MARKING SCHEME MATHEMATICS MODEL PAPER CLASS 10

SECTION –A
Time: 20 Minutes Marks: 15
Scoring Keys:
1. The quadratic equation in the following is:
A. 𝑥 4 + 11𝑥 2 + 9 = 0
B. 𝑥 3 + 11𝑥 2 + 9 = 0
C. 𝑥 3 + 11𝑥 + 9 = 0
D. 𝒙𝟐 + 𝟏𝟏𝒙 + 𝟗 = 𝟎
2. The solution set of 2𝑥 2 − 9𝑥 + 5 = 0 is:
−9±√41
A. { }
4

𝟗±√𝟒𝟏
B. { }
𝟒

−9±√41
C. { }
2

−9±√41
D. { }
2
1 1
3. 𝛼 + 𝛽 =
1
A. 𝛼𝛽
1
B. 𝛼+𝛽

𝛼𝛽
C. 𝛼+𝛽

𝜶+𝜷
D. 𝜶𝜷

4. The discriminant of equation 𝑥 2 + 6𝑥 + 2 = 0 is equal to:

A. 8
B. 28
C. 36
D. 44
5. Direct variation between 𝑝 and 𝑞 can be expressed as:
A. 𝑝 = 𝑞
1
B. 𝑝 = 𝑞

C. 𝒑 ∝ 𝒒
1
D. 𝑝 ∝ 𝑞
6. In continued proportion 𝑝: 𝑞 = 𝑞: 𝑟, 𝑟 is called as:
A. first proportional to 𝑝, 𝑞.
B. second proportional to 𝑝, 𝑞.
C. third proportional to 𝒑, 𝒒.
D. fourth proportional to 𝑝, 𝑞.
𝑥 2 +1
7. is an example of:
𝑥+1

A. proper fraction only

B. improper fraction only
C. both proper and rational fraction
D. both improper and irrational fraction
8. The set of the whole numbers (𝑊) in the following is:
A. {𝟎, 𝟏, 𝟐, 𝟑, … … … }
B. {0, ±2, ±4, … … … }
C. {1,2,3, … … … }
D. {0, ±1, ±2, ±3, … … … }
9. The range of 𝑅 = {(1,2), (2,2), (3,1), (4,4)} is:
A. {1,3,4}
B. {𝟏, 𝟐, 𝟒}
C. {2,3,4}
D. {1,2,3,4}
10. If 𝐴 = {1,2,3,4} and 𝐵 = {5,6,7,8}, then which of the following binary relations is a
function from 𝐵 to 𝐴?
A. 𝑅 = {(1,5), (2,6), (3,7), (4,8)}
B. 𝑅 = {(1,6), (2,5), (4,8), (4,7)}
C. 𝑹 = {(𝟓, 𝟏), (𝟔, 𝟐), (𝟕, 𝟑), (𝟖, 𝟒)}
D. 𝑅 = {(5,2), (6,1), (8,4), (8,3)}
11. The value that appears more times in a data is called:
A. mean
B. median
C. mode
D. variance
12. In the given set of data, 71, 73, 79, 77, 76, 75, 80, the median is:
A. 73
B. 76
C. 77
D. 79
13. In radians, 45° is equal to:
𝜋
A. 2
𝜋
B. 3
𝝅
C. 𝟒
𝜋
D. 6

14. 1 + 𝑐𝑜𝑡 2 𝜃 =
A. 𝑠𝑖𝑛2 𝜃
B. 𝑐𝑜𝑠 2 𝜃
C. 𝑡𝑎𝑛2 𝜃
D. 𝒄𝒐𝒔𝒆𝒄𝟐 𝜽
15. The number of circles that can pass through three non-collinear points is:
A. 0
B. 𝟏
C. 2
D. 3

KEY:
MCQs
No.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Key D B D B C C C A B C C B C D B
 SECTION-B
Time: 2 Hours 40 Minutes Marks: 36
1. Attempt any NINE of the following short questions. Each question carries 4
marks.

i. Derive quadratic formula for 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where 𝑎 ≠ 0, by using

completing square method.

Solution:

As general form of quadratic equation is

𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0

𝑎𝑥 2 + 𝑏𝑥 = −𝑐 Step-1 (1 Mark)

÷ 𝑖𝑛𝑔 𝑏𝑦 "𝑎"

𝑏 −𝑐
𝑥 2 + 2(𝑥) (2𝑎) = 𝑎

𝑏 2
𝐴𝑑𝑑𝑖𝑛𝑔 ( ) 𝑜𝑛 𝐵. 𝑆
2𝑎

2
𝑏 𝑏 2 𝑏 2 𝑐
𝑥 + 2(𝑥) ( ) + ( ) = ( ) −
2𝑎 2𝑎 2𝑎 𝑎

𝑢𝑠𝑖𝑛𝑔 𝑎2 + 2𝑎𝑏 + 𝑏 2 = (𝑎 + 𝑏)2 Step-2 (1 Mark)

𝑏 2 𝑏2 𝑐
(𝑥 + ) = −
2𝑎 4𝑎2 𝑎

𝑏 2 𝑏 2 −4𝑎𝑐
(𝑥 + 2𝑎) = 4𝑎2

𝑇𝑎𝑘𝑖𝑛𝑔 𝑆𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑜𝑛 𝐵. 𝑆

b b 2  4ac
x  Step-3 (1 Mark)
2a 4a 2

𝑏 √𝑏2 −4𝑎𝑐
𝑥 + 2𝑎 = ± 2𝑎

b b 2  4ac
x 
2a 2a

−𝑏±√𝑏 2 −4𝑎𝑐
𝑥= Step-4 (1 Mark)
2𝑎

𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.

ii. Solve 4. 22𝑥 − 10. 2𝑥 + 4 = 0.

Solution:

4. 22𝑥 − 10. 2𝑥 + 4 = 0

𝑊𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒:
 4. (2𝑥 )2 − 10. 2𝑥 + 4 = 0 ⟶ 

𝐿𝑒𝑡 𝑦 = 2𝑥 ⟶ 

 ⇒ 4(𝑦)2 − 10𝑦 + 4 = 0 Step-1

4𝑦 2 − 10𝑦 + 4 = 0

4𝑦 2 − 8𝑦 − 2𝑦 + 4 = 0

4𝑦(𝑦 − 2) − 2(𝑦 − 2) = 0 Step-2

(𝑦 − 2)(4𝑦 − 2) = 0

So either 𝑦 − 2 = 0 𝑜𝑟 4𝑦 − 2 = 0

𝑦=2 4𝑦 = 2

4 2
𝑦=
4 4
1
𝑦=
2

Re-putting Values:

𝑦 = 2𝑥 1
2𝑥 =
2
2𝑥 = 21 Step-3
2𝑥 = 2−1 Step-4

By Comparing:

𝑥=1 𝑥 = −1

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛. 𝑆𝑒𝑡 = {1, −1}.

iii. Find the cube roots of 64.

Solution:

𝐿𝑒𝑡 𝑥 𝑏𝑒 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒 𝑟𝑜𝑜𝑡 𝑜𝑓 64 𝑖. 𝑒.

3
𝑥 = √64
1⁄
𝑥 = 64 3

𝑇𝑎𝑘𝑖𝑛𝑔 𝐶𝑢𝑏𝑒 𝑜𝑛 𝐵. 𝑆 Step-1

1⁄
𝑥 3 = 643× 3

𝑥 3 − 64 = 0

𝑥 3 − 43 = 0

(𝑥 − 4)(𝑥 2 + 4𝑥 + 16) = 0 𝐵𝑦 𝑢𝑠𝑖𝑛𝑔 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 )

So either

𝑥 2 + 4𝑥 + 16 = 0 𝑜𝑟 𝑥−4=0
 𝐻𝑒𝑟𝑒 𝑎 = 1, 𝑏 = 4, 𝑐 = 16 𝑜𝑟 𝑥=4

−𝑏±√𝑏 2 −4𝑎𝑐
So 𝑥 = Step-2
2𝑎

−4±√42 −4(1)(6)
= 2(1)

−4±√16−64
= 2

−4±√−48
= Step-3
2

−4±4𝑖√3
= 2

−1±𝑖√3
= 4( )
2

−1+𝑖√3 −1−𝑖√3
𝑥 = 4( ),𝑥 = 4( )
2 2

𝑥 = 4𝑤, 𝑥 = 4𝑤 2 Step-4

𝐻𝑒𝑛𝑐𝑒 {4, 4𝑤, 4𝑤 2 } 𝑎𝑟𝑒 𝑐𝑢𝑏𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 64.

𝛼 𝛽
iv. If 𝛼, 𝛽 are roots of 𝑥 2 − 4𝑥 + 2 = 0, find the equation whose roots are 𝛽 , 𝛼.

Solution:
𝑏 −4 𝑐 2
𝐴𝑠 𝑓𝑜𝑟 𝑥 2 − 4𝑥 + 2 = 0, 𝛼 + 𝛽 = − 𝑎 = − = 4 and 𝛼𝛽 = 𝑎 = 1 = 2
1

𝛼 𝛽
Required equation whose roots are , 𝑖𝑠 𝑥 2 − 𝑆𝑥 + 𝑃 = 0 ⟶  Step-1
𝛽 𝛼

𝛼 𝛽 𝛼 𝛽
𝐹𝑜𝑟 𝑆 = 𝛽 + 𝛼 & 𝑃 =𝛽×𝛼

𝛼2 +𝛽 2
𝑆= & 𝑃=1 Step-3
𝛼𝛽

(𝛼+𝛽)2 −2𝛼𝛽
𝑆= Step-2
𝛼𝛽

42 −2(2) 16−4 12
𝑆= ⇒ ⇒
2 2 2

𝑆=6

𝑃𝑢𝑡 𝑆 = 6 𝑎𝑛𝑑 𝑃 = 1 𝑖𝑛 

So ⇒ 𝑥 2 − 6𝑥 + 1 = 0 𝑖𝑠 𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛. Step-4

𝑎+𝑏
v. Find the mean proportional of 𝑎2 − 𝑏 2 and 𝑎−𝑏.

Solution:
𝑎+𝑏
Let x be the mean proportional of 𝑎2 − 𝑏 2 and 𝑎−𝑏.
𝑎+𝑏
i.e. 𝑎2 − 𝑏 2 ∶ 𝑥: 𝑎−𝑏 are in continued proportion. Step-1
𝑎+𝑏
⇒ 𝑎2 − 𝑏 2 ∶ 𝑥 ∷ 𝑥: 𝑎−𝑏
𝐴𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑚𝑒𝑎𝑛 = 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑒𝑥𝑡𝑟𝑒𝑚𝑒
 𝑎+𝑏
𝑥 2 = 𝑎2 − 𝑏 2 × 𝑎−𝑏 Step-2
𝑎+𝑏
𝑥 2 = (𝑎 + 𝑏)(𝑎 − 𝑏) × 𝑎−𝑏 Step-3
𝑥 2 = (𝑎 + 𝑏)2
Taking square root on B.S.
√𝑥 2 = √(𝑎 + 𝑏)2 Step-4
𝑥 = ±(𝑎 + 𝑏)
4𝑥+2
vi. Resolve into partial fraction (𝑥+2)(2𝑥−1).

Solution:
4𝑥+2 𝐴 𝐵
𝐿𝑒𝑡 = 𝑥+2 + 2𝑥−1 ⟶ 
(𝑥+2)(2𝑥−1)
× 𝑖𝑛𝑔 𝐵. 𝑆 𝑏𝑦 (𝑥 + 2)(2𝑥 − 1) Step-1
4𝑥 + 2 = 𝐴(2𝑥 − 1) + 𝐵(𝑥 + 2) ⟶ 
1
𝑃𝑢𝑡 2𝑥 − 1 ⇒ 𝑥 = 2 𝑖𝑛 
1 1 1
4(2) + 2 = 𝐴 (2(2) − 1) + 𝐵(2 + 2)
5
4 = 𝐵(2) Step-2
8
𝐵=5
𝑃𝑢𝑡 𝑥 + 2 = 0 ⇒ 𝑥 = −2 in 

4(−2) + 2 = 𝐴(2(−2) − 1) + 𝐵(−2 + 2) Step-3

−8 + 2 = 𝐴(−4 − 1)
−6 = 𝐴(−5)
6
𝐴=5
6 8
𝑃𝑢𝑡 𝐴 = 5 𝑎𝑛𝑑 𝐵 = 5 𝑖𝑛 
6 8
4𝑥+2 5 5
= + Step-4
(𝑥+2)(2𝑥−1) 𝑥+2 2𝑥−1
6 8
= 5(𝑥+2) + 5(2𝑥−1) 𝐴𝑛𝑠.

vii. If 𝑈 = {1,2,3, … … ,10}, 𝐴 = {2,4,6,8,10} and 𝐵 = {1,3,5,7,9}, then verify

(𝐴 ∪ 𝐵)′ = 𝐴′ ∩ 𝐵 ′ .

Solution:

𝐻𝑒𝑟𝑒, 𝑈 = {1,2,3, … … ,10}, 𝐴 = {2,4,6,8,10} and 𝐵 = {1,3,5,7,9}

𝑇𝑜 𝑃𝑟𝑜𝑣𝑒: (𝐴 ∪ 𝐵)′ = 𝐴′ ∩ 𝐵 ′.
First:
𝐴 ∪ 𝐵 = {2,4,6,8,10} ∪ {1,3,5,7,9}
𝐴 ∪ 𝐵 = {1,2,3,4,5,6,7,8,9,10} Step-1
𝑳. 𝑯. 𝑺: (𝐴 ∪ 𝐵)′
𝑈 − (𝐴 ∪ 𝐵)′ = {1,2,3,4,5,6,7,8,9,10} − {1,2,3,4,5,6,7,8,9,10} Step-2
={}⟶
& 𝐴′ = 𝑈 − 𝐴 = {1,2,3,4,5,6,7,8,9,10} − {2,4,6,8,10}
𝐴′ = {1,3,5,7,9}
& 𝐵 ′ = 𝑈 − 𝐵 = {1,2,3,4,5,6,7,8,9,10} − {1,3,5,7,9} Step-3
𝐵 ′ = {2,4,6,8,10}
𝑹. 𝑯. 𝑺 = 𝐴′ ∩ 𝐵 ′
𝐴′ ∩ 𝐵 ′ = {1,3,5,7,9} ∩ {2,4,6,8,10}
={}⟶
From  & 
𝑳. 𝑯. 𝑺 = 𝑹. 𝑯. 𝑺 Step-4
i.e. (𝐴 ∪ 𝐵)′ = 𝐴′ ∩ 𝐵 ′ .

viii. A set of data contains the values as 105,80,90,75,100,105 𝑎𝑛𝑑 110. Show
 that 𝑀𝑜𝑑𝑒 > 𝑀𝑒𝑑𝑖𝑎𝑛 > 𝑀𝑒𝑎𝑛.

Proof:

Given Data is 105,80,90,75,100,105 𝑎𝑛𝑑 110.

To Find Mean:
𝑥1 +𝑥2 +𝑥3 +𝑥4 +𝑥5 +𝑥6 +𝑥7
𝑀𝑒𝑎𝑛 = 7

105+80+90+75+100+105+ 110
𝑀𝑒𝑎𝑛 = Step-1
7

𝑀𝑒𝑎𝑛 = 95 → 

To Find Median: First arrange data in ascending order i.e.

75,80,90,100,105,105,110 Step-2

𝑀𝑒𝑑𝑖𝑎𝑛 = 100 → 

To Find Mode: Mode is the most frequent value, So.

𝑀𝑜𝑑𝑒 = 105 →  Step-3

From , and ,

𝑀𝑜𝑑𝑒 > 𝑀𝑒𝑑𝑖𝑎𝑛 > 𝑀𝑒𝑎𝑛 Step-4

105 > 100 > 95

ix. An arc of a circle subtends an angle of 2 radians at the center. If the area of

sector formed is 64𝑐𝑚2 , find the radius of the circle.

Solution:

𝜃 = 2 𝑟𝑎𝑑𝑖𝑎𝑛

𝐴𝑟𝑒𝑎 = 64 𝑐𝑚2

𝑟 =?

We know that: Step-1

1
𝐴 = 2 𝑟 2𝜃

1
64 = 2 𝑟 2 (2) Step-2

64 = 𝑟 2 ⇒ 𝑟 2 = 64 Step-3

Taking square on B.S

𝑟 = 8𝑐𝑚 Step-4

x. Prove that: 𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛2 𝑥 = 𝑐𝑜𝑠 3 𝑥.

Proof:

𝑳. 𝑯. 𝑺 = 𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛2 𝑥

 𝑐𝑜𝑠𝑥(1 − 𝑠𝑖𝑛2 𝑥) = 1 Step-1

𝐴𝑠 cos2 𝜃 + sin2 𝜃 = 1 ⇒ cos2 𝜃 = 1 − sin2 𝜃 Step-2

= 𝑐𝑜𝑠𝑥(cos2 𝑥) Step-3

= cos3 𝑥

= 𝑹. 𝑯. 𝑺 Step-4

̅̅̅̅ and 𝐴𝐶
xi. 𝐴𝐵 ̅̅̅̅ are tangent segments to the circle with centre 𝑂. If

̅̅̅̅ = 6𝑐𝑚 and 𝑚𝑂𝐴

𝑚𝑂𝐵 ̅̅̅̅ = 10𝑐𝑚, then find 𝑚𝐴𝐵
̅̅̅̅ and 𝑚𝐴𝐶
̅̅̅̅ .

Solution:

Step-1

Since AOB is a right triangle.

    m AB    mOB 
2 2 2
 mOA Step-2

10    m AB    6 
2 2 2

100   m AB   36
2

 m AB   100  36
2
Step-3

 m AB   64
2

m AB  8cm
m AB  m AC  8cm Step-4

i. Prove that equal chords of a circle subtend equal angles at the center.

Prove for only one circle.

Proof:

Given:

A circle with center 𝑂. ̅̅̅̅

𝐴𝐵 and ̅̅̅̅
𝐶𝐷 are

two chords of the circle (which are not

diameters) such that ̅̅̅̅

𝐴𝐵 ≅ ̅̅̅̅
𝐶𝐷 or

̅̅̅̅ ≅ 𝑚𝐶𝐷
𝑚𝐴𝐵 ̅̅̅̅ .

Arcs subtend ∠1 and ∠2 at the center.

To Prove: ∠1 = ∠2

̅̅̅̅ =
Construction: We Join O to A, B, C and D respectively so that 𝑚𝑂𝐴
 ̅̅̅̅ = 𝑚𝑂𝐶
𝑚𝑂𝐵 ̅̅̅̅ = 𝑚𝑂𝐷
̅̅̅̅ = 𝑟𝑎𝑑𝑖𝑖 of a circle.

Proof:

Statements Reasons

𝐼𝑛 ∆𝑂𝐴𝐵 ↔ ∆𝑂𝐶𝐷

̅̅̅̅ ̅̅̅̅
𝑂𝐴 ≅ 𝑂𝐶 Radii of same circle.

̅̅̅̅ ≅ ̅̅̅̅
𝑂𝐵 𝑂𝐷 Radii of same circle.

̅̅̅̅
𝐴𝐵 ≅ ̅̅̅̅
𝐶𝐷
Given
∴ ∆𝑂𝐴𝐵 ≅ ∆𝑂𝐶𝐷 S. S. S ≅ S. S. S

∴ ∠1 ≅ ∠2 Corresponding angles of congruent

triangles.

Given & To Prove Construction Proof

01 Mark 01 Mark 02 Marks

 SECTION-C
Marks: 24
NOTE: Attempt any THREE of the following questions. Each question carries 8 marks.

2. In ∆𝐴𝐵𝐶, 𝑚𝐴𝐵 ̅̅̅̅ = 12𝑐𝑚, mB  100o . The projection of 𝐵𝐶

̅̅̅̅ = 8𝑐𝑚, 𝑚𝐵𝐶 ̅̅̅̅ on ̅̅̅̅
𝐴𝐵 is
̅̅̅̅
6𝑐𝑚. Find 𝑚𝐴𝐶 .

Solution:

Step-1 (04 Marks)

Since,
b 2  a 2  c 2  2cp Step-2 (01 Mark)
 b 2  12    8   2  8  5 
2 2

b 2  144  64  80 Step-3 (01 Mark)

b 2  288 Step-4 (01 Mark)
b  16.97cm Step-5 (01 Mark)

3. Prove that If two chords of a circle are congruent then they will be equidistant
from the center.

Given:

A circle with center 𝑂, ̅̅̅̅

𝐴𝐵 and ̅̅̅̅
𝐶𝐷 are two congruent chords of the circle.

̅̅̅̅ and 𝐶𝐷
To Prove: 𝐴𝐵 ̅̅̅̅ are equidistant from the

center 𝑂.

Construction: Join 𝑂 to 𝐴 and 𝐶. Also draw

̅̅̅̅ and 𝑂𝐹
perpendicular 𝑂𝐸 ̅̅̅̅ on the given chords 𝐴𝐵
̅̅̅̅
̅̅̅̅
and 𝐶𝐷 respectively.
Proof:

Statements Reasons

𝑆𝑖𝑛𝑐𝑒 ̅̅̅̅
𝑂𝐸 ⊥ ̅̅̅̅
𝐴𝐵 𝑎𝑛𝑑 ̅̅̅̅
𝑂𝐹 ⊥ ̅̅̅̅
𝐶𝐷 Construction

∴ ̅̅̅̅ 𝐸𝐵 𝑎𝑛𝑑 ̅̅̅̅

𝐴𝐸 ⊥ ̅̅̅̅ 𝐶𝐹 ⊥ ̅̅̅̅
𝐷𝐹 By the use of Theorem 9.3

𝑜𝑟 ̅̅̅̅ = 𝑚𝐸𝐵
𝑚𝐴𝐸 ̅̅̅̅ = 𝑚𝐷𝐹
̅̅̅̅ 𝑎𝑛𝑑 𝑚𝐶𝐹 ̅̅̅̅
 But ̅̅̅̅ ≅ 𝑚𝐶𝐷
m𝐴𝐵 ̅̅̅̅
Given

𝑜𝑟 ̅̅̅̅ + 𝑚𝐸𝐵
𝑚𝐴𝐸 ̅̅̅̅ + 𝑚𝐷𝐹
̅̅̅̅ = 𝑚𝐶𝐹 ̅̅̅̅ Segment addition postulate

̅̅̅̅ + 𝑚𝐴𝐸
𝑚𝐴𝐸 ̅̅̅̅ = 𝑚𝐶𝐹
̅̅̅̅ + 𝑚𝐶𝐹
̅̅̅̅ ̅̅̅̅̅𝑎𝑛𝑑 𝑚𝐷𝐹
̅̅̅̅ = 𝑚𝐴𝐸
∵ 𝑚𝐸𝐵 ̅̅̅̅
̅̅̅̅ = 𝑚𝐶𝐹

̅̅̅̅ = 2𝑚𝐶𝐹
2𝑚𝐴𝐸 ̅̅̅̅ Adding equal quantities.

̅̅̅̅ = 𝑚𝐶𝐹
𝑚𝐴𝐸 ̅̅̅̅ Dividing both sides by 2.

Or ̅̅̅̅ ̅̅̅̅ → 
𝐴𝐸 = 𝐶𝐹

Now, in ∆𝐴𝑂𝐸 ↔ ∆𝐶𝑂𝐹

̅̅̅̅ ̅̅̅̅
𝑂𝐴 = 𝑂𝐶 Radii of the same circle
From  proved above
̅̅̅̅
𝐴𝐸 = ̅̅̅̅
𝐶𝐹 →  Right angles

∠𝐴𝐸𝑂 ≅ ∠𝐶𝐹𝑂 𝐻. 𝑆 ≅ 𝐻. 𝑆

∴ ∆𝐴𝑂𝐸 ≅ ∆𝐶𝑂𝐹 Corresponding sides of the triangle.

∴ ̅̅̅̅
𝑂𝐸 = ̅̅̅̅ ̅̅̅̅ = 𝑚𝑂𝐹
𝑂𝐹 𝑜𝑟 𝑚𝑂𝐸 ̅̅̅̅

∴ ̅̅̅̅
𝐴𝐵 and ̅̅̅̅
𝐶𝐷 are equidistant from the center of

the circle.

Given To Prove Construction Proof

01 Mark 01 Mark 02 Marks 04 Marks

4. Prove that the angle in a semi-circle is a right angle.

Given:

A circle with center 𝑂, ̅̅̅̅

𝐴𝐵 is a diameter of the

circle and ∠𝐴𝐶𝐵 is the any angle in the semi-

circle.

To Prove: ∠𝐴𝐶𝐵 is a right angle i.e. 𝑚∠𝐴𝐶𝐵 =

90°.

Construction: Join 𝑂 to 𝐴 and 𝐶.

Proof:

Statements Reasons

𝐼𝑛 ∆𝑂𝐴𝐶,

̅̅̅̅ ≅ 𝑚𝑂𝐶
𝑚𝑂𝐴 ̅̅̅̅ Radii of the same circle.
 ∴ ∆𝑂𝐴𝐶 𝑖𝑠 𝑎𝑛 𝑖𝑠𝑜𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒. Definition of 𝑖𝑠𝑜𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒

If two sides of a triangle are equal, the

𝑎𝑛𝑑 𝑚∠𝑂𝐴𝐶 ≅ 𝑚∠𝑂𝐶𝐴 → 
angles which are opposite to them are
also equal.
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑖𝑛 𝑡ℎ𝑒 ∆𝑂𝐶𝐵

̅̅̅̅ ≅ 𝑚𝑂𝐶
𝑚𝑂𝐵 ̅̅̅̅ Radii of a circle.

∴ 𝑎𝑛𝑑 𝑚∠𝑂𝐵𝐶 ≅ 𝑚∠𝑂𝐶𝐵 → 

∴ 𝑚∠𝑂𝐴𝐶 + 𝑚∠𝑂𝐵𝐶 = 𝑚∠𝑂𝐶𝐴 + 𝑚∠𝑂𝐶𝐵 Adding  and 

𝑚∠𝑂𝐴𝐶 + 𝑚∠𝑂𝐵𝐶 = 𝑚∠𝐴𝐶𝐵 →  ∵ 𝑚∠𝑂𝐶𝐴 + 𝑚∠𝑂𝐶𝐵 = 𝑚∠𝐴𝐶𝐵

𝐵𝑢𝑡 𝑚∠𝑂𝐴𝐶 + 𝑚∠𝑂𝐵𝐶 + 𝑚∠𝐴𝐶𝐵 = 180° ∵ 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑜𝑓 𝑎

𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 180°.
𝑂𝑟 𝑚∠𝐴𝐶𝐵 + 𝑚∠𝐴𝐶𝐵 = 180° ∵ 𝑚∠𝑂𝐴𝐶 + 𝑚∠𝑂𝐵𝐶 = 𝑚∠𝐴𝐶𝐵
Adding two equal numbers.
⇒ 𝑚∠𝐴𝐶𝐵 = 90°

or ∠𝐴𝐶𝐵 𝑖𝑠 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒.

The angle inscribed in a semicircle is always a right angle(90°).

Given To Prove Construction Proof

01 Mark 01 Mark 02 Marks 04 Marks

5. Construct a triangle with sides 4 cm, 4.5 cm and 5 cm. Also draws its
circumcircle.

Steps of Construction:
i. Draw a line segment ̅̅̅̅ 𝐴𝐵 = 4𝑐𝑚.
ii. At point “B” draw an arc of radius 4.5cm.
iii. At point “A” draw an arc of radius 5cm.
iv. Both arcs intersect at point C.
v. Join A, B to C.
vi. ∆𝐴𝐵𝐶 is a required triangle.
vii. Draw right bisector of 𝐴𝐵 ̅̅̅̅ , 𝐵𝐶
̅̅̅̅ 𝑎𝑛𝑑 𝐶𝐴
̅̅̅̅.
viii. All right bisector can pass through O.
ix. Draw radius ̅̅̅̅ ̅̅̅̅ 𝑎𝑛𝑑 𝑂𝐶
𝑂𝐴, 𝑂𝐵 ̅̅̅̅ .
x. ̅̅̅̅, 𝑂𝐵
Draw a circle of radius 𝑂𝐴 ̅̅̅̅ 𝑜𝑟 𝑂𝐶
̅̅̅̅ , which is the required circumcircle of
the given triangle.
Construction Steps of Construction
04 Marks 04 Marks