Class VIII
Class VIII (Summative Assessment I)
MATHEMATICSMathematics
SUMMATIVE Assessment (SA I)
Model Practice Paper no. 01
Key features of Model Practice Papers:
 Model Practice Papers are based on the entire syllabus of the subjects
covered under CBSE Board Examinations.
 They are prepared exactly as per the papers set in the CBSE Board
Examinations.
 The questions are picked/ set in such a way that students get
acquainted with each and every concept of the syllabus.
 All the questions provided in the practice papers have detailed and
authentic solutions.
 Practice Papers will provide sufficient practice before the actual
examination for obtaining high scores.

It enables the student to use an effective study method and “have a
go” before the final examination.
1
Class VIII (Summative Assessment I)
Time allowed: 3 hours
Mathematics
Maximum Marks: 90
General Instructions:
(i)
The question paper consists of four Sections – A, B, C and D. Section – A consists
of 8 questions of 1 mark each; Section – B consists of 6 questions of 2 marks each;
Section – C consists of 10 questions of 3 marks each and Section-D consists of 10
questions of 4 marks each. Question No. 1 to 8 are multiple Choice Questions
where you are to select only one correct option out of four given options.
(ii)
All questions are compulsory.
(iii)
In questions on construction, the drawing should be neat and exactly as per the
given measurements. Use rule and compass only.
(iv)
There is no overall choice. However, internal choices have been given in some
questions.
2
Class VIII (Summative Assessment I)
Mathematics
SECTION – A
Q1.
Multiplicative inverse of
(a)
(c)
Q2.
(c)
Q4.
Q5.
Q6.
−5
−3
7
is
(b)
56
8
8
×
×
7
−3
5
8
×
3
7
(d) none of these
The value of (8–1 + 9–1) –1 ÷ (4–1 + 9–1)–1 is
(a)
Q3.
15
−5
26
(b)
17
19
−6
17
(d) none of these
17
The value of √0.3136 is
(a) 0.046
(b) 0.56
(c) 3
(d) none of these
3
The value of √1331 is
(a) 19
(b) 11
(c) 13
(d) None of these
A two digit number whose one’s digit is 𝑥 and ten’s digit is 𝑦 is
(a) 10𝑥 + 𝑦
(b) 10𝑦 + 𝑥
(c) 10𝑥 − 𝑦
(d) 10𝑦 − 𝑥
The measures of three angles of a quadrilateral are 30°, 145° and 18°. Find the
fourth angle.
(a) 107°
(b) 167°
(c) 197°
(d) None of these
3
Class VIII (Summative Assessment I)
Q7.
Mathematics
The area of a rhombus is 240 cm2 and one of the diagonals is 16 cm, then other
diagonal is
Q8.
(a) 20 cm
(b) 30 cm
(c) 16 cm
(d) None of these
The height of a cylinder whose radius 7 cm and the total surface area 968 cm 2 is
(a) 14 cm
(b) 15 cm
(c) 25 cm
(d) None of these
4
Class VIII (Summative Assessment I)
Mathematics
SECTION – B
Q9.
Find the absolute value of (
−1
3
÷
7
−3
).
OR
The product of two rational numbers is
−28
81
. If one of the number is
14
27
then find
the other number.
1 2
1 3
Q10. Simplify: [( ) − ( )
2
4
] × 23
Q11. Prove: (6, 8, 10) is a Pythagorean triplet.
Q12. Solve the following equations and check your answer:
(i) 4(𝑥 + 3) − 2(𝑥 − 1) = 3𝑥 + 3
(ii) 0.26𝑥 + 0.09𝑥 = 8 − 0.45𝑥
Q13. Find the measure of each exterior angle of a regular polygon of:
(i) 6 sides
(ii) 9 sides
Q14. Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively
and the distance between them is 14 cm. Find the area of the trapezium.
5
Class VIII (Summative Assessment I)
Mathematics
SECTION – C
Q15. Find: |
−4
3
+
5
8
17
|÷
24
Q16. Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
1
−3
Q17. Find value of 𝑎6 − (𝑏 + 1)3 when 𝑎 = 2, 𝑏 = 4 .
Q18. By what number should (
−2 3
3
)
9 −2
be divided so that the quotient is ( ) ?
4
Q19. Find the cost of erecting a fence around a square field whose area is 16 hectares
if fencing costs Rs 25 per metre.
OR
Show that 6292 is not a perfect square.
Q20. What least number must be added to 5607 to make the sum a perfect square?
Find this perfect square and its square root.
3
3
Q21. Find the value of √968 × √1375.
Q22. Write the cubes of first sixe natural numbers.
OR
Prove that the cube root of a negative perfect cube is negative.
Q23. Solve:
𝑥 2 −9
5+𝑥 2
=
−5
9
for any value of 𝑥.
6
Class VIII (Summative Assessment I)
Mathematics
Q24. A rectangular piece of paper 22 cm × 6 cm is folded without overlapping to
make a cylinder of height 6 cm. Find the volume of the cylinder.
OR
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete
revolutions to move once over to level playground. Find the area of the
playground in m2.
7
Class VIII (Summative Assessment I)
Mathematics
SECTION – D
1
1
and .
8
2
Q25. Find four rational numbers between
Q26. Simplify:
1 8
[(− )
3
Q27. Evaluate √1
÷ (−
56
169
1 5
1 5
3
3
) ] − [(− )
÷ (−
1 3
3
) ]
. [Using division method]
Q28. Find the cube of each of the following:
(i) (–6)
(ii) 3.3
(iii) 2
(iv)
1
2
3
4
Q29. Prema receives a certain amount of money on her retirement from her
employer. She gives half of this money and an additional sum of R 10,000 to her
daughter. She also gives one third of the money received and an additional sum
of R 3000 to her son. If the daughter gets twice as much as the son, find the
amount of money Prema received on her retirement.
OR
The sides (included of right angle) of a right triangle are in the ratio 3 : 4. A rectangle
is described on its hypotenuse, the hypotenuse being the longer side of the
rectangle. The breadth of the rectangle is four-fifth of its length. Find the shortest
side of the right triangle, if the perimeter of the rectangle is 180 cm.
8
Class VIII (Summative Assessment I)
Mathematics
Q30. In the adjoining figure, ABCD is a parallelogram, AO and BO are the bisectors of
∠A and ∠B respectively. Prove that ∠AOB = 90°.
Q31. Construct a quadrilateral ABCD in which BC = 6 cm, CD = 6 cm, DA = 4 cm,
∠𝐶 = 60° and ∠𝐷 = 75°.
OR
ABCD is a kite in which ∠OBC = 20𝑂 and ∠OCD = 35𝑂 , find
(i) ∠ ABC
(ii) ∠ ADC
(iii) ∠ BAD
Q32. Find the area of the field shown alongside. All dimensions are in metres.
Q33. Find the volume of wood required for making a closed box with external
measurements 14 cm by 9.5 cm by 6 cm, and wood is 7.5 mm thick.
Q34. To prove that the diagonals of a rhombus bisect each other at right angles.
9
Class VIII (Summative Assessment I)
Mathematics
SOLUTION OF MODEL PRACTICE PAPER NO. 01
SECTION – A
−5
×
A1.
(c) Multiplicative inverse of
A2.
(a) (8−1 + 9−1 )−1 ÷ (4−1 + 9−1 )−1
8
−3
7
is
8
−5
7
×
−3
1 1 −1
1 1 −1
9+8 −1
9+4 −1
=( + )
÷( + )
=(
) ÷( )
8 9
4 9
72
36
17 −1
=( )
72
13 −1
÷( )
36
=
72
17
÷
36
13
=
72
17
×
13
36
=
26
17
A3.
(b) √0.3136 = 0.56
A4.
(b) √1331 = √11 × 11 × 11 = 11
A5.
(b) A two digit number whose one’s digit is 𝑥 and ten’s digit is 𝑦 is (10𝑦 + 𝑥).
A6.
(b) Given three angles are 30𝑜 , 145𝑜 and 18𝑜 .
3
3
Let the measure of the fourth angle be 𝑥 𝑜 .
Since, 30𝑜 + 145𝑜 + 18𝑜 + 𝑥 𝑜 = 360𝑜
⟹
A7.
𝑥 𝑜 = 360𝑜 − 193𝑜
⟹
𝑥 𝑜 = 167𝑜
(b) Length of one diagonal = 16 cm
∴ Area of the rhombus =
So, 240 =
1
2
1
2
× Length of one diagonal × length of the other diagonal
× 16 × length of the other diagonal
Length of the other diagonal = 30 cm
A8.
(b) Total surface area = 2𝜋𝑟(ℎ + 𝑟), where ℎ = height of the cylinder
𝑟 = radius = 7 cm
∴
22
968 = 2 × 7 × 7 × (ℎ + 7)
⟹
ℎ = 15 cm
10
Class VIII (Summative Assessment I)
Mathematics
SECTION – B
A9.
|
−1
3
7
÷
−3
|=|
−1
3
×
−3
7
1
1
7
7
|=| |=
OR
According to question,
14
27
𝑎
⟹
A10.
𝑏
1 2
[( )
2
×
=
−
𝑎
𝑏
=
−28
81
𝑎
𝑏
is the other rational number then
−28
81
×
27
14
=
1 3
( ) ] × 23
4
=
−2
3
[
1
22
−
1
43
] × 23
1 1
16−1
15
15
= [4 − 64] × 8 = [ 64 ] × 8 = 64 × 8 = 8 .
A11. For three numbers to be Pythagorean triplet, they should be expressible as
(2𝑚, 𝑚2 − 1, 𝑚2 + 1)
Let 𝑚 = 3, then 2𝑚 = 2 × 3 = 6 , 𝑚2 − 1 = 32 − 1 = 8, 𝑚2 + 1 = 32 + 1 = 10
So, (6, 8, 10) is a Pythagorean triplet.
A12. (i) 4(𝑥 + 3) − 2(𝑥 − 1) = 3𝑥 + 3
⟹
4𝑥 + 12 − 2𝑥 + 2 = 3𝑥 + 3
⟹
4𝑥 − 2𝑥 − 3𝑥 = 3 − 12 − 2
⟹
−𝑥 = −11
⟹ 𝑥 = 11
(ii) 0.26𝑥 + 0.09𝑥 = 8 − 0.45𝑥
⟹
0.26𝑥 + 0.09𝑥 + 0.45𝑥 = 8
⟹
0.80𝑥 = 8
⟹
𝑥 = 10
11
Class VIII (Summative Assessment I)
Mathematics
360 𝑜
A13. (i) Each exterior angle of a regular polygon of 6 sides = (
) = 60𝑜
6
360 𝑜
(ii) Each exterior angle of a regular polygon of 9 sides = (
) = 40𝑜
9
A14. Area of the trapezium =
=
1
2
1
2
× (sum of parallel sides) × (distance between them)
1
× (27 + 19) × 14 = (2 × 46 × 14 ) cm2 = 322 cm2
12
Class VIII (Summative Assessment I)
Mathematics
SECTION – C
A15. We have,
|
−4
5
+
3
8
|÷
17
24
−32+15
17
| ÷ 24
24
−17
17 17 17
= | 24 | ÷ 24 = 24 ÷ 24
17 24
= 24 × 17 = 1
=|
A16. (i) 0 is a rational number but its reciprocal is not defined.
(ii) 1 and −1 are the rational numbers that are equal to their reciprocals.
(iii) 0 is the rational number that is equal to its negative.
3
1 6
−3
A17. 𝑎6 − (𝑏 + 1)3 = (2) − ( 4 + 1)
1
−3 1 3
= 6 − ( 4 + 1)
2
1
−3+4 3
= 6−( 4 )
2
1
1 3
= 6 − (4 )
2
1
1
1
1
1−1
= 6−
= 6− 6= 6 =0
3
2
2
2
2
2
(2 )
A18. Let the required number be 𝑥. Then
−2 3
( )
3
⟹
⟹
⟹
−2 3
9 −2
÷𝑥 =( )
4
( )
×
(−2)3
1
3
33
𝑥=−
×
1
𝑥
𝑥
(2)3
33
=(
4 2
9
)
42
= 92
⟹
𝑥=
(−2)3
33
2
×
(32 )
(22 )2
⟹
𝑥=−
23
33
×
92
42
4
×
3
4
2
⟹
𝑥=−
34−3
24−3
=−
3
2
13
Class VIII (Summative Assessment I)
Mathematics
A19. Area of the square field = (16 × 10000) m2 = 160000 m2
Length of each side of the field = √160000 m = 400 m
Perimeter of the field = (4 × 400) m = 1600 m
Cost of fencing = Rs (1600 × 25) = Rs 40000
OR
Prime factors of 6292
6292 = 2 × 2 × 11 × 11 × 13 = 22 × 112 × 13
Thus, 6292 cannot be expressed as a product of pairs of equal factors.
Hence, 6292 is not a perfect square.
A20. First we try to find out the square root of 5607.
74
7
56 07
–49 ↓
144
7 07
–5 76
131
We observe here that (74)2 < 5607 < (75)2
The required number to be added = (75)2 − 5607 = 5625 − 5607 = 18
Therefore, the required perfect square = 5607 + 18 = 5625
Hence, √5625 = 75.
A21. First writing 968 and 1375 as a product of prime factors, we have
968 = 2 × 2 × 2 × 11 × 11
1375 = 5 × 5 × 5 × 11
∴
3
3
3
√968 × √1375 = √968 × 1375
3
= √2 × 2 × 2 × 11 × 11 × 5 × 5 × 5 × 11
3
= √2 × 2 × 2 × 5 × 5 × 5 × 11 × 11 × 11
= 2 × 5 × 11 = 110
14
Class VIII (Summative Assessment I)
Mathematics
A22. Cubes of first 6 natural are as following
Natural
Numbers
1
Cube
2
8
3
27
4
64
5
125
6
216
1
OR
Let negative number be (– 8)
Resolving (–8) into prime factors, we have
–8 = (–2) × (–2) × (–2)
∴
3
3
√– 8 = √(– 2) × (– 2) × (– 2) = –2
The cube root of a negative perfect cube is negative.
A23. We have,
𝑥 2 −9
5+𝑥 2
=
proved
−5
9
Cross- multiplying, we get
9(𝑥 2 − 9) = −5(5 + 𝑥 2 )
⟹
9𝑥 2 − 81 = −25 − 5𝑥 2
⟹
9𝑥 2 + 5𝑥 2 = 81 − 25
⟹
14𝑥 2 = 56
⟹
𝑥2 = 4
⟹
𝑥 = ±2
15
Class VIII (Summative Assessment I)
Mathematics
A24. Clearly, the length of the given rectangle is the perimeter of the base of the
given cylinder and breadth of the rectangle is the height of the cylinder.
∴
height = 6 cm
⟹ ℎ = 6 cm
6 cm
6 cm
Let 𝑟 be the radius of its base.
22
7
Then, 2𝜋𝑟 = 22 cm ⟹ 2 × 7 × 𝑟 = 22 cm ⟹ 𝑟 = 2 cm
∴
volume of the cylinder = 𝜋𝑟 2 ℎ cubic units
22 7 7
= ( 7 × 2 × 2 × 6) cm2 = 231 cm2
Hence, the volume of the cylinder is 231 cm2.
OR
Given, diameter of the roller = 84 cm
⟹ Its radius =
84
2
= 42 cm =
42
100
Length of the roller = 120 cm =
m
120
100
m=
6
5
m
Now, area of the playground leveled by the roller in one complete revolution
= lateral surface area of the roller
= 2𝜋𝑟ℎ
22
42
6
= (2 × 7 × 100 × 5) m2
6
6
= (44 × 100 × 5) m2
∴
6
6
Area levelled in 500 complete revolutions = (44 × 100 × 5) × 500 m2
= 1584 m2
Hence, area of the playground = 1584 m2
16
Class VIII (Summative Assessment I)
Mathematics
SECTION – D
A25. First rational number between
1
and
8
1
2
1 1
(
=
2 8
+
1
2
)
1 1+4
1 5
5
= 2 ( 8 ) = 2 × 8 = 16,
1
Second rational number between
8
5
and
16
=
1 1
(
2 8
+
5
),
16
1 2+5
1 7
7
= 2 ( 16 ) = 2 × 16 = 32
Third rational number between
5
16
and
1
=
2
1
(
5
2 16
+
1
2
)
1 5+8
1 13 13
= 2 ( 16 ) = 2 × 16 = 32,
Fourth rational number between
13
32
and
1
2
=
1 13
(
2 32
+
1
2
)
1 13+16
1 29 29
= 2 ( 32 ) = 2 × 32 = 64
Hence, four rational numbers between
A26.
1 8
[(− )
3
⟹
÷ (−
3
1 3
(− )
⟹
(−1)3 ×
⟹
⟹
1 5
3
3
) ] − [(− )
1 8−5
(− )
⟹
⟹
1 5
3
− (−
− (−
1
8
÷ (−
and
1
2
are
7
,
5
,
13 29
,
32 16 32 64
.
1 3
3
) ]
1 5−3
3
)
1 2
3
)
1
1
2
3 − (−1) × 2
3
3
1
1
− 3− 2
3
3
1
1
− 27 − 9
−1−3
−4
⟹
27
27
17
Class VIII (Summative Assessment I)
Mathematics
56
225
A27. We have 1 169 = 169
56
=
169
√1
∴
√
225
169
=
√225
√169
We find √225 and √169 using division method separately, as given below.
15
13
1
2 25
–1
4
1 69
–1
25
1 25
–1 25
23
69
–69
0
0
∴
√225 = 15 and √169 = 13
Thus,
√1
56
169
=
√
225
169
=
15
2
√225
=
=1 .
13
13
√169
A28. (i) (−6)3 = (−6) × (−6) × (−6) = −216
(ii) (3.3)3 = (3.3) × (3.3) × (3.3) = 35.937
(iii)
1 3
(2 )
2
=
3 3
(iv) ( ) =
4
5 3
( )
2
=
5
5
5
5×5×5
2
2
2
2 ×2 ×2
( )×( )×( )=
3
3
3
3 ×3 × 3
4
4
4
4 ×4 ×4
( )×( )×( )=
=
=
125
8
27
64
A29. Let the amount of money received by Prema be Rs. 𝑥 .
𝑥 + 20000
𝑥
Money given to her daughter = Rs. (2 + 10000) = Rs. (
)
2
𝑥 + 9000
𝑥
Money given to her son = Rs. (3 + 3000) = Rs. (
)
3
By the given condition, daughter’s share = twice the son’s share
∴
𝑥 + 20000
2
𝑥 + 9000
2𝑥+18000
= 2(
)=
3
3
18
Class VIII (Summative Assessment I)
Mathematics
⇒
3(𝑥 + 20000) = 2(2𝑥 + 18000)
⇒
3𝑥 + 60000 = 4𝑥 + 36000
⇒
4𝑥 − 3𝑥 = 36000 − 60000
⇒
𝑥 = 24000
∴
Amount of money received by Prema = Rs. 24000
OR
Let sides of right triangle are 3𝑥 and 4𝑥 .
Then, hypotenuse = √sum of squares of sides
= √(3𝑥)2 + (4𝑥)2 = √25𝑥 2 = 25𝑥
∴
Breadth of the rectangle =
4
5
(5𝑥) = 4𝑥
[By given]
Perimeter of the rectangle = 2(length + breadth) = 2(5𝑥 + 4𝑥)
By given, 2(5𝑥 + 4𝑥) = 180
10𝑥 + 8𝑥 = 180
⟹
𝑥 = 10
The shortest side of the right triangle = 4𝑥 = 4 × 10 = 40.
A30. We know that the sum of two adjacent angles of a parallelogram is 180𝑂
∴
∠𝐴 + ∠𝐵 = 180𝑂
…(i)
Since AO and BO are the bisectors of ∠𝐴 and ∠𝐵 respectively. We have
1
1
∠𝑂𝐴𝐵 = 2 ∠𝐴 and ∠𝐴𝐵𝑂 = 2 ∠𝐵
From ∆𝑂𝐴𝐵, we have ∠𝑂𝐴𝐵 + ∠𝐴𝑂𝐵 + ∠𝐴𝐵𝑂 = 180𝑂
⟹
⟹
⟹
⟹
1
2
1
2
1
2
∠𝐴 + ∠𝐴𝑂𝐵 +
1
2
∠𝐵 =180𝑂
(∠𝐴 + ∠𝐵 ) + ∠𝐴𝑂𝐵 =180𝑂
× 180𝑂 + ∠𝐴𝑂𝐵 =180𝑂
[from (i)]
90𝑂 + ∠𝐴𝑂𝐵 =180𝑂
Hence, ∠𝐴𝑂𝐵 = 90𝑂
19
Class VIII (Summative Assessment I)
Mathematics
A31. First we draw a rough sketch of the quadrilateral ABCD, and write down its
dimensions as shown.
Steps of Construction:
(i) Draw BC = 6 cm.
(ii) Construct ∠𝐵𝐶𝐷 = 60𝑂 and cut off CD = 6 cm.
(iii) Construct ∠𝐶𝐷𝐴 = 75𝑂 and cut off DA = 4 cm.
(iv) Join AB.
Then ABCD is the required quadrilateral.
OR
20
Class VIII (Summative Assessment I)
Mathematics
(i) Since diagonal BD bisects ∠ ABC, therefore
∠ABC = 2 ∠ OBC = 2 × 20𝑜 = 40𝑜
(ii) We know that the diagonals of a kite intersect at right angles
∠ DOC = 90𝑜
∠ DOC = 180𝑜 − (∠𝐷𝑂𝐶 + ∠ 𝑂𝐶𝐷)
= 180𝑜 − (90𝑜 + 35𝑜 ) = 180𝑜 − 125𝑜 = 55𝑜
As diagonal BD bisects ∠ ADC, so ∠ ADC = 2 ∠ ODC = 2× 55𝑜 = 110𝑜
(iii) In ∆ OBC, ∠OCB = 180𝑜 − (∠𝐵𝑂𝐶 + ∠ 𝑂𝐵𝐶)
= 180𝑜 − (90𝑜 + 20𝑜 ) = 180𝑜 − 110𝑜 = 70𝑜
∴ ∠ BCD = ∠ OCB + ∠ OCD = 70𝑜 + 35𝑜 = 105𝑜
⟹
∠ BAD = 105𝑜
[By property of a kite, ∠ BAD = ∠ BCD]
21
Class VIII (Summative Assessment I)
Mathematics
A32. Area of field ABCDE = Area of ∆APB + Area of trap. PBCR + Area of ∆RCD
+ Area of ∆DQE + Area of ∆EQA
∴
Area of ∆APB =
1
2
× AP × PB =
Area of trap. PBCR =
Area of ∆RCD =
Area of ∆DQE =
Area of ∆ EQA =
1
2
1
2
1
2
1
2
1
2
× 30 × 20 = 300 m2
(BP + RC) × PR =
× RC × RD =
× DQ × EQ =
× EQ × AQ =
1
2
1
2
1
2
1
2
(20 + 40) × 40 m2 = 1200 m2
× 40 × 30 = 600 m2
× 50 × 30 = 750 m2
× 30 × 50 = 750 m2
Adding these, area of the field = (300 + 1200 + 600 + 750 + 750) m2 = 3600 m2.
A33. External dimensions are 14 cm, 9.5 cm, and 6 cm. Thickness of wood = 7.5 mm.
To get the internal dimensions, subtract double the thickness from each external
dimension.
∴
Internal length = 14 cm – 15 mm = 14 cm – 1.5 cm = 12.5 cm
Internal breadth = 9.5 cm – 1.5 cm = 8 cm,
22
Class VIII (Summative Assessment I)
Mathematics
Internal height = 6 cm – 1.5 mm = 4.5 cm
External volume of the box = (14 × 9.5 × 6) cm3 = 798 cm3
Internal volume of the box = (12.5 × 8 × 4.5) cm3 = 450 cm3
∴
Volume of wood = External volume – Internal volume
= 798 cm3 – 450 cm3 = 348 cm3
A34. Given- OA = OC; OB = OD; ∠ AOB = ∠ COB = 90𝑜
Proof- In ∆’s OAB and ODC
⟹
∠OAB = ∠ OCD
[Where, 𝐴𝐵 ∥ 𝐶𝐷, alternate angles]
∠OBA = ∠ ODC
[Where, 𝐴𝐵 ∥ 𝐶𝐷, alternate angles]
AB = CD
[Opposite sides of a rhombus]
∆𝑂𝐴𝐵 ≅ ∆𝑂𝐷𝐶
[From Angle-Side-Angle]
OA = OC, OB = OD
[By c.p.c.t.]
Again in triangles OAB and OBC
OA = OC
Proved
OB = OB
[Common]
AB = BC
[Sides of a rhombus]
∆𝑂𝐴𝐵 ≅ ∆𝑂𝐵𝐶
[From Side-Side-Side]
⟹
∠ AOB = ∠ COB
[By c.p.c.t.]
But
∠ AOB + ∠ COB = 180𝑜
[Linear Pair]
∴
∠ AOB = ∠ COB = 90𝑜
Proved.
23