Class 10 CBSE Mathematics Question Bank - 1000 Two Marks Questions

Class 10 CBSE Mathematics Question Bank

100 Important Questions | 2 Marks Each | Chapter Wise

Total Questions: 100 | 10 Chapters | 2 Marks Each

Solution:
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0

Therefore, HCF = 45.

Solution:
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0

HCF = 51.

Solution:
12 = 2² × 3
18 = 2 × 3²

HCF = 2 × 3 = 6
LCM = 2² × 3² = 36.

Solution:
140 = 14 × 10
= 2 × 7 × 2 × 5

Therefore,
140 = 2² × 5 × 7.

Solution:
Suppose 3√5 is rational.
Then √5 = (3√5)/3 would also be rational. But √5 is irrational. This is a contradiction. Therefore, 3√5 is irrational.

Solution:
A number ending in 0 must have factors 2 and 5. But 6ⁿ = (2 × 3)ⁿ contains no factor 5. Hence 6ⁿ cannot end with digit 0.

Solution:
404 = 96 × 4 + 20
96 = 20 × 4 + 16
20 = 16 × 1 + 4
16 = 4 × 4 + 0

HCF = 4.

Solution:
We know, HCF × LCM = Product of two numbers 6 × LCM = 18 × 30 LCM = 540/6 LCM = 90.

Solution:
160 = 2⁵ × 5. The denominator has only prime factors 2 and 5. Therefore, 13/160 has a terminating decimal expansion.

Solution:
30 = 2 × 3 × 5. The denominator contains the prime factor 3 other than 2 and 5. Therefore, 7/30 has a non-terminating recurring decimal expansion.

Solution:
x² − 5x + 6
= x² − 2x − 3x + 6
= x(x − 2) − 3(x − 2)
= (x − 2)(x − 3) Therefore, x = 2, 3.

Solution:
For ax² + bx + c, Sum of zeroes = −b/a = 7/2 Product of zeroes = c/a = 3/2.

Solution:
α + β = 8
αβ = 15.

Solution:
Required polynomial: (x − 3)(x + 4) = x² + x − 12. Therefore, x² + x − 12.

Solution:
Since 3 is a zero, 3² + 3k − 12 = 0 9 + 3k − 12 = 0 3k = 3 k = 1.

Solution:
For reciprocal zeroes, αβ = 1. But αβ = k. Therefore, k = 1.

Solution:
3x² − 12x = 3x(x − 4) Therefore, x = 0 or x = 4. Zeroes are 0 and 4.

Solution:
α + β = −b/a = −(−5)/4 = 5/4.

Solution:
For x² − 6x + k, Sum of zeroes = −b/a = −(−6)/1 = 6. Hence verified.

Solution:
Required polynomial: x² − (sum)x + product = x² − 5x + 6. Answer: x² − 5x + 6.

Solution:
Adding both equations: 2x = 8 x = 4 Therefore, y = 7 − 4 = 3. x = 4, y = 3.

Solution:
x + y = 5 x = 5 − y Substitute: 2(5 − y) + 3y = 13 10 + y = 13 y = 3 x = 2. Answer: x = 2, y = 3.

Solution:
a₁/a₂ = 2/4 = 1/2 b₁/b₂ = 3/6 = 1/2 c₁/c₂ = 5/10 = 1/2 All three ratios are equal. Therefore, the equations have infinitely many solutions.

Solution:
x + y = 5 x = 5 − y 3(5 − y) + 2y = 12 15 − y = 12 y = 3 x = 2. x = 2, y = 3.

Solution:
For infinitely many solutions: 2/4 = 3/k 1/2 = 3/k k = 6. k = 6.

Solution:
3x + y = 9 y = 9 − 3x 5x − 2(9 − 3x) = 4 11x = 22 x = 2 y = 3. x = 2, y = 3.

Solution:
For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Graphically, the two lines are parallel.

Solution:
Adding: 6x = 12 x = 2 Then, y = 11 − 8 = 3. x = 2, y = 3.

Solution:
For infinitely many solutions: 1/3 = 2/k = 5/15 Thus, 2/k = 1/3 k = 6.

Solution:
Let numbers be x and y. x + y = 25 x − y = 7 Adding: 2x = 32 x = 16 y = 9. Numbers are 16 and 9.

Solution:
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
Therefore, x = 2, 3.

Solution:
x² − 9 = 0
(x − 3)(x + 3) = 0
x = 3, −3.

Solution:
D = b² − 4ac
= (−5)² − 4(2)(3)
= 25 − 24
D = 1.

Solution:
D = 4² − 4(1)(5)
= 16 − 20 = −4.
Since D < 0, the equation has no real roots.

Solution:
2x² − 7x + 3 = 0
(2x − 1)(x − 3) = 0
x = 1/2, 3.

Solution:
For equal roots, D = 0.
k² − 36 = 0
k² = 36
k = ±6.

Solution:
x² − 10x + 25 = (x − 5)²
Therefore,
x = 5, 5.

Solution:
D = 0
36 − 4k = 0
4k = 36
k = 9.

Solution:
3x² − x − 2 = 0
(3x + 2)(x − 1) = 0
x = −2/3, 1.

Solution:
The roots are given by:

x = (−b ± √(b² − 4ac)) / 2a.

Solution:
a = 3, d = 4
a₁₀ = a + 9d
= 3 + 36
= 39.

Solution:
d = 17 − 12
d = 5.

Solution:
a = 5, d = 3
a₁₅ = 5 + 14(3)
= 5 + 42
= 47.

Solution:
Sₙ = n(n + 1)/2
S₁₀ = 10 × 11 / 2
= 55.

Solution:
a = 2, d = 3, n = 20
Sₙ = n/2[2a + (n−1)d]
= 10[4 + 57]
= 610.

Solution:
The nth term of an AP is:
aₙ = a + (n − 1)d.

Solution:
a = 7, d = 3
a₂₀ = 7 + 19(3)
= 64.

Solution:
81 = 5 + (n−1)4
76 = 4(n−1)
n−1 = 19
n = 20.

Solution:
a = 4, d = 3, n = 15
S = 15/2[8 + 42]
= 15/2 × 50
= 375.

Solution:
3 + (n−1)4 = 47
n = 12
Middle terms are 6th and 7th terms.
a₆ = 23, a₇ = 27.
Middle terms: 23 and 27.

Solution:
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Solution:
In a right-angled triangle,
(Hypotenuse)² = (Base)² + (Perpendicular)².

Solution:
Corresponding angles of similar triangles are equal. Their corresponding sides are proportional.

Solution:
For similar triangles, corresponding sides are proportional.
Therefore,
BC/EF = 3/5.

Solution:
If two angles of one triangle are respectively equal to two angles of another triangle, then the triangles are similar.

Solution:
h² = 6² + 8²
= 36 + 64 = 100
h = 10 cm.

Solution:
6/3 = 8/4 = 10/5 = 2.
Corresponding sides are proportional.
Therefore, the triangles are similar.

Solution:
Ratio of areas = square of ratio of corresponding sides.
= (2/3)²
= 4:9.

Solution:
If the corresponding sides of two triangles are proportional, then the triangles are similar by SSS similarity criterion.

Solution:
x² + 5² = 13²
x² = 169 − 25 = 144
x = 12 cm.

Solution:
Distance = √[(3−0)² + (4−0)²]
= √25
= 5 units.

Solution:
Midpoint = ((2+6)/2, (4+8)/2)
= (4,6).

Solution:
d = √[(4+2)² + (−5−3)²]
= √(36+64)
= 10 units.

Solution:
Midpoint = ((−3+7)/2, (5−1)/2)
= (2,2).

Solution:
Using section formula:
x = (1×8 + 2×2)/3 = 4
y = (1×9 + 2×3)/3 = 5
Point = (4,5).

Solution:
d = √[(3)² + (4)²]
= 5 units.

Solution:
x = (1×7 + 2×1)/3 = 3
y = (1×8 + 2×2)/3 = 4
(3,4).

Solution:
Slope of first two points = 1.
Slope of last two points = 1.
Since slopes are equal, the points are
collinear.

Solution:
Let point be (x,0).
Equating distances:
(x−2)² + 9 = (x−4)² + 25
Solving gives x = 6.
Point = (6,0).

Solution:
Base = 4, Height = 3
Area = 1/2 × 4 × 3
= 6 square units.

Solution:
sin θ = Perpendicular / Hypotenuse.

Solution:
sin 30° = 1/2
cos 60° = 1/2.

Solution:
tan 45° = 1
sin 30° = 1/2
Therefore,
3/2.

Solution:
By Pythagoras theorem:
P² + B² = H².
Dividing by H²:
sin²θ + cos²θ = 1.

Solution:
sec 60° = 1/cos60° = 2
cosec30° = 1/sin30° = 2.

Solution:
Take perpendicular = 3, base = 4.
Hypotenuse = 5.
sin θ = 3/5.
Answer = 3/5.

Solution:
cot45° = 1
tan45° = 1
Sum = 2.

Solution:
Perpendicular = 5, hypotenuse = 13.
Base = 12.
cos θ = 12/13.
Answer = 12/13.

Solution:
= 2(1/2) + 3(1/2)
= 1 + 3/2
= 5/2.

Solution:
Using identity:
sin²θ + cos²θ = 1
Therefore,
Answer = 1.

Solution:
The angle made by the line of sight with the horizontal when the object is above the observer is called angle of elevation.

Solution:
The angle made by the line of sight with the horizontal when the object is below the observer is called angle of depression.

Solution:
tan45° = h/10
1 = h/10
h = 10 m.

Solution:
tan45° = h/20
1 = h/20
h = 20 m.

Solution:
Trigonometric ratios establish relationships between angles and sides of right triangles. Hence inaccessible heights and distances can be calculated.

Solution:
sin30° = h/10
1/2 = h/10
h = 5 m.

Solution:
tan60° = 12/x
√3 = 12/x
x = 4√3 m.

Solution:
sin30° = 30/l
1/2 = 30/l
l = 60 m.

Solution:
tan60° = h/10
√3 = h/10
h = 10√3 m.

Solution:
tan45° = h/25
1 = h/25
h = 25 m.

Solution:
A tangent is a line that touches a circle at exactly one point.

Solution:
No tangent can be drawn from a point lying inside a circle.
Answer: 0.

Solution:
Exactly one tangent can be drawn from a point on the circle.
Answer: 1.

Solution:
Two tangents can be drawn from an external point to a circle.
Answer: 2.

Solution:
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution:
Tangents drawn from an external point to a circle are equal.
Therefore,
PA = PB.

Solution:
The shortest distance from the centre to a tangent is the radius. Therefore, the radius through the point of contact is perpendicular to the tangent.

Solution:
Each tangent = 8 cm.
Total length = 8 + 8
= 16 cm.

Solution:
The radius is perpendicular to the tangent.
Therefore, the angle is
90°.

Solution:
Tangents drawn from an external point to a circle are equal in length.

Therefore, PA = PB.

Note:
Class 10 CBSE Mathematics Question Bank में प्रश्न 1–100 तक कुल 100 प्रश्न दिए गए हैं। प्रत्येक प्रश्न 2 अंक का है। किसी भी प्रश्न पर क्लिक करके उसका समाधान देखा जा सकता है।

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Solution:
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
Therefore, HCF = 45.

Solution:
72 = 2³ × 3², 120 = 2³ × 3 × 5.
LCM = 2³ × 3² × 5 = 360.

Solution:
Assume √7 = p/q, where p and q are coprime integers.
Then p² = 7q², so 7 divides p. Let p = 7k.
Thus q² = 7k², so 7 divides q also, contradicting that p and q are coprime.
Hence, √7 is irrational.

Solution:
If 3 + 2√5 were rational, then 2√5 would be rational and hence √5 would be rational. This is impossible because √5 is irrational.
Therefore, 3 + 2√5 is irrational.

Solution:
26 = 2 × 13 and 91 = 7 × 13.
HCF = 13 and LCM = 182.

Solution:
3125 = 5⁵. Its prime factors contain only 5.
Therefore, the decimal expansion of 13/3125 is terminating.

Solution:
90 = 2 × 3² × 5. Since the denominator contains prime factor 3, the decimal expansion is non-terminating recurring.

Solution:
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
HCF = 51.

Solution:
HCF × LCM = Product of numbers.
12 × LCM = 7200
LCM = 600.

Solution:
A number ending in 0 must have factors 2 and 5. But 6n = 2 × 3 × n does not necessarily contain factor 5. Hence, for arbitrary natural n, it cannot be concluded that it ends in 0.

Solution:
Assume √3 = p/q in lowest form. Then p² = 3q², so 3 divides p. Hence 3 also divides q, contradicting coprimality. Therefore, √3 is irrational.

Solution:
250 = 2 × 5³. Since the denominator contains only 2 and 5, the decimal expansion is terminating.

Solution:
18 = 2 × 3², 24 = 2³ × 3, 30 = 2 × 3 × 5.
LCM = 2³ × 3² × 5 = 360.

Solution:
404 = 96 × 4 + 20
96 = 20 × 4 + 16
20 = 16 × 1 + 4
16 = 4 × 4 + 0
HCF = 4.

Solution:
Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of the prime factors.

Solution:
x² − 7x + 12 = (x − 3)(x − 4).
Zeroes = 3 and 4.

Solution:
For ax² + bx + c,
Sum = −b/a = 5/3
Product = c/a = 2/3.

Solution:
α + β = 9 and αβ = 14.

Solution:
Polynomial = (x − 4)(x + 3)
= x² − x − 12.
Required polynomial = x² − x − 12.

Solution:
Putting x = 1:
2 + 5 + k = 0
Therefore, k = −7.

Solution:
x² − 16 = (x − 4)(x + 4).
Zeroes = 4, −4.

Solution:
Sum of zeroes = −p.
3 + 4 = −p
Therefore, p = −7.

Solution:
(x + 2)(x − 5) = x² − 3x − 10.
Required polynomial = x² − 3x − 10.

Solution:
Required polynomial = x² − (α+β)x + αβ
= x² − 6x + 5.

Solution:
2x² − 8x = 2x(x − 4).
Zeroes = 0 and 4.

Solution:
α + β = 4, αβ = 1.
α² + β² = (α+β)² − 2αβ = 16 − 2 = 14.

Solution:
4 + 2k − 6 = 0
2k = 2
k = 1.

Solution:
Sum = −b/a = −3/5.

Solution:
Product = c/a = 9/7.

Solution:
Sum of zeroes = 5.
Other zero = 5 − 2 = 3.

Solution:
Adding: 2x = 8 ⇒ x = 4.
Then y = 3.
x = 4, y = 3.

Solution:
Adding equations: 3x = 9 ⇒ x = 3.
y = 2.
(x,y) = (3,2).

Solution:
a₁/a₂ = 2/4 = 1/2 and b₁/b₂ = 3/6 = 1/2.
Both equations represent the same line.
Therefore, there are infinitely many solutions.

Solution:
From x+y=5, x=5−y.
3(5−y)+2y=12 ⇒ y=3.
x=2.
(2,3).

Solution:
For infinitely many solutions:
k/4 = 3/6 = 6/12 = 1/2.
Therefore, k = 2.

Solution:
Adding: 8x = 16 ⇒ x=2.
10−2y=4 ⇒ y=3.
(2,3).

Solution:
For no solution:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
The two lines are parallel and distinct.

Solution:
From second equation, y=2x−4.
4x+3(2x−4)=18 ⇒ 10x=30 ⇒ x=3.
y=2.
(3,2).

Solution:
For no solution: 3/6 = k/10 ≠ 5/8.
Thus k/10 = 1/2 ⇒ k = 5.

Solution:
3x+2y=30 and x−y=3.
x=y+3.
3y+9+2y=30 ⇒ y=21/5, x=36/5.

Solution:
2x=14 ⇒ x=7.
y=3.
x=7, y=3.

Solution:
Adding: 10x=20 ⇒ x=2.
14+2y=19 ⇒ y=5/2.
(2, 5/2).

Solution:
A pair has a unique solution when
a₁/a₂ ≠ b₁/b₂.
The two lines intersect at exactly one point.

Solution:
x=8−2y.
16−4y+5y=19 ⇒ y=3.
x=2.
(2,3).

Solution:
2/4 = 3/k = 5/10.
1/2 = 3/k ⇒ k=6.

Solution:
y=5−2x.
4x−3(5−2x)=5 ⇒ 10x=20 ⇒ x=2.
y=1.
(2,1).

Solution:
Taking points (0,6) and (6,0), joining them gives a straight line.
It is a linear graph.

Solution:
The coordinates of the point of intersection satisfy both equations.
Hence, the pair has one unique solution.

Solution:
Adding: 5x=16 ⇒ x=16/5.
Then y=(10−48/5)/4 = 1/10.
(16/5, 1/10).

Solution:
Example:
x + 2y = 5
2x + 4y = 12
Here a₁/a₂ = b₁/b₂ but c₁/c₂ is different.
Therefore, no solution.

Solution:
(x−2)(x−3)=0.
Therefore, x = 2, 3.

Solution:
D = b²−4ac = 16−24 = −8.

Solution:
D = 36−36=0.
Hence, roots are real and equal.

Solution:
2x²−7x+3=(2x−1)(x−3).
x = 1/2, 3.

Solution:
For ax²+bx+c=0,
x = (−b ± √(b²−4ac))/(2a).

Solution:
(x−3)(x+3)=0.
Roots = 3, −3.

Solution:
D = 4−60 = −56 < 0.
Therefore, it has no real roots.

Solution:
For equal roots, D=0.
k²−36=0 ⇒ k= ±6.

Solution:
(x+3)(x+4)=0.
x = −3, −4.

Solution:
α+β = −b/a = 8.

Solution:
Product = c/a = −2/3.

Solution:
(2x−5)(2x+5)=0.
x = ±5/2.

Solution:
4−2k+2=0 ⇒ 6−2k=0.
k=3.

Solution:
D = 36−40 = −4 < 0.
Hence, no real roots.

Solution:
(x−5)²=0.
x = 5, 5.

Solution:
D = 16 + 20 = 36.

Solution:
3x(x−4)=0.
x = 0, 4.

Solution:
(x−2)(x−5)=0
Therefore, x²−7x+10=0.

Solution:
D=0 ⇒ k²−64=0.
Therefore, k=±8.

Solution:
(x+4)(x−3)=0.
x = −4, 3.

Solution:
a=3, d=4.
a₁₀ = a+9d = 3+36 = 39.

Solution:
d = 15−18 = −3.

Solution:
a=5,d=3.
a₁₅=5+14×3=47.

Solution:
a=2,d=3,n=10.
S₁₀ = 10/2[4+27] = 155.

Solution:
a=7,d=5.
a₂₀=7+19×5=102.

Solution:
S=20×21/2 = 210.

Solution:
a=100,d=−5.
a₂₅=100+24(−5)=−20.

Solution:
S₁₅=15/2[8+14×3]=375.

Solution:
81=5+(n−1)4.
76=4(n−1) ⇒ n−1=19.
n=20.

Solution:
79=3+(n−1)4 ⇒ n=20.
There are 20 terms, so middle terms are 10th and 11th.
They are 39 and 43.

Solution:
a=10,d=−2,n=12.
S=6[20−22]=−12.

Solution:
43=7+(n−1)3.
36=3(n−1) ⇒ n−1=12.
n=13.

Solution:
Sum of first n odd numbers = n².
S₂₅ = 625.

Solution:
a=−5,d=3.
a₁₂=−5+11×3=28.

Solution:
S₃₀=15[2+29×3]
=15×89=1335.

Solution:
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Solution:
If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.

Solution:
Corresponding angles of similar triangles are equal.

Solution:
Corresponding sides are proportional.
Therefore, BC/EF = 3/5.

Solution:
The criterion is SSS similarity criterion.

Solution:
Ratio of areas = square of ratio of corresponding sides.
= (2/3)² = 4:9.

Solution:
Their areas are proportional to the squares of their corresponding sides.

Solution:
If two angles of one triangle are respectively equal to two angles of another triangle, the triangles are similar.

Solution:
Area ratio = 4²:7² = 16:49.

Solution:
Square of hypotenuse = sum of squares of the other two sides.
c² = a² + b².

Solution:
c²=6²+8²=36+64=100.
c=10 cm.

Solution:
b²=13²−5²=169−25=144.
b=12 cm.

Solution:
The line divides the other two sides proportionally, so the corresponding ratios are equal.

Solution:
Yes. Similar triangles have equal corresponding angles and proportional sides, but their sizes may be different. Hence they need not be congruent.

Solution:
The ratio of perimeters equals the ratio of corresponding sides.
Therefore, 5:8.

Solution:
Scale factor = DE/AB = 9/6 = 3/2.

Solution:
Corresponding altitudes are proportional to their corresponding sides.

Solution:
Corresponding medians are proportional to the corresponding sides.

Solution:
Side ratio = √(25/49) = 5:7.

Solution:
AB/DE = BC/EF.
4/6 = 5/EF.
EF = 7.5 cm.

Solution:
d=√(3²+4²)=5 units.

Solution:
Midpoint = ((2+6)/2,(4+8)/2) = (4,6).

Solution:
d=√[(5−2)²+(7−3)²]=√25=5 units.

Solution:
Midpoint = (2,2).
Answer: (2,2).

Solution:
d=√(6²+8²)=10 units.

Solution:
Using section formula:
((1×8+2×2)/3,(1×9+2×3)/3)
= (4,5).

Solution:
Point = ((2×7+1×1)/3,(2×8+1×2)/3)
= (5,6).

Solution:
Area = 1/2 × 4 × 3 = 6 square units.

Solution:
d=√(25+144)=√169=13 units.

Solution:
Midpoint = ((−4+2)/2,(−6+4)/2)
= (−1,−1).

Solution:
If points are (x₁,y₁), (x₁,y₂), distance = |y₂−y₁|.

Solution:
Distance = |x₂−x₁|.

Solution:
Base=3, height=4.
Area=1/2×3×4=6 square units.

Solution:
Let point be (x,0). Equating distances gives:
(x−2)²+9=(x−4)²+25.
x=6.
Point = (6,0).

Solution:
For (x₁,y₁) and (x₂,y₂):
d = √[(x₂−x₁)²+(y₂−y₁)²].

Solution:
sin θ = Perpendicular / Hypotenuse.

Solution:
cos θ = Base / Hypotenuse.

Solution:
tan θ = Perpendicular / Base.

Solution:
sin 30° = 1/2.

Solution:
cos 60° = 1/2.

Solution:
tan 45° = 1.

Solution:
sin 60° = √3/2.

Solution:
cos 30° = √3/2.

Solution:
tan 60° = √3.

Solution:
sin 90° = 1.

Solution:
cos 90° = 0.

Solution:
tan 0° = 0.

Solution:
By Pythagoras theorem, P²+B²=H².
Dividing by H² gives:
sin²θ + cos²θ = 1.

Solution:
From sin²θ+cos²θ=1, divide by cos²θ:
tan²θ+1=sec²θ.

Solution:
Divide sin²θ+cos²θ=1 by sin²θ:
1+cot²θ=cosec²θ.

Solution:
sec 60° = 1/cos60° = 2.

Solution:
cosec30° = 1/sin30° = 2.

Solution:
cot45° = 1.

Solution:
Take perpendicular=3 and base=4. Hypotenuse=5.
sinθ = 3/5.

Solution:
cosθ = √(1−25/169) = 12/13.

Solution:
sinθ=5/13.
tanθ=(5/13)/(12/13)=5/12.

Solution:
By identity, sin²θ+cos²θ=1.

Solution:
1+1=2.

Solution:
sec²45°=2 and tan²45°=1.
Answer = 1.

Solution:
Base=7, perpendicular=24, hypotenuse=25.
cosecθ = 25/24.

Solution:
The angle formed by the line of sight with the horizontal when an object is above the observer is called the angle of elevation.

Solution:
The angle formed by the line of sight with the horizontal when an object is below the observer is called the angle of depression.

Solution:
tanθ = 10/(10√3)=1/√3.
Therefore, θ=30°.

Solution:
tan45°=20/shadow.
Shadow = 20 m.

Solution:
tanθ=15/15=1.
Therefore, θ=45°.

Solution:
tan30°=h/20.
h=20/√3 = 20√3/3 m.

Solution:
h=10tan60°=10√3 m.

Solution:
Height=10sin60°=5√3 m.

Solution:
sin30°=8/L.
L=16 m.

Solution:
tan45°=30/d.
d=30 m.

Solution:
Height=100sin30°=50 m.

Solution:
h=25tan45°=25 m.

Solution:
tanθ=12/(4√3)=√3.
θ=60°.

Solution:
The horizontal lines are parallel. The angles form alternate interior angles, so they are equal.

Solution:
tan30°=50/d.
d=50√3 = 50√3 m.

Solution:
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution:
No tangent can be drawn from a point inside a circle.

Solution:
Exactly one tangent can be drawn.

Solution:
Exactly two tangents can be drawn.

Solution:
Tangents drawn from an external point to a circle are equal.
Therefore, PA = PB.

Solution:
The angle is 90°.

Solution:
Distance from centre to tangent equals radius.
Answer = 7 cm.

Solution:
Joining the centre to points of contact forms two right triangles with equal hypotenuse and one common side. By RHS congruence, the tangent lengths are equal.

Solution:
PA²=OP²−OA²=169−25=144.
PA=12 cm.

Solution:
PA=√(100−36)=√64=8 cm.

Solution:
Maximum number = 2.

Solution:
The tangent length increases as the external point moves farther away.

Solution:
No. A tangent touches a circle at exactly one point.

Solution:
Angle between tangents = 30°+30° = 60°.

Solution:
The line joining the centre to the point of contact is perpendicular to the tangent.

Solution:
C=2πr=2×22/7×7=44 cm.

Solution:
Area=πr²=22/7×196=616 cm².

Solution:
Area=1/2×22/7×49=77 cm².

Solution:
Arc length = 90/360×2π×14 = 22 cm.

Solution:
Area=90/360×22/7×196=154 cm².

Solution:
2πr=44 ⇒ r=7 cm.
Diameter=14 cm.

Solution:
Area=1/4×616=154 cm².

Solution:
Perimeter=πr+2r=22+14=36 cm.

Solution:
Area=π(R²−r²)=π(49−9)=40π cm².

Solution:
Fraction=120/360=1/3.

Solution:
CSA=2πrh=2×22/7×7×10=440 cm².

Solution:
V=πr²h=22/7×49×10=1540 cm³.

Solution:
V=a³=5³=125 cm³.

Solution:
TSA=6a²=6×36=216 cm².

Solution:
V=lbh=8×5×3=120 cm³.

Solution:
V=4/3πr³=4/3π×27=36π cm³.

Solution:
SA=4πr²=4×22/7×49=616 cm².

Solution:
V=2/3πr³=2/3π×216=144π cm³.

Solution:
CSA=πrl=22/7×7×10=220 cm².

Solution:
V=1/3πr²h=1/3π×9×4=12π cm³.

Solution:
l=√(r²+h²)=√(36+64)=10 cm.

Solution:
r=7 cm.
V=22/7×49×5=770 cm³.

Solution:
TSA=2πr(h+r)
=2×22/7×7×12=528 cm².

Solution:
Diameter=2×3=6 cm.
Volume=4/3π×27=36π cm³.

Solution:
Volume=10³=1000 cm³.
TSA=6×10²=600 cm².