Maths-
General
Easy

Question

In the figure, ABC is an isosceles triangle with AB = AC. If D is the midpoint of BC, prove that:
(i) ΔABD⩭ ΔACD
(ii) straight angle A D C equals straight angle A D B equals90°

Hint:

Find the congruence rule and from that find the angles.

The correct answer is: 90°


    (i) In ΔABD and ΔACD, we have
    AB = AC (isosceles triangle)
    AD = AD (common side)
    BD = DC (Since D is the midpoint of BC)
    Hence ΔABD⩭  ΔACD by SSS congruence rule.
    (ii) Since  ΔABD⩭  ΔACD by SSS congruence rule
    We have, straight angle ADC equals straight angle ADB
    straight angle ADC plus straight angle ADB equals 180 to the power of ring operator (linear pair)
    therefore straight angle ADC equals straight angle ADB equals 180 over 2 equals 90 to the power of ring operator

    Related Questions to study

    General
    Maths-

    State the property of congruence and write symbolically the congruency in the correct order.
    (i) 
    (ii) 

    (i) ASA congruence rule (Angle - Side -  Angle)
    Two angles are said to be congruent if two angles and the included side of one
    triangle is equal to two angles and the included side of another triangle.
    Here,
    straight angle C A B equals straight angle R P Q comma straight angle C B A equals straight angle R Q P and AB = PQ
    Comparing these correspondence between sides and angles, we have
    ΔABC ⩭ ΔPQR
    (ii) RHS congruence rule (Right angle - Hypotenuse - Side)
    Two right angles are congruent if the hypotenuse and one side of first triangle are
    respectively equal to the hypotenuse and one side of the second triangle.
    Here,
    straight angle X equals straight angle L equals 90 to the power of ring operator, XY = LN= 3 cm and YZ = NM = 5cm
    Comparing these correspondence between sides and eagles, we have
    ΔXYZ ⩭ ΔLNM

    State the property of congruence and write symbolically the congruency in the correct order.
    (i) 
    (ii) 

    Maths-General
    (i) ASA congruence rule (Angle - Side -  Angle)
    Two angles are said to be congruent if two angles and the included side of one
    triangle is equal to two angles and the included side of another triangle.
    Here,
    straight angle C A B equals straight angle R P Q comma straight angle C B A equals straight angle R Q P and AB = PQ
    Comparing these correspondence between sides and angles, we have
    ΔABC ⩭ ΔPQR
    (ii) RHS congruence rule (Right angle - Hypotenuse - Side)
    Two right angles are congruent if the hypotenuse and one side of first triangle are
    respectively equal to the hypotenuse and one side of the second triangle.
    Here,
    straight angle X equals straight angle L equals 90 to the power of ring operator, XY = LN= 3 cm and YZ = NM = 5cm
    Comparing these correspondence between sides and eagles, we have
    ΔXYZ ⩭ ΔLNM
    General
    Maths-

    In the figure, ray AZ bisects straight angle D A B as well as straight angle D C B.
    (i) State the three pairs of equal parts in triangles BAC and DAC.
    (ii) Is ΔBAC ⩭ ΔDAC ? Give reasons.
    (iii) Is AB = AD? Justify your answer.
    (iv) Is CD = CB? Give reasons.

    (i) From the figure, we have
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell straight angle D A C equals straight angle B A C end cell row cell straight angle D C A equals straight angle B C A end cell end table
    AC = AC (common side)
    (ii) By, ASA congruence criterion, the triangles are congruent.
    That is, straight angle D A C equals straight angle B A C
    straight angle D C A equals straight angle B C A
    AC = AC
    (iii) Since the triangles are congruent, AB = AD because both sides originate from
    equal angles in 2 different triangles.
    (iv) Since the triangles are congruent, CD = CB because both sides originate from
    equal angles in 2 different triangles.

    In the figure, ray AZ bisects straight angle D A B as well as straight angle D C B.
    (i) State the three pairs of equal parts in triangles BAC and DAC.
    (ii) Is ΔBAC ⩭ ΔDAC ? Give reasons.
    (iii) Is AB = AD? Justify your answer.
    (iv) Is CD = CB? Give reasons.

    Maths-General
    (i) From the figure, we have
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell straight angle D A C equals straight angle B A C end cell row cell straight angle D C A equals straight angle B C A end cell end table
    AC = AC (common side)
    (ii) By, ASA congruence criterion, the triangles are congruent.
    That is, straight angle D A C equals straight angle B A C
    straight angle D C A equals straight angle B C A
    AC = AC
    (iii) Since the triangles are congruent, AB = AD because both sides originate from
    equal angles in 2 different triangles.
    (iv) Since the triangles are congruent, CD = CB because both sides originate from
    equal angles in 2 different triangles.
    General
    Maths-

    If line segments AB and CD bisect each other at O, then using SAS congruence rule prove that ΔAOC  ⩭ ΔBOD.

    From the figure, we have
    AO = OB
    And, CO = OD
    To prove: straight triangle A O C approximately equal to straight triangle B O D (By SAS congruence rule)
    AO = OB (given)
    CO = OD (given)
    straight angle A O C equals straight angle B O D(vertically opposite angles)
    therefore straight triangle A O C approximately equal to straight triangle by SAS congruence rule.

    If line segments AB and CD bisect each other at O, then using SAS congruence rule prove that ΔAOC  ⩭ ΔBOD.

    Maths-General
    From the figure, we have
    AO = OB
    And, CO = OD
    To prove: straight triangle A O C approximately equal to straight triangle B O D (By SAS congruence rule)
    AO = OB (given)
    CO = OD (given)
    straight angle A O C equals straight angle B O D(vertically opposite angles)
    therefore straight triangle A O C approximately equal to straight triangle by SAS congruence rule.
    parallel
    General
    Maths-

    What least number must be multiplied to 6912 so that the product becomes a perfect Cube?

    Explanation :-
    6912 = 2 × 3456
    = 2 × 2 × 1728
    = 2 × 2 × 2 × 864
    = 2 × 2 × 2 × 2 × 432
    = 2 × 2 × 2 × 2 × 2 × 216
    = 2 × 2 × 2 × 2 × 2 × 2 ×108
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 54
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×3 × 9
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
    2 to the power of 8 space cross times space 3 cubed
    we know 2 to the power of 8  is perfect square and to make 3 cubed perfect square multiply it with 3
    We need to multiply 6912 with 3 to make

    What least number must be multiplied to 6912 so that the product becomes a perfect Cube?

    Maths-General
    Explanation :-
    6912 = 2 × 3456
    = 2 × 2 × 1728
    = 2 × 2 × 2 × 864
    = 2 × 2 × 2 × 2 × 432
    = 2 × 2 × 2 × 2 × 2 × 216
    = 2 × 2 × 2 × 2 × 2 × 2 ×108
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 54
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×3 × 9
    = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
    2 to the power of 8 space cross times space 3 cubed
    we know 2 to the power of 8  is perfect square and to make 3 cubed perfect square multiply it with 3
    We need to multiply 6912 with 3 to make
    General
    Maths-

    Rohit, Peter and Santosh walk around a circular park. They take 1 third h comma 2 over 5 h 𝑎𝑛𝑑 5 over 12 hto complete one round. What is the total time taken by them to complete a round in minutes.

    Time taken by Rohith to walk around the circular path = 1 third h
    Time taken by Peter to walk around the circular path = 2 over 5 h
    Time taken by Santhosh to walk around the circular path = 5 over 12 h
    Total time taken = 1 third h plus 2 over 5 h plus 5 over 12 h equals open parentheses 1 third plus 2 over 5 plus 5 over 12 close parentheses h
    On taking the LCM of denominators 3,5 and 12, we get LCM as 60.
    So, 1 third equals fraction numerator 1 cross times 20 over denominator 3 cross times 20 end fraction equals 20 over 60 and
    2 over 5 equals fraction numerator 2 cross times 12 over denominator 5 cross times 12 end fraction equals 24 over 60 and
    5 over 12 equals fraction numerator 5 cross times 5 over denominator 12 cross times 5 end fraction equals 25 over 60
    On adding the terms , we get open parentheses 20 over 60 plus 24 over 60 plus 25 over 60 close parentheses h equals 69 over 60 cross times h
    To convert it into minutes, multiply by 60.
    not stretchy rightwards double arrow 69 over 60 cross times 60 equals 69 minutes.
    They took 69 minutes to complete a round in a circular park.

    Rohit, Peter and Santosh walk around a circular park. They take 1 third h comma 2 over 5 h 𝑎𝑛𝑑 5 over 12 hto complete one round. What is the total time taken by them to complete a round in minutes.

    Maths-General
    Time taken by Rohith to walk around the circular path = 1 third h
    Time taken by Peter to walk around the circular path = 2 over 5 h
    Time taken by Santhosh to walk around the circular path = 5 over 12 h
    Total time taken = 1 third h plus 2 over 5 h plus 5 over 12 h equals open parentheses 1 third plus 2 over 5 plus 5 over 12 close parentheses h
    On taking the LCM of denominators 3,5 and 12, we get LCM as 60.
    So, 1 third equals fraction numerator 1 cross times 20 over denominator 3 cross times 20 end fraction equals 20 over 60 and
    2 over 5 equals fraction numerator 2 cross times 12 over denominator 5 cross times 12 end fraction equals 24 over 60 and
    5 over 12 equals fraction numerator 5 cross times 5 over denominator 12 cross times 5 end fraction equals 25 over 60
    On adding the terms , we get open parentheses 20 over 60 plus 24 over 60 plus 25 over 60 close parentheses h equals 69 over 60 cross times h
    To convert it into minutes, multiply by 60.
    not stretchy rightwards double arrow 69 over 60 cross times 60 equals 69 minutes.
    They took 69 minutes to complete a round in a circular park.
    General
    Maths-

    Determine y so that y minus 2 1 third equals open parentheses negative 3 7 over 12 close parentheses

    To find y minus 2 1 third equals open parentheses negative 3 7 over 12 close parentheses
    So here, 2 1 third equals fraction numerator left parenthesis 2 cross times 3 right parenthesis plus 1 over denominator 3 end fraction equals 7 over 3 and negative 3 7 over 12 equals fraction numerator left parenthesis 12 cross times negative 3 right parenthesis plus 7 over denominator 12 end fraction equals fraction numerator negative 29 over denominator 12 end fraction
    y minus 2 1 third equals open parentheses negative 3 7 over 12 close parentheses can be written as y minus 7 over 3 equals fraction numerator negative 29 over denominator 12 end fraction
    On adding 7 over 3 on both the sides, we have y equals fraction numerator negative 29 over denominator 12 end fraction plus 7 over 3
    On taking the LCM of denominators 12 and 3, we get LCM as 12.
    So,7 over 3 equals fraction numerator 7 cross times 4 over denominator 3 cross times 4 end fraction equals 28 over 12
    On adding, we get y equals fraction numerator negative 29 over denominator 12 end fraction plus 28 over 12 equals fraction numerator negative 1 over denominator 12 end fraction
    Hence y equals fraction numerator negative 1 over denominator 12 end fraction

    Determine y so that y minus 2 1 third equals open parentheses negative 3 7 over 12 close parentheses

    Maths-General
    To find y minus 2 1 third equals open parentheses negative 3 7 over 12 close parentheses
    So here, 2 1 third equals fraction numerator left parenthesis 2 cross times 3 right parenthesis plus 1 over denominator 3 end fraction equals 7 over 3 and negative 3 7 over 12 equals fraction numerator left parenthesis 12 cross times negative 3 right parenthesis plus 7 over denominator 12 end fraction equals fraction numerator negative 29 over denominator 12 end fraction
    y minus 2 1 third equals open parentheses negative 3 7 over 12 close parentheses can be written as y minus 7 over 3 equals fraction numerator negative 29 over denominator 12 end fraction
    On adding 7 over 3 on both the sides, we have y equals fraction numerator negative 29 over denominator 12 end fraction plus 7 over 3
    On taking the LCM of denominators 12 and 3, we get LCM as 12.
    So,7 over 3 equals fraction numerator 7 cross times 4 over denominator 3 cross times 4 end fraction equals 28 over 12
    On adding, we get y equals fraction numerator negative 29 over denominator 12 end fraction plus 28 over 12 equals fraction numerator negative 1 over denominator 12 end fraction
    Hence y equals fraction numerator negative 1 over denominator 12 end fraction
    parallel
    General
    Maths-

    Find the simplest form of open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 1 third plus 1 over 6 close parentheses divided by open parentheses fraction numerator negative 2 over denominator 5 end fraction close parentheses

    To find open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 1 third plus 1 over 6 close parentheses divided by open parentheses fraction numerator negative 2 over denominator 5 end fraction close parentheses
    Determine the LCM of the denominators of rational numbers inside the brackets.
    LCM of 2, 3 and 6 = 6
    So, fraction numerator negative 1 over denominator 2 end fraction equals fraction numerator negative 1 cross times 3 over denominator 2 cross times 3 end fraction equals fraction numerator negative 3 over denominator 6 end fraction and
    1 third equals fraction numerator 1 cross times 2 over denominator 3 cross times 2 end fraction equals 2 over 6
    On adding the terms inside the bracket, we have fraction numerator negative 3 over denominator 6 end fraction plus 2 over 6 plus 1 over 6 equals 0
    Divide fraction numerator negative 2 over denominator 5 end fraction with , we get fraction numerator 0 over denominator fraction numerator negative 2 over denominator 5 end fraction end fraction equals 0
    Hence open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 1 third plus 1 over 6 close parentheses divided by open parentheses fraction numerator negative 2 over denominator 5 end fraction close parentheses equals 0

    Find the simplest form of open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 1 third plus 1 over 6 close parentheses divided by open parentheses fraction numerator negative 2 over denominator 5 end fraction close parentheses

    Maths-General
    To find open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 1 third plus 1 over 6 close parentheses divided by open parentheses fraction numerator negative 2 over denominator 5 end fraction close parentheses
    Determine the LCM of the denominators of rational numbers inside the brackets.
    LCM of 2, 3 and 6 = 6
    So, fraction numerator negative 1 over denominator 2 end fraction equals fraction numerator negative 1 cross times 3 over denominator 2 cross times 3 end fraction equals fraction numerator negative 3 over denominator 6 end fraction and
    1 third equals fraction numerator 1 cross times 2 over denominator 3 cross times 2 end fraction equals 2 over 6
    On adding the terms inside the bracket, we have fraction numerator negative 3 over denominator 6 end fraction plus 2 over 6 plus 1 over 6 equals 0
    Divide fraction numerator negative 2 over denominator 5 end fraction with , we get fraction numerator 0 over denominator fraction numerator negative 2 over denominator 5 end fraction end fraction equals 0
    Hence open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 1 third plus 1 over 6 close parentheses divided by open parentheses fraction numerator negative 2 over denominator 5 end fraction close parentheses equals 0
    General
    Maths-

    What is the result of 2 minus 11 over 39 plus 5 over 26

    To find 2 minus 11 over 39 plus 5 over 26
    Taking the LCM of 39 and 26, we have 78 as LCM.
    So, 2 equals 2 over 1 equals fraction numerator 2 cross times 78 over denominator 1 cross times 78 end fraction equals 156 over 78 and
    11 over 39 equals fraction numerator 11 cross times 2 over denominator 39 cross times 2 end fraction equals 22 over 78 and
    5 over 26 equals fraction numerator 5 cross times 3 over denominator 26 cross times 3 end fraction equals 15 over 78
    So, 2 minus 11 over 39 plus 5 over 26 equals 156 over 78 minus 22 over 78 plus 15 over 78
    not stretchy rightwards double arrow 149 over 78

    What is the result of 2 minus 11 over 39 plus 5 over 26

    Maths-General
    To find 2 minus 11 over 39 plus 5 over 26
    Taking the LCM of 39 and 26, we have 78 as LCM.
    So, 2 equals 2 over 1 equals fraction numerator 2 cross times 78 over denominator 1 cross times 78 end fraction equals 156 over 78 and
    11 over 39 equals fraction numerator 11 cross times 2 over denominator 39 cross times 2 end fraction equals 22 over 78 and
    5 over 26 equals fraction numerator 5 cross times 3 over denominator 26 cross times 3 end fraction equals 15 over 78
    So, 2 minus 11 over 39 plus 5 over 26 equals 156 over 78 minus 22 over 78 plus 15 over 78
    not stretchy rightwards double arrow 149 over 78
    General
    Maths-

    What is the difference between the greatest and least numbers 5 over 9 to the power of blank comma 1 over 9 and 11 over 9

    Given numbers are 5 over 9 to the power of blank comma 1 over 9 and 11 over 9
    Greatest number among the given numbers = 11 over 9
    Least number among the given numbers = 1 over 9
    Difference of greatest and least number = 11 over 9 space minus space 1 over 9
    not stretchy rightwards double arrow 10 over 9

    What is the difference between the greatest and least numbers 5 over 9 to the power of blank comma 1 over 9 and 11 over 9

    Maths-General
    Given numbers are 5 over 9 to the power of blank comma 1 over 9 and 11 over 9
    Greatest number among the given numbers = 11 over 9
    Least number among the given numbers = 1 over 9
    Difference of greatest and least number = 11 over 9 space minus space 1 over 9
    not stretchy rightwards double arrow 10 over 9
    parallel
    General
    Maths-

    Which is the correct descending order of negative 2 comma fraction numerator negative 4 over denominator 5 end fraction comma fraction numerator negative 11 over denominator 20 end fraction comma 3 over 4

    Given numbers are negative 2 comma fraction numerator negative 4 over denominator 5 end fraction comma fraction numerator negative 11 over denominator 20 end fraction comma 3 over 4
    Taking the LCM of 5,4 and 20. we get the LCM as 20.
    So, negative 2 equals fraction numerator negative 2 over denominator 1 end fraction equals fraction numerator negative 2 cross times 20 over denominator 1 cross times 20 end fraction equals fraction numerator negative 40 over denominator 20 end fraction and
    and fraction numerator negative 4 over denominator 5 end fraction equals fraction numerator negative 4 cross times 4 over denominator 5 cross times 4 end fraction equals fraction numerator negative 16 over denominator 20 end fraction
    3 over 4 equals fraction numerator 3 cross times 5 over denominator 4 cross times 5 end fraction equals 15 over 20
    Now we have fraction numerator negative 40 over denominator 20 end fraction comma fraction numerator negative 16 over denominator 20 end fraction comma fraction numerator negative 11 over denominator 20 end fraction comma 15 over 20
    Hence fractions in descending order are  15 over 20 greater than fraction numerator negative 11 over denominator 20 end fraction greater than fraction numerator negative 16 over denominator 20 end fraction greater than fraction numerator negative 40 over denominator 20 end fraction
    That is, 3 over 4 greater than fraction numerator negative 11 over denominator 20 end fraction greater than fraction numerator negative 4 over denominator 5 end fraction greater than negative 2
    Hence option B is the correct order.

    Which is the correct descending order of negative 2 comma fraction numerator negative 4 over denominator 5 end fraction comma fraction numerator negative 11 over denominator 20 end fraction comma 3 over 4

    Maths-General
    Given numbers are negative 2 comma fraction numerator negative 4 over denominator 5 end fraction comma fraction numerator negative 11 over denominator 20 end fraction comma 3 over 4
    Taking the LCM of 5,4 and 20. we get the LCM as 20.
    So, negative 2 equals fraction numerator negative 2 over denominator 1 end fraction equals fraction numerator negative 2 cross times 20 over denominator 1 cross times 20 end fraction equals fraction numerator negative 40 over denominator 20 end fraction and
    and fraction numerator negative 4 over denominator 5 end fraction equals fraction numerator negative 4 cross times 4 over denominator 5 cross times 4 end fraction equals fraction numerator negative 16 over denominator 20 end fraction
    3 over 4 equals fraction numerator 3 cross times 5 over denominator 4 cross times 5 end fraction equals 15 over 20
    Now we have fraction numerator negative 40 over denominator 20 end fraction comma fraction numerator negative 16 over denominator 20 end fraction comma fraction numerator negative 11 over denominator 20 end fraction comma 15 over 20
    Hence fractions in descending order are  15 over 20 greater than fraction numerator negative 11 over denominator 20 end fraction greater than fraction numerator negative 16 over denominator 20 end fraction greater than fraction numerator negative 40 over denominator 20 end fraction
    That is, 3 over 4 greater than fraction numerator negative 11 over denominator 20 end fraction greater than fraction numerator negative 4 over denominator 5 end fraction greater than negative 2
    Hence option B is the correct order.
    General
    Maths-

    The sum of two rational numbers is -5. If one of the numbers is fraction numerator negative 3 over denominator 5 end fraction What is the other number?

    Given that one rational number is fraction numerator negative 3 over denominator 5 end fraction and let the other one be x.
    Their sum = -5, that is fraction numerator negative 3 over denominator 5 end fraction plus x equals negative 5
    not stretchy rightwards double arrow x equals negative 5 plus 3 over 5

    not stretchy rightwards double arrow x equals fraction numerator negative 22 over denominator 5 end fraction

    The sum of two rational numbers is -5. If one of the numbers is fraction numerator negative 3 over denominator 5 end fraction What is the other number?

    Maths-General
    Given that one rational number is fraction numerator negative 3 over denominator 5 end fraction and let the other one be x.
    Their sum = -5, that is fraction numerator negative 3 over denominator 5 end fraction plus x equals negative 5
    not stretchy rightwards double arrow x equals negative 5 plus 3 over 5

    not stretchy rightwards double arrow x equals fraction numerator negative 22 over denominator 5 end fraction
    General
    Maths-

    In the given figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7cm, find the area of the shaded region

    It is given that OA = 7 cm
    i.e. Radius of bigger circle = 7 cm
    Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC
    Now, OD is radius of bigger circle and diameter of smaller circle
    i.e. radius of smaller circle = half of radius of bigger circle
    r1 = 3.5 cm
    Area of smaller circle = straight pi (r1)2
    = 3.14 × (3.5)2 = 38.465 cm2
    Area of semi-circle OACB = 1 half × r2
    = 1 half × 3.14 × 72 = 76.93 cm2
    AB = diameter of bigger circle = 2 × 7 = 14 cm
    OC = radius of bigger circle = 7 cm
    Area of triangle ABC = 1 half × b × h  = 1 half × AB × OC
    = 1 half × 14 × 7 = 49 cm2
    Now, Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC
    = 38.465 cm2 + 76.93 cm2 - 49 cm2
    = 66.395 cm2

    In the given figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7cm, find the area of the shaded region

    Maths-General
    It is given that OA = 7 cm
    i.e. Radius of bigger circle = 7 cm
    Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC
    Now, OD is radius of bigger circle and diameter of smaller circle
    i.e. radius of smaller circle = half of radius of bigger circle
    r1 = 3.5 cm
    Area of smaller circle = straight pi (r1)2
    = 3.14 × (3.5)2 = 38.465 cm2
    Area of semi-circle OACB = 1 half × r2
    = 1 half × 3.14 × 72 = 76.93 cm2
    AB = diameter of bigger circle = 2 × 7 = 14 cm
    OC = radius of bigger circle = 7 cm
    Area of triangle ABC = 1 half × b × h  = 1 half × AB × OC
    = 1 half × 14 × 7 = 49 cm2
    Now, Area of shaded region = Area of smaller circle + Area of semi-circle OACB – Area of triangle ABC
    = 38.465 cm2 + 76.93 cm2 - 49 cm2
    = 66.395 cm2
    parallel
    General
    Maths-

    What is the number that is to be subtracted from fraction numerator negative 7 over denominator 8 end fraction 𝑡𝑜 𝑔𝑒𝑡 fraction numerator negative 13 over denominator 12 end fraction

    Given rational number = fraction numerator negative 7 over denominator 8 end fraction and let the other rational number to be subtracted from
    fraction numerator negative 7 over denominator 8 end fraction =  x
    On subtraction, we get the result to be fraction numerator negative 13 over denominator 12 end fraction
    That is, we have fraction numerator negative 7 over denominator 8 end fraction minus x equals fraction numerator negative 13 over denominator 12 end fraction
    On rearranging the equation, we get  x equals 13 over 12 plus fraction numerator negative 7 over denominator 8 end fraction
    On taking LCM of 12 and 8 , we have 24 as the LCM.
    not stretchy rightwards double arrow x equals fraction numerator 13 cross times 2 over denominator 12 cross times 2 end fraction plus fraction numerator negative 7 cross times 3 over denominator 8 cross times 3 end fraction
    not stretchy rightwards double arrow x equals 26 over 24 minus 21 over 24
    not stretchy rightwards double arrow x equals 5 over 24

    What is the number that is to be subtracted from fraction numerator negative 7 over denominator 8 end fraction 𝑡𝑜 𝑔𝑒𝑡 fraction numerator negative 13 over denominator 12 end fraction

    Maths-General
    Given rational number = fraction numerator negative 7 over denominator 8 end fraction and let the other rational number to be subtracted from
    fraction numerator negative 7 over denominator 8 end fraction =  x
    On subtraction, we get the result to be fraction numerator negative 13 over denominator 12 end fraction
    That is, we have fraction numerator negative 7 over denominator 8 end fraction minus x equals fraction numerator negative 13 over denominator 12 end fraction
    On rearranging the equation, we get  x equals 13 over 12 plus fraction numerator negative 7 over denominator 8 end fraction
    On taking LCM of 12 and 8 , we have 24 as the LCM.
    not stretchy rightwards double arrow x equals fraction numerator 13 cross times 2 over denominator 12 cross times 2 end fraction plus fraction numerator negative 7 cross times 3 over denominator 8 cross times 3 end fraction
    not stretchy rightwards double arrow x equals 26 over 24 minus 21 over 24
    not stretchy rightwards double arrow x equals 5 over 24
    General
    Maths-

    The circle touch each other internally. The sum of their area is 116 pie cm2 and the distance between their centres is 6 cm. Find the radii of the circles.

    It is given that the distance between their centres is 6 cm
    rightwards double arrow r2 - r1 = 6 rightwards double arrow r2 = 6 + r1
    The sum of their area of two circles =  116 straight pi cm2
    rightwards double arrowstraight pi (r1)2straight pi (r2)2 = 116 straight pi
    rightwards double arrow straight pi [ (r1)2 + (6 + r1)2 ] = 116 straight pi
    rightwards double arrow  (r1)2 + 36 +  (r1)2 + 12 r = 116
    rightwards double arrow 2 (r1)2 + 12 r + 36 – 116 = 0
    rightwards double arrow 2 (r1)2 + 12 r – 80 = 0 rightwards double arrow (r1)2 + 6 r – 40 = 0
    Using splitting the middle term,
    We get, (r1)2 + 10 r – 4r – 40 = 0
    (r1 + 10 )( r1 – 4) = 0 rightwards double arrow r1 = 4 , - 10
    Since radius is always positive, rightwards double arrow r1 = 4 cm
    rightwards double arrowr2 = 4 + 6 = 10 cm

    The circle touch each other internally. The sum of their area is 116 pie cm2 and the distance between their centres is 6 cm. Find the radii of the circles.

    Maths-General
    It is given that the distance between their centres is 6 cm
    rightwards double arrow r2 - r1 = 6 rightwards double arrow r2 = 6 + r1
    The sum of their area of two circles =  116 straight pi cm2
    rightwards double arrowstraight pi (r1)2straight pi (r2)2 = 116 straight pi
    rightwards double arrow straight pi [ (r1)2 + (6 + r1)2 ] = 116 straight pi
    rightwards double arrow  (r1)2 + 36 +  (r1)2 + 12 r = 116
    rightwards double arrow 2 (r1)2 + 12 r + 36 – 116 = 0
    rightwards double arrow 2 (r1)2 + 12 r – 80 = 0 rightwards double arrow (r1)2 + 6 r – 40 = 0
    Using splitting the middle term,
    We get, (r1)2 + 10 r – 4r – 40 = 0
    (r1 + 10 )( r1 – 4) = 0 rightwards double arrow r1 = 4 , - 10
    Since radius is always positive, rightwards double arrow r1 = 4 cm
    rightwards double arrowr2 = 4 + 6 = 10 cm
    General
    Maths-

    A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

    It is given that diameter of the wheel = 77 cm
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction = 38.5
    Circumference of the wheel = 2 straight pi r
    = 2 × 22 over 7 × 38.5 = 242 cm
    It is given that Speed of the bucket = 1.1 m/s
    Time taken = 1 min 28 sec = 88 sec
    Distance covered by the wheel = speed × time
    = 1.1 × 88 = 96.8 m
    = 96.8 × 100 cm
    = 9680 cm
    No of revolutions =  fraction numerator text  distance covered by wheel  end text over denominator text  circumference of wheel  end text end fraction
    =  9680 over 242 = 40 revolutions

    A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s, calculate the number of complete revolutions of the wheel makes in raising the bucket.

    Maths-General
    It is given that diameter of the wheel = 77 cm
    rightwards double arrow Radius = fraction numerator text  diameter  end text over denominator 2 end fraction = 38.5
    Circumference of the wheel = 2 straight pi r
    = 2 × 22 over 7 × 38.5 = 242 cm
    It is given that Speed of the bucket = 1.1 m/s
    Time taken = 1 min 28 sec = 88 sec
    Distance covered by the wheel = speed × time
    = 1.1 × 88 = 96.8 m
    = 96.8 × 100 cm
    = 9680 cm
    No of revolutions =  fraction numerator text  distance covered by wheel  end text over denominator text  circumference of wheel  end text end fraction
    =  9680 over 242 = 40 revolutions
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