Bihar Board Class 10 · Maths · Matric exam 2027
Bihar Board Class 10 Maths Question Paper 2025
The board’s 2025 question paper, set B — all 138 questions as printed; 100 objective answers from the board’s own key; full model answers for 3 written questions, how the answer begins for 35 more. Each question names its chapter, where that chapter’s most-asked questions are.
Full marks 100 · Time 3 h 15 min
Section A · Objective questions (100)
The distance between the points ( 8 sin 60°, 0 ) and ( 0, 8 cos 60°) is
- (A) 8 ✓
- (B) 25
- (C) 64
- (D) 1/8
Answer (the board’s key): (A)
If O ( 0, 0 ) be the origin and co-ordinates of the point P be ( x, y ) then the distance OP is
- (A) sqrt(x^2 - y^2)
- (B) sqrt(x^2 + y^2) ✓
- (C) x^2 - y^2
- (D) none of these
Answer (the board’s key): (B)
The distance of the point ( 12, 14 ) from the y-axis is
- (A) 12 ✓
- (B) 14
- (C) 13
- (D) 15
Answer (the board’s key): (A)
The ordinate of the point ( - 6, - 8 ) is
- (A) - 6
- (B) - 8 ✓
- (C) 6
- (D) 8
Answer (the board’s key): (B)
In which quadrant does the point ( 3, - 4 ) lie ?
- (A) First
- (B) Second
- (C) Third
- (D) Fourth ✓
Answer (the board’s key): (D)
Which of the following points lies in second quadrant ?
- (A) ( 3, 2 )
- (B) ( - 3, 2 ) ✓
- (C) ( 3, - 2 )
- (D) ( - 3, - 2 )
Answer (the board’s key): (B)
The co-ordinates of the mid-point of the line segment joining the points ( 4, - 4 ) and ( - 4, 4 ) are
- (A) ( 4, 4 )
- (B) ( 0, 0 ) ✓
- (C) ( 0, - 4 )
- (D) ( - 4, 0 )
Answer (the board’s key): (B)
The mid-point of line segment AB is ( 2, 4 ) and the co-ordinates of point A are ( 5, 7 ), then the co-ordinates of point B are
- (A) ( 2, - 2 )
- (B) ( 1, - 1 )
- (C) ( - 2, - 2 )
- (D) ( - 1, 1 ) ✓
Answer (the board’s key): (D)
The co-ordinates of the ends of a diameter of a circle are ( 10, - 6 ) and ( - 6, 10 ). Then the co-ordinates of the centre of the circle are
- (A) ( - 2, - 2 )
- (B) ( 2, 2 ) ✓
- (C) ( - 2, 2 )
- (D) ( 2, - 2 )
Answer (the board’s key): (B)
The co-ordinates of the vertices of a triangle are ( 4, 6 ), ( 0, 4 ) and ( 5, 5 ) then the co-ordinates of the centroid of the triangle are
- (A) ( 5, 3 )
- (B) ( 3, 4 )
- (C) ( 4, 4 )
- (D) ( 3, 5 ) ✓
Answer (the board’s key): (D)
If A ( 0, 1 ), B ( 0, 5 ) and C ( 3, 4 ) are the vertices of any triangle ABC, then the area ( in square unit ) of triangle ABC is
- (A) 16
- (B) 12
- (C) 6 ✓
- (D) 4
Answer (the board’s key): (C)
tan 10° . tan 23° . tan 80° . tan 67° =
- (A) 0
- (B) 1 ✓
- (C) sqrt(3)
- (D) 1/sqrt(3)
Answer (the board’s key): (B)
If the ratio of areas of two similar triangles is 100 : 144 then the ratio of their corresponding sides is
- (A) 10 : 8
- (B) 12 : 10
- (C) 10 : 12 ✓
- (D) 10 : 13
Answer (the board’s key): (C)
A line which intersects a circle in two distinct points is called
- (A) Chord
- (B) Secant ✓
- (C) Tangent
- (D) None of these
Answer (the board’s key): (B)
The corresponding sides of two similar triangles are in the ratio 4 : 9. What will be the ratio of the areas of the triangles ?
- (A) 9 : 4
- (B) 16 : 81 ✓
- (C) 81 : 16
- (D) 2 : 3
Answer (the board’s key): (B)
triangle ABC ~ triangle DEF and BC = 3 cm, EF = 4 cm. If the area of triangle ABC is 54 cm^2, then the area of triangle DEF is
- (A) 56 cm^2
- (B) 96 cm^2 ✓
- (C) 196 cm^2
- (D) 49 cm^2
Answer (the board’s key): (B)
In any triangle ABC, angle A = 90°, BC = 13 cm, AB = 12 cm ; then the value of AC is
- (A) 3 cm
- (B) 4 cm
- (C) 5 cm ✓
- (D) 6 cm
Answer (the board’s key): (C)
In triangle DEF and triangle PQR it is given that angle D = angle Q and angle R = angle E, then which of the following is correct ?
- (A) angle F = angle P ✓
- (B) angle F = angle Q
- (C) angle D = angle P
- (D) angle E = angle P
Answer (the board’s key): (A)
triangle ABC and triangle DEF are such that AB/DE = BC/EF = CA/DF and angle A = 40°, angle B = 80° ; then the measure of angle F is
- (A) 30°
- (B) 45°
- (C) 60° ✓
- (D) 40°
Answer (the board’s key): (C)
The number of common tangents of two intersecting circles is
- (A) 1
- (B) 2 ✓
- (C) 3
- (D) infinitely many
Answer (the board’s key): (B)
From an external point P, two tangents PA and PB are drawn on a circle. If PA = 8 cm then PB =
- (A) 6 cm
- (B) 8 cm ✓
- (C) 12 cm
- (D) 16 cm
Answer (the board’s key): (B)
If PA and PB are the tangents drawn from an external point P to a circle with centre at O and angle APB = 80° then angle POA =
- (A) 40°
- (B) 50° ✓
- (C) 80°
- (D) 60°
Answer (the board’s key): (B)
What is the angle between the tangent drawn at any point of a circle and the radius passing through the point of contact?
- (A) 30°
- (B) 45°
- (C) 60°
- (D) 90° ✓
Answer (the board’s key): (D)
The ratio of the radii of two circles is 3 : 4 ; then the ratio of their areas is
- (A) 3 : 4
- (B) 4 : 3
- (C) 9 : 16 ✓
- (D) 16 : 9
Answer (the board’s key): (C)
The area of the sector of a circle of radius 42 cm and central angle 30° is
- (A) 515 cm^2
- (B) 416 cm^2
- (C) 462 cm^2 ✓
- (D) 406 cm^2
Answer (the board’s key): (C)
The ratio of the circumferences of two circles is 5 : 7 ; then the ratio of their radii is
- (A) 7 : 5
- (B) 5 : 7 ✓
- (C) 25 : 49
- (D) 49 : 25
Answer (the board’s key): (B)
7 sec^2 A - 7 tan^2 A =
- (A) 49
- (B) 7 ✓
- (C) 14
- (D) 0
Answer (the board’s key): (B)
If x = a cos theta and y = b sin theta then b^2x^2 + a^2y^2 =
- (A) a^2b^2 ✓
- (B) ab
- (C) a^4b^4
- (D) a^2 + b^2
Answer (the board’s key): (A)
The angle of elevation of the top of a tower at a distance of 10 m from its base is 60° ; then the height of the tower is
- (A) 10 m
- (B) 10sqrt(3) m ✓
- (C) 15sqrt(3) m
- (D) 20sqrt(3) m
Answer (the board’s key): (B)
A kite is at a height 30 m from the earth and its string makes an angle 60° with the earth. Then the length of the string is
- (A) 30sqrt(2) m
- (B) 35sqrt(3) m
- (C) 20sqrt(3) m ✓
- (D) 45sqrt(2) m
Answer (the board’s key): (C)
For what value of k are the roots of the quadratic equation kx^2 - 6x + 1 = 0 real and equal?
- (A) 6
- (B) 8
- (C) 9 ✓
- (D) 10
Answer (the board’s key): (C)
If one of the zeros of the polynomial p(x) is 2 then which of the following is a factor of p(x) ?
- (A) x - 2 ✓
- (B) x + 2
- (C) x - 1
- (D) x + 1
Answer (the board’s key): (A)
If alpha and beta be the zeros of the polynomial cx^2 + ax + b then the value of alpha . beta is
- (A) a/c
- (B) -a/c
- (C) b/c ✓
- (D) -b/c
Answer (the board’s key): (C)
Which of the following is a quadratic equation ?
- (A) (x + 3)(x - 3) = x^2 - 4x^3
- (B) (x + 3)^2 = 4(x + 4) ✓
- (C) (2x - 2)^2 = 4x^2 + 7
- (D) 4x + 1/4x = 4x
Answer (the board’s key): (B)
Which of the following is not a quadratic equation ?
- (A) 5x - x^2 = x^2 + 3
- (B) x^3 - x^2 = (x - 1)^3
- (C) (x + 3)^2 = 3(x^2 - 5)
- (D) (sqrt(2)x + 3)^2 = 2x^2 + 5 ✓
Answer (the board’s key): (D)
The discriminant of the quadratic equation 2x^2 - 7x + 6 = 0 is
- (A) 1 ✓
- (B) -1
- (C) 27
- (D) 37
Answer (the board’s key): (A)
Which of the following points lies on the graph of x = 2 ?
- (A) (2, 0)
- (B) (2, 1)
- (C) (2, 2)
- (D) all of these ✓
Answer (the board’s key): (D)
If P + 1, 2P + 1, 4P - 1 are in A.P. then the value of P is
- (A) 1
- (B) 2 ✓
- (C) 3
- (D) 4
Answer (the board’s key): (B)
The common difference of arithmetic progression 1, 5, 9, ... is
- (A) 2
- (B) 3
- (C) 4 ✓
- (D) 5
Answer (the board’s key): (C)
Which term of the arithmetic progression 5, 8, 11, 14, ... is 38 ?
- (A) 10th
- (B) 11th
- (C) 12th ✓
- (D) 13th
Answer (the board’s key): (C)
The ratio of the volumes of two spheres is 64 : 125. Then the ratio of their surface areas is
- (A) 25 : 8
- (B) 25 : 16
- (C) 16 : 25 ✓
- (D) none of these
Answer (the board’s key): (C)
The radii of two cylinders are in the ratio 4 : 5 and their heights are in the ratio 6 : 7. Then the ratio of their volumes is
- (A) 96 : 125
- (B) 96 : 175 ✓
- (C) 175 : 96
- (D) 20 : 63
Answer (the board’s key): (B)
What is the total surface area of a hemisphere of radius R ?
- (A) pi R^2
- (B) 2pi R^2
- (C) 3pi R^2 ✓
- (D) 4pi R^2
Answer (the board’s key): (C)
If the curved surface area of a cone is 880 cm^2 and its radius is 14 cm, then its slant height is
- (A) 10 cm
- (B) 20 cm ✓
- (C) 40 cm
- (D) 30 cm
Answer (the board’s key): (B)
If the length of the diagonal of a cube is 2sqrt(3) cm, then the length of its edge is
- (A) 2 cm ✓
- (B) 2sqrt(3) cm
- (C) 3 cm
- (D) 4 cm
Answer (the board’s key): (A)
If the edge of a cube is doubled then the total surface area will become how many times of the previous total surface area ?
- (A) Two times
- (B) Four times ✓
- (C) Six times
- (D) Twelve times
Answer (the board’s key): (B)
The ratio of the total surface area of a sphere and that of a hemisphere having the same radius is
- (A) 2 : 1
- (B) 4 : 9
- (C) 3 : 2
- (D) 4 : 3 ✓
Answer (the board’s key): (D)
If the curved surface area of a hemisphere is 1232 cm^2 then its radius is
- (A) 7 cm
- (B) 14 cm ✓
- (C) 21 cm
- (D) 28 cm
Answer (the board’s key): (B)
If cos theta + cos^2 theta = 1 then the value of sin^2 theta + sin^4 theta is
- (A) - 1
- (B) 1 ✓
- (C) 0
- (D) 2
Answer (the board’s key): (B)
(1 + tan^2 A) / (1 + cot^2 A) =
- (A) sec^2 A
- (B) - 1
- (C) cot^2 A
- (D) tan^2 A ✓
Answer (the board’s key): (D)
sin(90° - A) =
- (A) sin A
- (B) cos A ✓
- (C) tan A
- (D) sec A
Answer (the board’s key): (B)
If alpha = beta = 60° then the value of cos(alpha - beta) is
- (A) 1/2
- (B) 1 ✓
- (C) 0
- (D) 2
Answer (the board’s key): (B)
If theta = 45° then the value of sin theta + cos theta is
- (A) 1/sqrt(2)
- (B) sqrt(2) ✓
- (C) 1/2
- (D) 1
Answer (the board’s key): (B)
If A = 30° then the value of (2 tan A) / (1 - tan^2 A) is
- (A) 2 tan 30°
- (B) tan 60° ✓
- (C) 2 tan 60°
- (D) tan 30°
Answer (the board’s key): (B)
If tan theta = 12/5 then the value of sin theta is
- (A) 5/12
- (B) 12/13 ✓
- (C) 5/13
- (D) 12/5
Answer (the board’s key): (B)
(cos 59° / sin 31°) * (tan 80° / cot 10°) =
- (A) 1/sqrt(2)
- (B) 1 ✓
- (C) sqrt(3)/2
- (D) 1/2
Answer (the board’s key): (B)
If tan 25° * tan 65° = sin A then the value of A is
- (A) 25°
- (B) 65°
- (C) 90° ✓
- (D) 45°
Answer (the board’s key): (C)
If cos theta = x then tan theta =
- (A) sqrt(1+x^2)/x
- (B) sqrt(1-x^2)/x ✓
- (C) sqrt(1-x^2)
- (D) x/sqrt(1-x^2)
Answer (the board’s key): (B)
(1-cos^4 theta) =
- (A) cos^2 theta(1-cos^2 theta)
- (B) sin^2 theta(1+cos^2 theta) ✓
- (C) sin^2 theta(1-sin^2 theta)
- (D) sin^2 theta(1+sin^2 theta)
Answer (the board’s key): (B)
What is the form of a point lying on y-axis ?
- (A) (y, 0)
- (B) (2, y)
- (C) (0, x)
- (D) None of these ✓
Answer (the board’s key): (D)
Which of the following quadratic polynomials has zeroes 3 and -10 ?
- (A) x^2 + 7x - 30 ✓
- (B) x^2 - 7x - 30
- (C) x^2 + 7x + 30
- (D) x^2 - 7x + 30
Answer (the board’s key): (A)
If the sum of zeroes of a quadractic polynomial is 3 and their product is -2 then that quadratic polynomial is
- (A) x^2 - 3x - 2 ✓
- (B) x^2 - 3x + 3
- (C) x^2 - 2x + 3
- (D) x^2 + 3x - 2
Answer (the board’s key): (A)
If p(x) = x^4 - 2x^3 + 17x^2 - 4x + 30 is divided by q(x) = x + 2 then the degree of the quotient is
- (A) 6
- (B) 3 ✓
- (C) 4
- (D) 5
Answer (the board’s key): (B)
How many solutions will x + 2y + 3 = 0, 3x + 6y + 9 = 0 have ?
- (A) One solution
- (B) No solution
- (C) Infinitely many solutions ✓
- (D) None of these
Answer (the board’s key): (C)
If the graphs of two linear equations are parallel then the number of solutions will be
- (A) 1
- (B) 2
- (C) infinitely many
- (D) none of these ✓
Answer (the board’s key): (D)
The pair of linear equations 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 is
- (A) consistent ✓
- (B) inconsistent
- (C) dependent
- (D) none of these
Answer (the board’s key): (A)
If alpha and beta are roots of the quadratic equation 3x^2 - 5x + 2 = 0 then the value of alpha^2 + beta^2 is
- (A) 13/9 ✓
- (B) 9/13
- (C) 5/3
- (D) 3/5
Answer (the board’s key): (A)
If one root of the quadratic equation 2x^2 - 7x - p = 0 is 2 then the value of p is
- (A) 4
- (B) - 4
- (C) - 6 ✓
- (D) 6
Answer (the board’s key): (C)
If one root of the quadratic equation 2x^2 - x - 6 = 0 is -3/2 then its another root is
- (A) - 2
- (B) 2 ✓
- (C) 3/2
- (D) 3
Answer (the board’s key): (B)
What is the nature of the roots of the quadratic equation 2x^2 - 6x + 3 = 0 ?
- (A) real and unequal ✓
- (B) real and equal
- (C) not real
- (D) none of these
Answer (the board’s key): (A)
The length of the class intervals of the classes, 2 - 5, 5 - 8, 8 - 11, ... is
- (A) 2
- (B) 3 ✓
- (C) 4
- (D) 3.5
Answer (the board’s key): (B)
If the mean of four consecutive odd numbers is 6 then the largest number is
- (A) 4.5
- (B) 9 ✓
- (C) 21
- (D) 15
Answer (the board’s key): (B)
The mean of first 6 even natural numbers is
- (A) 4
- (B) 6
- (C) 7 ✓
- (D) none of these
Answer (the board’s key): (C)
1 + cot^2 theta =
- (A) sin^2 theta
- (B) cosec^2 theta ✓
- (C) tan^2 theta
- (D) sec^2 theta
Answer (the board’s key): (B)
The mode of 8, 7, 9, 3, 9, 5, 4, 5, 7, 5 is
- (A) 5 ✓
- (B) 7
- (C) 8
- (D) 9
Answer (the board’s key): (A)
If P(E) = 0.02 then P(E') is equal to
- (A) 0.02
- (B) 0.002
- (C) 0.98 ✓
- (D) 0.97
Answer (the board’s key): (C)
Two dice are thrown at the same time. What is the probability that the difference of the numbers appearing on top is zero ?
- (A) 1/36
- (B) 1/6 ✓
- (C) 5/18
- (D) 5/36
Answer (the board’s key): (B)
The probability of getting heads on both the coins in throwing two coins is
- (A) 1/2
- (B) 1/3
- (C) 1/4 ✓
- (D) 1
Answer (the board’s key): (C)
A month is selected at random in a year. The probability of it being June or September is
- (A) 3/4
- (B) 1/12
- (C) 1/6 ✓
- (D) 1/4
Answer (the board’s key): (C)
The probability of getting a number 4 or 5 in throwing a die is
- (A) 1/2
- (B) 1/3 ✓
- (C) 1/6
- (D) 2/3
Answer (the board’s key): (B)
If 5th term of an A.P. is 11 and common difference is 2 then what is its first term ?
- (A) 1
- (B) 2
- (C) 3 ✓
- (D) 4
Answer (the board’s key): (C)
The sum of an A.P. with n terms is n^2 + 2n + 1 then its 6th term is
- (A) 29
- (B) 19
- (C) 15
- (D) none of these ✓
Answer (the board’s key): (D)
Which of the following is in an A.P. ?
- (A) 1, 7, 9, 16, ...
- (B) x^2, x^3, x^4, x^5, ...
- (C) x, 2x, 3x, 4x, ... ✓
- (D) 2^2, 4^2, 6^2, 8^2, ...
Answer (the board’s key): (C)
Which of the following is not in an A.P. ?
- (A) 1, 2, 3, 4, ...
- (B) 3, 6, 9, 12, ...
- (C) 2, 4, 6, 8, ...
- (D) 2^2, 4^2, 6^2, 8^2, ... ✓
Answer (the board’s key): (D)
The sum of first 20 terms of the A.P. 1, 4, 7, 10, ... is
- (A) 500
- (B) 540
- (C) 590 ✓
- (D) 690
Answer (the board’s key): (C)
Which of the following values is equal to 1 ?
- (A) sin^2 60° + cos 60°
- (B) sin 90° x cos 90°
- (C) sin^2 60°
- (D) sin 45° x 1/cos 45° ✓
Answer (the board’s key): (D)
cos^2 A(1 + tan^2 A) =
- (A) sin^2 A
- (B) cosec^2 A
- (C) 1 ✓
- (D) tan^2 A
Answer (the board’s key): (C)
tan 30° =
- (A) 1/sqrt(3)
- (B) sqrt(3)/2
- (C) 1/sqrt(3) ✓
- (D) 1
Answer (the board’s key): (C)
cos 60° =
- (A) 1/2 ✓
- (B) sqrt(3)/2
- (C) 1/sqrt(2)
- (D) 1
Answer (the board’s key): (A)
sin^2 90^o - tan^2 45^o =
- (A) 1
- (B) 1/2
- (C) 1/sqrt(2)
- (D) 0 ✓
Answer (the board’s key): (D)
Which of the following fractions has terminating decimal expansion ?
- (A) 14 / (2^0 x 3^2)
- (B) 9 / (5^1 x 7^2)
- (C) 8 / (2^2 x 3^2)
- (D) 15 / (2^2 x 5^3) ✓
Answer (the board’s key): (D)
In the form of p/(2^n * 5^m), 0.505 can be written as
- (A) 101/(2^1 * 5^2)
- (B) 101/(2^1 * 5^3)
- (C) 101/(2^2 * 5^2)
- (D) 101/(2^3 * 5^2) ✓
Answer (the board’s key): (D)
If in division algorithm a = bq + r, b = 4, q = 5 and r = 1, then what is the value of a ?
- (A) 20
- (B) 21 ✓
- (C) 25
- (D) 31
Answer (the board’s key): (B)
The zeroes of the polynomial 2x^2 - 4x - 6 are
- (A) 1, 3
- (B) -1, 3 ✓
- (C) 1, -3
- (D) -1, -3
Answer (the board’s key): (B)
The degree of the polynomial (x^3 + x^2 + 2x + 1) (x^2 + 2x + 1) is
- (A) 3
- (B) 4
- (C) 5 ✓
- (D) 6
Answer (the board’s key): (C)
Which of the following is not a polynomial ?
- (A) x^2 - 7
- (B) 2x^2 + 7x + 6
- (C) 1/2 x^2 + 1/2 x + 4
- (D) x + 4/x ✓
Answer (the board’s key): (D)
Which of the following quadratic polynomials has zeroes 2 and -2 ?
- (A) x^2 + 4
- (B) x^2 - 4 ✓
- (C) x^2 - 2x + 4
- (D) x^2 + sqrt(8)
Answer (the board’s key): (B)
If alpha and beta are the zeroes of the polynomial t^2 + 7t + 10 then the value of alpha + beta is
- (A) 7
- (B) 10
- (C) -7 ✓
- (D) -10
Answer (the board’s key): (C)
(sin 30^o + cos 30^o) - (sin 60^o + cos 60^o) =
- (A) -1
- (B) 0 ✓
- (C) 1
- (D) 2
Answer (the board’s key): (B)
If one zero of the quadratic polynomial (k - 1)x^2 + kx + 1 is - 4 then the value of k is
- (A) - 5/4
- (B) 5/4 ✓
- (C) - 4/3
- (D) 4/3
Answer (the board’s key): (B)
Section B · Written questions (38)
Find the co-ordinates of the point which divides line segment joining the points (- 1, 7) and (4, - 3) in the ratio 2 : 3 internally.
Medha Ma’am
AskMedha’s AI teacher
Model answer · section formula, internal division
- 1.The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are ((m1x2 + m2x1)/(m1+m2), (m1y2 + m2y1)/(m1+m2)). This is known as the section formula. Here A(x1, y1) = (-1, 7), B(x2, y2) = (4, -3) and m1 : m2 = 2 : 3. Using the section formula, the x-coordinate is x = (2×4 + 3×(-1))/(2+3) = (8-3)/5 = 5/5 = 1.1 mark
- 2.Using the section formula, the y-coordinate is y = (m1y2 + m2y1)/(m1+m2) = (2×(-3) + 3×7)/(2+3) = (-6+21)/5 = 15/5 = 3. Therefore, the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 : 3 internally is (1, 3).1 mark
- Write the section formula fully before substituting values.
- Clearly state m1 : m2 = 2 : 3 along with A and B coordinates.
- Write the final point as an ordered pair (1, 3) with both coordinates.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
Find the area of the triangle whose vertices are (- 5, - 1), (3, - 5) and (5, 2).
Medha Ma’am
AskMedha’s AI teacher
Model answer · area of triangle, coordinates
- 1.The area of the triangle formed by the points (x1, y1), (x2, y2) and (x3, y3) is the numerical value of the expression 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)].1 mark
- 2.Given: vertices A(–5, –1), B(3, –5) and C(5, 2). Formula: Area = 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] Substitution: Area = 1/2 [–5(–5 – 2) + 3(2 – (–1)) + 5(–1 – (–5))] Calculation: = 1/2 [–5(–7) + 3(3) + 5(4)] = 1/2 [35 + 9 + 20] = 1/2 (64) = 32 Answer: the area of the triangle = 32 square units.1 mark
- Write the area formula exactly as printed before substituting values.
- Show each step: substitution, calculation, and final value separately.
- End with the answer stated as '32 square units' including the unit.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
The diagonal of a cube is 9sqrt(3) cm. Find the total surface area of cube.
Medha Ma’am
AskMedha’s AI teacher
Model answer · surface area of cube from diagonal
- 1.The diagonal of a cube of side a is given by the formula diagonal = √3 × a (a√3).0.5 marks
- 2.Here the diagonal of the cube is given as 9√3 cm. Using a√3 = 9√3, we get a = 9 cm. So the side of the cube is 9 cm.0.5 marks
- 3.The total surface area of a cube of side a is given by the formula 6 × (side)², i.e., 6a².0.5 marks
- 4.Total surface area = 6a² = 6 × (9)² = 6 × 81 = 486 So, the total surface area of the cube is 486 cm².0.5 marks
- Write the formula diagonal = a√3 before finding the side.
- Show the step a√3 = 9√3 to get a = 9 cm as working.
- End with the final value 486 cm² including the unit.
Marks split from a structure derived from past papers (checked twice); not Bihar Board’s official scheme
If tan theta = 5/12 then find the value of sin theta + cos theta.
How the model answer begins
1. Since tan theta = 5/12, consider a right-angled triangle in which the side opposite to theta (perpendicular) = 5 and the side adjacent to theta (base) = 12. By Pythagoras theorem, hypotenuse = root(perpendicular^2 +…
The full answer from Medha Ma’am →If sin 3A = cos(A - 26°), where 3A is an acute angle, then find the value of A.
How the model answer begins
1. We are given that sin 3A = cos (A – 26°), where 3A is an acute angle. Since sin A = cos (90° – A) for all values of angle A lying between 0° and 90°, we can write sin 3A = cos (90° – 3A). Substituting this in the given…
The full answer from Medha Ma’am →The sum of two numbers is 50 and one number is 7/3 times of the other; then find the numbers.
How the model answer begins
1. Let the first number be x and the second number be y. Since the sum of the two numbers is 50, we get the equation x + y = 50.
The full answer from Medha Ma’am →A ladder 7 m long makes an angle of 30° with the wall. Find the height of the point on the wall where the ladder touches the wall.
How the model answer begins
1. The ladder (7 m) is the hypotenuse and makes 30° with the wall, so the height h on the wall is the side next to that angle: cos 30° = h/7.
The full answer from Medha Ma’am →E is a point on the extended part of the side AD of a parallelogram ABCD and BE intersects CD at F; then prove that triangle ABE ~ triangle CFB.
How the model answer begins
1. In parallelogram ABCD, AB is parallel to CD, and angle A is equal to angle C because opposite angles of a parallelogram are equal. Since F lies on CD, angle C is the same as angle FCB, so angle A is equal to angle FCB.
The full answer from Medha Ma’am →ABC is an isosceles right triangle with angle C as right angle. Prove that AB^2 = 2AC^2.
How the model answer begins
1. According to the Pythagoras Theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In triangle ABC, angle C is a right angle, so AB is the hypotenuse.…
The full answer from Medha Ma’am →E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC, prove that triangle ABD ~ triangle ECF.
How the model answer begins
1. Since AB = AC, ∠ABC = ∠ACB. E lies on CB produced, so ∠ABD is the same angle as ∠ABC and ∠ECF the same as ∠ACB; hence ∠ABD = ∠ECF.
The full answer from Medha Ma’am →Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Then prove that triangle ABC ~ triangle PQR.
How the model answer begins
1. (The question's 'PR' is read as QR — the median PM is drawn to QR.) AD and PM are medians, so BC = 2BD and QR = 2QM. From AB/PQ = BC/QR = AD/PM, AB/PQ = BD/QM = AD/PM, so ΔABD ~ ΔPQM by SSS similarity and ∠B = ∠Q.
The full answer from Medha Ma’am →triangle ABC and triangle DEF are similar and their areas are 9 cm^2 and 64 cm^2 respectively. If DE = 5.1 cm then find AB.
How the model answer begins
1. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Since triangle ABC and triangle DEF are similar, AB and DE are corresponding sides, so ar(ABC)/ar(DEF) =…
The full answer from Medha Ma’am →Divide x^3 + 1 by x + 1.
How the model answer begins
1. Long division: write x^3 + 1 as x^3 + 0x^2 + 0x + 1. First term x^3 ÷ x = x^2; subtract x^2(x + 1) = x^3 + x^2 to leave −x^2 + 0x + 1. Next term −x^2 ÷ x = −x; subtract −x(x + 1) = −x^2 − x to leave x + 1. Next term x ÷…
The full answer from Medha Ma’am →Using Euclid's division algorithm, find the H.C.F. of 504 and 1188.
How the model answer begins
1. Since 1188 > 504, we apply Euclid's division lemma, a = bq + r, to 1188 and 504, to get 1188 = 504 × 2 + 180. Since the remainder 180 ≠ 0, we apply the division lemma to 504 and 180, to get 504 = 180 × 2 + 144.
The full answer from Medha Ma’am →Find the discriminant of the quadratic equation 2x^2 + 5x - 3 = 0 and find the nature of the roots also.
How the model answer begins
1. Comparing the quadratic equation 2x^2 + 5x - 3 = 0 with ax^2 + bx + c = 0, we get a = 2, b = 5, c = -3. The discriminant of this quadratic equation is given by D = b^2 - 4ac.
The full answer from Medha Ma’am →If the radius of base of a cone is 7 cm and its height is 24 cm then find its curved surface area.
How the model answer begins
1. Given the radius of the base of the cone r = 7 cm and the height h = 24 cm. The slant height of the cone is l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.
The full answer from Medha Ma’am →The length of the minute hand for a clock is 7 cm. Find the area swept by it in 40 minutes.
How the model answer begins
1. The angle subtended by the minute hand in 60 minutes is 360°, so the angle subtended in 40 minutes is θ = (360/60) × 40 = 240°.
The full answer from Medha Ma’am →Prove that tan 7° . tan 60° . tan 83° = sqrt(3).
Using quadratic formula find the roots of the equation 2x^2 - 2sqrt(2)x + 1 = 0.
How the model answer begins
1. Comparing the given equation 2x^2 - 2\sqrt2 x + 1 = 0 with ax^2 + bx + c = 0, we get a = 2, b = -2\sqrt2 and c = 1.
The full answer from Medha Ma’am →Find the sum of 3 + 11 + 19 + ... + 67.
How the model answer begins
1. The given list 3, 11, 19, ..., 67 is an arithmetic progression (AP) with first term a = 3, common difference d = 11 - 3 = 8, and last term l = 67.
The full answer from Medha Ma’am →If 5th and 9th terms of an A.P. are 43 and 79 respectively, find the A.P.
How the model answer begins
1. Using the formula for the nth term of an arithmetic progression, an = a + (n - 1)d, we get the equations for the given terms. For the 5th term: a + 4d = 43 For the 9th term: a + 8d = 79 Subtracting the first equation…
The full answer from Medha Ma’am →Prove that sqrt((1 + cos theta) / (1 - cos theta)) = (1 + cos theta) / sin theta.
How the model answer begins
1. To prove sqrt((1+cos theta)/(1-cos theta)) = (1+cos theta)/sin theta, we start with LHS = sqrt((1+cos theta)/(1-cos theta)). Rationalizing the denominator by multiplying numerator and denominator inside the square root…
The full answer from Medha Ma’am →Prove that tan 9° . tan 27° = cot 63° . cot 81°.
How the model answer begins
1. We know that tan (90° – A) = cot A. Writing tan 9° as tan (90° – 81°), we get tan 9° = cot 81°.
The full answer from Medha Ma’am →If cos A = 4/5 then find the values of cot A and cosec A.
How the model answer begins
1. Given cos A = 4/5, consider a right triangle ABC, right-angled at B, with ∠A the angle whose cosine is taken. Since cos A = AB/AC = 4/5, let AB = 4k and AC = 5k, where k is a positive number. By the Pythagoras theorem,…
The full answer from Medha Ma’am →Find two consecutive positive integers, sum of whose squares is 365.
How the model answer begins
1. Let the two consecutive positive integers be x and x + 1.
The full answer from Medha Ma’am →The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Write the equation for this statement.
How the model answer begins
1. Let the larger number be x and the smaller y; the difference of their squares is 180, so x² − y² = 180.
The full answer from Medha Ma’am →In a triangle PQR, two points S and T are on the sides PQ and PR respectively such that PS/SQ = PT/TR and angle PST = angle PRQ, then prove that deltaPQR is an isosceles triangle.
How the model answer begins
1. It is given that PS/SQ = PT/TR. So, by the converse of the Basic Proportionality Theorem (Thales Theorem), ST is parallel to QR.
The full answer from Medha Ma’am →Prove that 5 - sqrt(3) is an irrational number.
How the model answer begins
1. Let us assume, to the contrary, that 5 - sqrt(3) is a rational number. That is, we can find coprime a and b (b ≠ 0) such that 5 - sqrt(3) = a/b.
The full answer from Medha Ma’am →For what value of k points (1, 1), (3, k) and (-1, 4) are collinear?
How the model answer begins
1. If the area of a triangle is 0 square units, then its vertices will be collinear. So for the points (1, 1), (3, k) and (-1, 4) to be collinear, the area of the triangle formed by them must be zero.
The full answer from Medha Ma’am →Find such a point on y-axis which is equidistant from the points (6, 5) and (-4, 3).
How the model answer begins
1. We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A(6, 5) and B(– 4, 3), i.e., PA = PB.
The full answer from Medha Ma’am →Draw the graphs of the pair of linear equations x + 3y - 6 = 0 and 2x - 3y - 12 = 0 and solve them.
How the model answer begins
1. The given equation x + 3y - 6 = 0 can be written as x + 3y = 6. To draw its graph, we find at least two solutions of this equation. When x = 0, 3y = 6, so y = 2. When y = 0, x = 6. Thus we get two solutions of the…
The full answer from Medha Ma’am →If one angle of a triangle is equal to one angle of the other triangle and the sides included between these angles are proportional then prove that the triangles are similar.
How the model answer begins
1. Given: in ΔABC and ΔDEF, ∠A = ∠D and AB/DE = AC/DF. To prove: ΔABC ~ ΔDEF.
The full answer from Medha Ma’am →A two digit number is four times of the sum of its digits and twice the product of its digits. Find the number.
How the model answer begins
1. Let the ten's digit of the required two-digit number be x and the unit's digit be y. Then the first number may be written in the expanded form as 10x + y.
The full answer from Medha Ma’am →Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure both parts.
How the model answer begins
1. To divide a line segment of length 7.6 cm in the ratio 5 : 8, first draw a line segment AB of length 7.6 cm using a ruler.
The full answer from Medha Ma’am →Prove that (sec theta - tan theta) / (sec theta + tan theta) = 1 + 2 tan^2 theta - 2 sec theta . tan theta.
How the model answer begins
1. To prove that (sec θ – tan θ)/(sec θ + tan θ) = 1 + 2 tan² θ – 2 sec θ . tan θ, we start with the left hand side LHS = (sec θ – tan θ)/(sec θ + tan θ). We multiply and divide this expression by (sec θ – tan θ), so that…
The full answer from Medha Ma’am →The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
How the model answer begins
1. Let the radius of the first circle be r1 = 19 cm. The circumference of a circle = 2πr. So, the circumference of the first circle = 2πr1 = 2π(19) = 38π cm.
The full answer from Medha Ma’am →Find the mean of the following distribution : [Table provided]
How the model answer begins
1. The class mark (xi) of each class interval is obtained by taking the average of the upper and lower limits of that class, that is, xi = (upper limit + lower limit) / 2. Using this formula, the class mark xi is…
The full answer from Medha Ma’am →The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
How the model answer begins
1. The slant height of the frustum of the cone is given as l = 4 cm. Let r1 and r2 be the radii of the two circular ends of the frustum. We need to find the curved surface area of the frustum, which requires the values of…
The full answer from Medha Ma’am →
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