RD Sharma Class 9 Solutions: Complete Guide [2026]

Last Updated: May 02, 2026 | This article has been updated with the latest information for 2026.

RD Sharma Class 9 Solutions Chapter 15 MCQS

Read more: RD Sharma — Complete Guide

RD Sharma Class 9 Solutions Chapter 15 MCQs are essential resources for students preparing for their 2026 board examinations. These comprehensive solutions, crafted by our subject experts, specifically target Chapter 15 – Areas of Parallelograms and Triangles. Students can access detailed step-by-step solutions for all multiple-choice questions, ensuring thorough preparation for the academic year 2026. You can find complete RD Sharma Class 9 Solutions Chapter 15 MCQs directly below.

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Chapter 15 MCQs – Complete Solutions

Class 9 ke students ke liye RD Sharma Chapter 15 ke MCQs bohot important hain. Ye chapter “Areas of Parallelograms and Triangles” par based hai aur 2026 board exams mein frequently poocha jaata hai. Hamare experts ne har question ko step-by-step solve kiya hai taaki aapko complete understanding mil sake.

Question 1.
Agar ek circle ke chord ki length 16 cm hai aur wo centre se 15 cm ki distance par hai, toh circle ka radius kya hoga?
(a) 15 cm
(b) 16 cm
(c) 17 cm
(d) 34 cm
Solution:
Circle ke chord AB ki length = 16 cm
Centre se distance OL = 15 cm
Areas of Parallelograms and Triangles With Solutions PDF <a href=RD Sharma Class 9 Solutions” width=”181″ height=”162″ />
OA ko radius maante hain, toh right triangle OAL mein,
OA² = OL² + AL²
16
= (15)² + (16/2)² = 15² + 8²
= 225 + 64 = 289 = (17)²
∴ OA = 17 cm
Hence circle ka radius = 17 cm (c)

Question 2.
Ek circle ka radius 6 cm hai. Centre se chord tak perpendicular distance kya hogi agar chord ki length 8 cm hai?
<a href=RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles” width=”284″ height=”71″ />
Solution:
Circle ka radius (r) = 6 cm
Perpendicular distance from centre = ?
Chord ki length = 8 cm
RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles
AB chord hai, OL distance hai
Right triangle OAL mein
OA² = AL² + OL²
RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles

Question 3.
Agar O ek circle ka centre hai radius r ke saath aur AB ek chord hai jo centre se r/2 distance par hai, toh ∠BAO ka value kya hoga?
(a) 60°
(b) 45°
(c) 30°
(d) 15°
Solution:
r circle ka radius hai centre O ke saath
AB chord hai, centre se r/2 distance par
RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles

Question 4.
ABCD ek cyclic quadrilateral hai jisme ∠ADB = 30° aur ∠DCA = 80°, toh ∠DAB ka value kya hai?
(a) 70°
(b) 100°
(c) 125°
(d) 150°
Solution:
ABCD cyclic quadrilateral hai jisme ∠DCA = 80° aur ∠ADB = 30°
RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles
∵∠ADB = ∠ACB (Same segment ke angles)
∴ ∠ACB = 30°
∴ ∠BCD = 80° + 30° = 110°
∵ ABCD cyclic quadrilateral hai
∴∠BAD + ∠BCD = 180°
⇒ ∠BAD + 110°= 180°
⇒ ∠BAD = 180°- 110° = 70°
ya ∠DAB = 70° (a)

Question 5.
14 cm length ka chord centre se 6 cm distance par hai. Same circle mein dusra chord centre se 2 cm distance par hai. Dusre chord ki length kya hogi?
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 18 cm
Solution:
Circle mein AB chord = 14 cm
Centre se distance OL = 6 cm
r radius maante hain, toh OA² = AL² + OL²
⇒ r² = (7)² + (6)² = 49 + 36 = 85
RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles
Same circle mein dusre chord CD ki length = ?
Centre se distance = 2 cm
∴ r² = OM² + MD²
⇒ 85 = (2)² + DM²
⇒ 85 = 4 + DM²
⇒ DM² = 85-4 = 81 = (9)²
∴ DM = 9
∴ CD = 2 x DM = 2 x 9 = 18 cm
∴Dusre chord ki length = 18 cm (d)

Question 6.
Ek circle ka chord 10 cm hai. Is circle ka radius kya hona chahiye?
(a) 5 cm
(b) 5 cm se zyada
(c) 5 cm ke barabar ya zyada
(d) 5 cm se kam
Solution:
Circle ke chord ki length = 10 cm
Circle ke radius ki length chord ke half se zyada hoti hai
10/2 = 5 cm se zyada (b)

Question 7.
ABC triangle mein B right angle hai, AC = 5 cm aur AB = 4 cm. O centre se ek circle draw kiya gaya hai OC radius ke saath. C aur B se passing chord ki length kya hogi?
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Solution:
Right triangle ABC mein, ∠B = 90°
AC = 5 cm, AB = 4 cm
Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions
∴ BC² = AC² -AB²
= 5² – 4² = 25 – 16
= 9 = (3)²
∴ BC = 3 cm
∴ Chord BC ki length = 3 cm (a)

Question 8.
Agar AB, BC aur CD equal chords hain ek circle ke jiska centre O hai, aur AD diameter hai toh ∠AOB kya hoga?
(a) 60°
(b) 90°
(c) 120°
(d) none of these
Solution:
Circle mein chords AB = BC = CD
RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles
O circle ka centre hai
∴ ∠AOB = determine nahi kar sakte (d)

Question 9.
C arc AB ka mid-point hai ek circle ka jisme m(AB) = 183°. Agar region jo arc ACB aur line segment AB se bounded hai use S denote karte hain, toh centre O circle ka kaha lie karta hai?
(a) S ke interior mein
(b) S ke exterior mein
(c) segment AB par
(d) AB par aur AB ko bisect karta hai
Solution:
Arc AB = 183°
∴ AB circle ka diameter hai centre O ke saath aur C arc AB ka mid point hai
Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles
Line segment AB = S
∴ Centre AB par lie karega (c)

Question 10.
Ek circle mein, major arc minor arc se 3 times hai. Corresponding central angles aur degree measures kya honge?
(a) 90° aur 270°
(b) 90° aur 90°
(c) 270° aur 90°
(d) 60° aur 210°
Solution:
Circle mein, major arc = 3 times minor arc yani arc ACB = 3 arc ADB
∴ Reflex ∠AOB = 3∠AOB
Lekin angle at O = 360°
Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles
Agar ∠AOB = x
Toh reflex ∠ADB = x
x + 3x = 360°
⇒ 4x = 360°
⇒ x = 360°/4 = 90°
∴ 3x = 90° x 3 = 270°
Angles hain 270° aur 90° (c)

Question 11.
Agar A aur B circle ke do points hain jisme m(AB) = 260°. Arc BA ka angle jo circle ke kisi point par subtend hota hai, possible value kya ho sakti hai?
(a) 100°
(b) 75°
(c) 50°
(d) 25°
Solution:
A aur B circle ke do points hain jisme reflex ∠AOB = 260°
Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions
∴ ∠AOB = 360° – 260° = 100°
C circle ka ek point hai
∴ AC aur BC join karne par,
∠ACB = 1/2 ∠AOB = 1/2 x 100° = 50° (c)

Question 12.
Equilateral triangle ABC ek circle mein inscribed hai centre O ke saath. ∠BOC ka measure kya hai?
(a) 30°
(b) 60°
(c) 90°
(d) 120°
Solution:
∆ABC equilateral triangle hai jo circle mein inscribed hai centre O ke saath
RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles
∴ ∠BOC ka measure = 2∠BAC
= 2 x 60° = 120° (d)

Question 13.
Agar circle ke do diameters ek dusre ko right angles par intersect karte hain, toh unke end points ko join karne se bana quadrilateral kya hoga?
(a) rhombus
(b) rectangle
(c) parallelogram
(d) square
Solution:
Circle ke do diameter AB aur CD ek dusre ko right angles par intersect karte hain
RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles
AD, DB, BC aur CA join karke quadrilateral ABCD banta hai.
∵ Diagonals equal hain aur ek dusre ko right angles par bisect karte hain
∴ ACBD ek square hai (d)

Question 14.
ABC ek circle ka arc hai aur ∠ABC = 135°, toh arc AB ka ratio circumference ke saath kya hai?
(a) 1 : 4
(b) 3 : 4
(c) 3 : 8
(d) 1 : 2
Solution:
Arc ABC ek circle ka aur ∠ABC = 135°
OA aur OC join karte hain
Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions
∴ Arc ABC ka angle centre par = 2 x ∠ABC = 2 x 135° = 270°
Circle ke centre par angle = 360°
∴ Circumference ke saath ratio = 270° : 360° = 3:4 (b)

Question 15.
Circle ka chord uske radius ke equal hai. Ye chord minor arc par kya angle subtend karta hai?
(a) 60°
(b) 75°
(c) 120°
(d) 150°
Solution:
Circle ka chord = circle ka radius
Figure mein OA = OB = AB
RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles
∴ ∠AOB = 60°
(Equilateral triangle ka har angle = 60°) (a)

Question 16.
PQRS ek cyclic quadrilateral hai jisme PR circle ka diameter hai. Agar ∠QPR = 67° aur ∠SPR = 72°, toh ∠QRS kya hai?
(a) 41°
(b) 23°
(c) 67°
(d) 18°
Solution:
PQRS cyclic quadrilateral hai centre O ke saath aur ∠QPR = 67°
∠SPR = 72°
RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles
∴ ∠QPS = 67° + 72° = 139°
∵ ∠QPS + ∠QRS = 180° (Cyclic quadrilateral ke opposite angles ka sum)
⇒ 139° + ∠QRS = 180°
⇒ ∠QRS = 180° – 139° = 41° (a)

Question 17.
Agar A, B, C circle ke teen points hain centre O ke saath jisme ∠AOB = 90° aur ∠BOC = 120°, toh ∠ABC kya hai?
(a) 60°
(b) 75°
(c) 90°
(d) 135°
Solution:
A, B aur C circle ke teen points hain centre O ke saath
∠AOB = 90° aur ∠BOC = 120°
RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles
∴ ∠AOC = 360° – (120° + 90°)
= 360° -210°= 150°
Lekin ∠AOC centre par arc AC se bana hai aur ∠ABC circle ke remaining part par
∴ ∠ABC = 1/2 ∠AOC
= 1/2 x 150° = 75° (b)

Question 18.
Circle ka greatest chord kya kehlaata hai?
(a) radius
(b) secant
(c) diameter
(d) none of these
Solution:
Circle ka greatest chord uska diameter kehlaata hai. (c)

Question 19.
Circle ke minor segment mein bana angle kaisa hota hai?
(a) acute
(b) obtuse
(c) right angle
(d) none of these
Solution:
Circle ke minor segment mein bana angle obtuse angle hota hai. (b)

Question 20.
Teen non-collinear points se kitne circles draw ho sakte hain?
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
Teen non-collinear points se sirf ek hi circle pass ho sakta hai. (a)

Question 21.
Figure mein, agar circle ke chords AB aur CD ek dusre ko right angles par intersect karte hain, toh x + y kya hai?
(a) 45°
(b) 60°
(c) 75°
(d) 90°
RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles
Solution:
Circle mein, AB aur CD do chords hain jo P par right angle par intersect karte hain yani ∠CPB = 90°
Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles
∠CAB aur ∠CDB same segment mein hain
∴ ∠CDB = ∠CAB = x
Ab triangle PDB mein,
External ∠CPB = ∠D + ∠DBP
⇒ 90° = x + y (∵ CD ⊥ AB)
Hence x + y = 90° (d)

Question 22.
Figure mein, agar ∠ABC = 45°, toh ∠AOC kya hai?
(a) 45°
(b) 60°
(c) 75°
(d) 90°
RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles
Solution:
∵ arc AC subtends
∠AOC centre par aur ∠ABC
circle ke remaining part par
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles
∴ ∠AOC = 2∠ABC
= 2 x 45° = 90°
Hence ∠AOC = 90° (d)

Question 23.
Figure mein, chords AD aur BC ek dusre ko right angles par point P par intersect karte hain. Agar ∠DAB = 35°, toh ∠ADC kya hai?
(a) 35°
(b) 45°
(c) 55°
(d) 65°
RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles
Solution:
Do chords AD aur BC right angles par P par intersect karte hain, ∠DAB = 35°
AB aur CD join kiye gaye hain
Triangle ABP mein,
External ∠APC = ∠B + ∠A
RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles
⇒ 90° = ∠B + 35°
∠B = 90° – 35° = 55°
∵ ∠ABC aur ∠ADC same segment mein hain
∴ ∠ADC = ∠ABC = 55° (c)

Question 24.
Figure mein, O circle ka centre hai aur ∠BDC = 42°. ∠ACB ka measure kya hai?
(a) 42°
(b) 48°
(c) 58°
(d) 52°
Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions
Solution:
Figure mein, O circle ka centre hai
∠BDC = 42°
∠ABC = 90° (Semicircle mein angle)
aur ∠BAC aur ∠BDC circle ke same segment mein hain.
∴ ∠BAC = ∠BDC = 42°
Ab triangle ABC mein,
∠A + ∠ABC + ∠ACB = 180° (Triangle ke angles ka sum)
RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles
⇒ 42° + 90° + ∠ACB = 180°
⇒ 132° + ∠ACB = 180°
⇒ ∠ACB = 180° – 132° = 48° (b)

Question 25.
Centre O wale circle mein, AB aur CD do diameters hain jo ek dusre ke perpendicular hain. Chord AC ki length kya hai?
Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles
Solution:
AB aur CD circle ke do diameters hain centre O ke saath
Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions

Question 26.
Radius r ke do equal circles aise intersect karte hain ki har ek dusre ke centre se pass hota hai. Common chord ki length kya hai?
RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles
Solution:
Do equal circles ek dusre ke centre se pass hote hain aur A aur B par intersect karte hain
r har circle ka radius hai
RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles

Question 27.
Agar AB ek circle ka chord hai, P aur Q circle ke do points hain jo A aur B se different hain, toh kya relation hai?
(a) ∠APB = ∠AQB
(b) ∠APB + ∠AQB = 180° ya ∠APB = ∠AQB
(c) ∠APB + ∠AQB = 90°
(d) ∠APB + ∠AQB = 180°
Solution:
AB circle ka chord hai,
P aur Q do points hain A aur B ke alawa
Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions
∵ ∠APB aur ∠AQB circle ke same segment mein hain
∴ ∠APB = ∠AQB (a)

Question 28.
AB aur CD centre O wale circle ke do parallel chords hain jisme AB = 6 cm aur CD = 12 cm. Chords same side par hain centre ke aur unke beech distance 3 cm hai. Circle ka radius kya hai?
RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles
Solution:
AB aur CD centre O wale circle ke do parallel chords hain
r circle ka radius hai, AB = 6 cm, CD = 12 cm
aur unke beech distance = 3 cm
OC aur OA join karte hain, LM = 3 cm
RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles
OM = x maante hain, toh OL = x + 3
RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles

Question 29.
17 cm radius wale circle mein, do parallel chords diameter ke opposite sides par drawn hain. Chords ke beech distance 23 cm hai. Agar ek chord ki length 16 cm hai toh dusre ki length kya hai?
(a) 34 cm
(b) 15 cm
(c) 23 cm
(d) 30 cm
Solution:
Circle ka radius = 17 cm
Do parallel chords ke beech distance = 23 cm
AB || CD aur LM = 23 cm
OA aur OC join karte hain,
∴ OA = OC = 17 cm
OL = x maante hain, toh OM = (23 – x) cm
AB = 16 cm
Ab right triangle OAL mein,
OA² = OL² + AL²
⇒ (17)² = x² + AL²
⇒ 289 = x² + AL²
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles

Question 30.
Figure mein, O circle ka centre hai jisme ∠AOC = 130°, toh ∠ABC kya hai?
(a) 130°
(b) 115°
(c) 65°
(d) 165°
RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles
Solution:
O circle ka centre hai aur ∠AOC = 130°
Reflex ∠AOC = 360° – 130° = 230°
RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles
Ab arc ADB centre par ∠AOC subtend karta hai aur circle ke remaining part par ∠ABC
∴ ∠ABC = 1/2 reflex ∠AOC
= 1/2 x 230°= 115° (b)

Ye sab solutions 2026 ke academic session ke liye updated hain. Hamare experts ne har question ko carefully solve kiya hai taaki students ko complete understanding mil sake. RD Sharma Class 9 Solutions Chapter 15 MCQs ko practice karne se students apne CBSE board exams mein better performance de sakte hain.

Chapter 15 “Areas of Parallelograms and Triangles” geometry ka important part hai. Ye MCQs students ko different types ke questions practice karne mein help karte hain. 2026 mein board exam pattern ke according, ye questions frequently asked jaate hain. Students ko chahiye ki wo regular practice karen aur har concept ko thoroughly understand karen.

In solutions ko access karne ke liye students ko koi special software ya app ki zarurat nahi hai. Ye solutions mobile-friendly hain aur kisi bhi device par easily access ho sakte hain. Students apne convenience ke according study kar sakte hain aur doubt clear kar sakte hain.

RD Sharma textbook CBSE board ke students ke liye highly recommended hai. Ye book comprehensive coverage provide karti hai aur step-by-step solutions students ki understanding improve karte hain. 2026 session ke liye updated content ensure karta hai ki students latest syllabus ke according prepare hoin.

Frequently Asked Questions

Can I access RD Sharma Class 9 Solutions Chapter 15 MCQs on my smartphone?

Haan bilkul! Aap RD Sharma Class 9 Solutions Chapter 15 MCQs ko kisi bhi device par access kar sakte hain – smartphone, tablet, ya computer. Ye solutions mobile-friendly format mein available hain jo 2026 ke students ke liye convenient hai.

What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 MCQs?

RD Sharma Class 9 Solutions Chapter 15 MCQs practice karne se students ko kaafi fayde hain. Ye solutions accuracy ke saath banaye gaye hain jo students ke academic performance improve karte hain. Regular practice se concept clarity aati hai aur board exams mein better marks mil sakte hain. Hamare experts ne 2026 syllabus ke according ye solutions prepare kiye hain.

Is RD Sharma enough for Class 12 Maths?

RD Sharma ek excellent book hai jo thousands of practice questions provide karti hai. Class 9 se hi RD Sharma practice karna students ke liye beneficial hai kyunki ye strong foundation build karta hai. Ye book comprehensive coverage deti hai aur step-by-step solutions students ki mathematical understanding enhance karte hain.

Students ko advice dete hain ki wo regular practice karen aur har question ko properly understand karne ki koshish karen. 2026 board exams ke liye ye solutions highly beneficial hain. Agar koi doubt hai toh students comment section mein freely puch sakte hain. Hamare experts ready hain help karne ke liye.

Complete RD Sharma Class 9 solutions ke liye students NCERT official website par bhi reference le sakte hain. Ye ensure karta hai ki students ko authentic aur updated information mile. 2026 mein success achieve karne ke liye consistent practice aur proper understanding zaroori hai.

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