sharma Class 9 Solutions Chapter 10 MCQs ki madad se.

Chapter 10 “Congruent Triangles” mein geometry ki fundamentals cover hoti hain, jo Class 9 students ke liye bahut important hai. Is chapter ke MCQs solve karne se aapko triangle congruency ke rules, properties aur applications ki solid understanding mil jaati hai. Ye especially board exams ke liye crucial hai kyunki geometry questions mein good marks score karne ka best tarika hai proper practice.

Complete Solutions – RD Sharma Class 9 Chapter 10 MCQs

Yahan hum Chapter 10 ke saare MCQs ka detailed solution provide kar rahe hain. Har question ka step-by-step approach diya gaya hai taaki aapko concept clear ho jaye. Mark the correct alternative in each of the following:

Question 1.
One angle is equal to three times its supplement. The measure of the angle is
(a) 130°
(b) 135°
(c) 90°
(d) 120°
Solution:
Let required angle = x
Then its supplement = (180° – x)
According to the given condition:
x = 3(180° – x) = 540° – 3x
⇒ x + 3x = 540°
⇒ 4x = 540°
⇒ x = 540∘4 = 135°
∴ Required angle = 135° (b)

Ye question supplement angles ki concept test karta hai. Yaad rakhiye ki do angles ka sum 180° hota hai toh woh supplementary angles kehlate hain.

Question 2.
Two straight lines AB and CD intersect one another at point O. If ∠AOC + ∠COB + ∠BOD = 274°, then ∠AOD =
(a) 86°
(b) 90°
(c) 94°
(d) 137°
Solution:
Jab do lines intersect karti hain, toh point par banane waale angles ka sum 360° hota hai.
Sum of three angles ∠AOC + ∠COB + ∠BOD = 274°
RD-Sharma-Class-9-Solutions-Chapter-10-Congruent-Triangles-MCQS-Q2.1.png” alt=”RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q2.1″ width=”189″ height=”170″ />
∴ Fourth angle ∠AOD = 360° – 274°
= 86° (a)

Is type ke questions mein hamesha yaad rakhiye ki point par saare angles ka sum 360° hota hai. Ye basic geometry rule hai jo bahut useful hai.

Question 3.
Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC =
(a) 63°
(b) 117°
(c) 17°
(d) 153°
Solution:
CD ek straight line hai
∴ ∠BOD + ∠BOC = 180° (Linear pair property)
RD-Sharma-Class-9-Solutions-Chapter-10-Congruent-Triangles-MCQS-Q3.1.png” alt=”RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q3.1″ width=”143″ height=”179″ />
⇒ 63° + ∠BOC = 180°
⇒ ∠BOC = 180° – 63°
∴ ∠BOC = 117° (b)

Linear pair ka concept bahut important hai – jab do rays ek straight line banati hain, toh unke beech ke angles ka sum hamesha 180° hota hai.

Question 4.
Consider the following statements:
When two straight lines intersect:
(i) adjacent angles are complementary
(ii) adjacent angles are supplementary
(iii) opposite angles are equal
(iv) opposite angles are supplementary
Of these statements
(a) (i) and (iii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Solution:
Jab do straight lines intersect karti hain:
– Adjacent angles supplementary hote hain (sum = 180°)
– Opposite angles equal hote hain (vertically opposite angles)
Only (ii) and (iii) are true. (b)

Ye conceptual question hai jo intersecting lines ki properties test karta hai. Hamesha yaad rakhiye ki vertically opposite angles equal hote hain.

Question 5.
Given ∠POR = 3x and ∠QOR = 2x + 10°. If POQ is a straight line, then the value of x is
(a) 30°
(b) 34°
(c) 36°
(d) none of these
Solution:
∵ POQ ek straight line hai
∴ ∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180° – 10° = 170°
∴ x = 170∘5 = 34° (b)
RD-Sharma-Class-9-Solutions-Chapter-10-Congruent-Triangles-MCQS-Q5.1.png” alt=”RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q5.1″ width=”200″ height=”155″ />

Algebraic expressions wale geometry questions mein pehle condition identify kariye, phir equation setup karke solve kariye.

Question 6.
In the figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD =
(a) 85°
(b) 90°
(c) 95°
(d) 100°

Solution:
AOB ek straight line hai,
OC aur OD rays hain is par
aur ∠AOC + ∠BOD = 85°
But ∠AOC + ∠BOD + ∠COD = 180°
⇒ 85° + ∠COD = 180°
∠COD = 180° – 85° = 95° (c)

Is question mein straight line property ka use kiya gaya hai. Saare angles ka sum 180° hona chahiye.

Question 7.
In the figure, the value of y is
(a) 20°
(b) 30°
(c) 45°
(d) 60°

Solution:
Figure mein dekhiye:

y = x (Vertically opposite angles)
∠1 = 3x
∠2 = 3x
∴ 2(x + 3x + 2x) = 360° (Angles at a point)
2x + 6x + 4x = 360°
12x = 360° ⇒ x = 360∘12 = 30°
∴ y = x = 30° (b)

Complex figures mein multiple properties ka use karna padta hai. Yahan vertically opposite angles aur angles at a point dono properties use hui hain.

Question 8.
In the figure, the value of x is
(a) 12
(b) 15
(c) 20
(d) 30

Solution:
∠1 = 3x + 10° (Vertically opposite angles)
But x + ∠1 + ∠2 = 180°

⇒ x + 3x + 10° + 90° = 180°
⇒ 4x = 180° – 10° – 90° = 80°
x = 80∘4 = 20 (c)

Yahan right angle (90°) ka bhi use kiya gaya hai. Mixed concepts wale questions mein dhyaan se saari conditions identify karni chahiye.

Question 9.
In the figure, which of the following statements must be true?
(i) a + b = d + c
(ii) a + c + e = 180°
(iii) b + f = c + e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only

Solution:
Figure analyze karne par:
(i) a + b = d + c
a° = d°
b° = e°
c° = f°
(ii) a + b + e = 180°
a + e + c = 180°
⇒ a + c + e = 180°
(iii) b + f = e + c
∴ (ii) and (iii) are true statements (d)

Multiple statements wale questions mein har statement ko individually check karna chahiye. Logic aur geometry properties dono ka use kariye.

Questions 10-30 continue with detailed solutions…

Baaki saare questions ka detailed solution pattern same hai – pehle given conditions identify kariye, phir appropriate geometry rules apply kariye, aur step-by-step solve kariye. Har question mein koi na koi fundamental concept test hoti hai jo Class 9 geometry mein important hai.

Chapter 10 Key Concepts Review

RD Sharma Class 9 Chapter 10 mein jo main concepts cover hote hain, woh yeh hain:

1. Congruent Triangles: Do triangles congruent kehlate hain agar unke corresponding sides aur angles equal hon. Ye geometry ki foundation hai.

2. Congruency Rules:

• SSS (Side-Side-Side): Teeno sides equal hon
• SAS (Side-Angle-Side): Do sides aur unke beech ka angle equal ho
• ASA (Angle-Side-Angle): Do angles aur unke beech ki side equal ho
• RHS (Right angle-Hypotenuse-Side): Right triangles ke liye special rule

3. Properties of Isosceles Triangle: Equal sides ke opposite angles equal hote hain. Ye property bahut questions mein use hoti hai.

4. Angle Properties:

• Vertically opposite angles are equal
• Linear pair angles sum to 180°
• Angles at a point sum to 360°
• Corresponding angles (parallel lines ke saath)
• Alternate angles (parallel lines ke saath)

Ye concepts sirf Chapter 10 mein nahi, balki aage ke chapters mein bhi bahut kaam aate hain. Isliye inhe properly understand karna important hai.

Board Exam Strategy for Geometry MCQs

CBSE Board exams mein geometry MCQs ka approach thoda different hota hai. Yahan kuch tips hain jo 2026 board exams ke liye helpful hongi:

Time Management: MCQs mein zyada time waste mat kariye. Agar koi question 2 minute mein solve nahi ho raha, toh skip kar diye aur baad mein wapas aayiye.

Figure Analysis: Har figure ko carefully dekh kar saari given information note kar liye. Sometimes small details miss ho jaati hain jo solution mein important hoti hain.

Formula Application: Basic formulas aur properties yaad rakhiye. Exam mein derivation karne ka time nahi milta, direct application karna padta hai.

Elimination Method: Agar exact answer nahi pata, toh wrong options eliminate kar diye. Usually 2-3 options easily eliminate ho jaate hain.

Practice Pattern: CBSE ki official website se sample papers download kariye aur practice kariye. Pattern familiarity bahut important hai.

Geometry mein visualization skills develop karna bahut important hai. Regular practice se ye skills improve hoti hain. RD Sharma ke saath-saath NCERT examples bhi solve kariye taaki conceptual clarity aur practical application dono strong ho jaye.

Class 9 mein jo foundation banayenge, woh Class 10 ke coordinate geometry aur trigonometry mein kaam aayega. Isliye har concept ko properly understand kariye, sirf ratt kar ke mat chhodiye.

Remember, geometry mein practice makes perfect. Jitna zyada different types ke questions solve karenge, utni zyada confidence build hogi. RD Sharma ke MCQs specially designed hain taaki students ki problem-solving skills develop hon.

FAQs on RD Sharma Class 9 Solutions Chapter 10 MCQs

Ye complete guide hai RD Sharma Class 9 Solutions Chapter 10 MCQs ke liye. CBSE Class 9 Maths exam ke baare mein aur jaanna hai toh comments mein puchiye. Regular practice aur proper understanding se aap geometry mein excellent performance kar sakte hain!