SSLC Math's Solution Chapter2 Circles PDF-Download

SSLC Math's Solution Chapter2 Circles PDF-DownloadSSLC Math's Solution Chapter2 Circles PDF-Download

An equation in two variables, often x2 + y2 = r2, must be satisfied by a collection of points (x,y) to be considered a circle in mathematics. These points are positioned at specific intervals from the point, which is at the centre (h,k). You must find the area using the circle's equation as a starting point. Solved examples for the topic of circles from the chapter "circles" of the SSLC 10th maths textbook. Examples of problems that have been solved include sums with thorough solutions.

The circumference of a circle, which is a closed, looping line, defines the shape as a planar figure. A circle's perimeter is known as its circumference or arc of definition. Another way to characterise a circle is as being on its circumference.

Board SCERT, Kerala
Text Book SCERT Based
Class SSLC 
Subject Maths Solution
Chapter Chapter 2
Chapter Name Circles
Category Kerala SSLC


Kerala Syllabus SSLC Class 10 Maths Textbook Solution Chapter 2 Circles



Chapter 2  Circles Textbook Solution

Qn 1.

Suppose we draw a circle with the bottom side of the triangles in the picture as diameter. Find out whether the top corner of each triangle is inside the circle, on the circle or outside the circle.

  • Answer)

              

    Angle of the first triangle =110°
    As 110° > 90° the top comer will be inside the circle
    Angle of second triangle = 90°
    ∴ The top comer will be on the circle.
    Angle of the third triangle = 70°
    70° > 90°
    ∴ The top comer will be outside the circle


Qn 2.

For each diagonal of the quadrilateral shown, check whether the other two corners are inside, on or outside the circle with that diagonal as diameter

  • Answer)


Qn 3.

If circles are drawn with each side of a triangle of sides 5 centimeters, 12 centimeters and 13 centimeters, as diameters, then with respect to each circle, where would be the third vertex?

  • Answer)

    As the sides are 5, 12, 13 cm and also
    52 + 122 = 25 + 144 = 169 = 132
    ∴ ABC is a right triangle

    Taking BC as diameter and drawing a circle, ∠A (<90°), A will be outside the circle.
    Taking AB as diameter and drawing a circle ∠C (<90°) C will be outside the circle.
    Taking AC as diameter and drawing a circle, ∠B = 90°, B will be on the circle.


Qn 4.

In the picture, a circle is drawn with a line as diameter and a smaller circle with half the line as diameter. Prove that any chord of the larger circle through the point where the circles meet is bisected by the small circle.

  • Answer)

    ∠ADO = ∠APB = 90°
    (angle subtended by diameter is always 90°)
    ⇒ OD\\PB

    AO = OB (Radius of bigger circle)
    (If in a triangle, the line drawn from midpoint of one side, is parallel to another side, then the line will bisect the third side)
    Therefore AD = DP
    (AB’s midpoint is ‘O’ and OD\\PB)


Qn 5.

Use a calculator to determine up to two decimal places, the perimeter and the area of the circle in the picture.

  • Answer)


Qn 6.

The two circles in the picture cross each other at A and B. The points P and Q are the other ends of the diameters through A.

i. Prove that P, B, Q lie on a line.
ii. Prove that PQ is parallel to the line joining the centers of the circles and is twice as long as this line.

  • Answer)

    ii)

     

     


Qn 7.

Prove that the two circles are drawn on the two equal sides of an isosceles triangle as diameters pass through the midpoint of the third side.

  • Answer)


Qn 8.

Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point.

Prove that this is true for any quadrilateral with adjacent sides equal, as in the picture.

  • Answer)

    ABCD is a rhombus so diameter are perpendicular bisectors.
    ∠AOD = 90°
    O be on the circle having diameter AD.
    ∠AOB = 90°, therefore
    O be on the circle having diameter AB.
    ∠BOC= 90°, therefore
    O be on the circle having diameter BC
    ∠DOC= 90°, therefore
    O be on the circle having diameter DC
    O be the common point on the circle.
    ∠A0D = ∠AOB and
    ∠COD = ∠BOC and,
    AD = AB,
    AO be the common side.

    Δ AOD, Δ AOB are equal triangles.
    OD = OB
    Both the circles can passed through O.
    A BCD is an isosceles triangle.
    Those circles having diameters CD and BC are passing through midpoint of BD.
    ∴ O be common for the four circles. (Diameter)


Qn 9.

A triangle is drawn by joining a point on a semicircle to the ends of the diameter. Then semicircles are drawn with the other two sides as diameter.

Prove that the sum of the areas of the blue and red crescents in the second picture is equal to the area of the triangle.

  • Answer)

    Area of triangle be A

    Area of small crescent be ‘a’

    Area of larger crescent be ‘b’

    Area of semicircle excluding one crescent be ‘x’ and ‘y’.

    Sides of the right triangle is p, q, r

    Area of semicircle with diameter ‘r’ = Area of semicircle with diameter P + Area of semicircle with diameter q.

    A + x + y = a + x + b + y

                A = a + b

    That is area of triangle is equal to two areas of crescents.


Additional Excercise 1.1

Qn 1.

In all the pictures given below, O is the centre of the circle and A, B, C are points on it. Calculate all angles of Δ ABC and Δ OBC in each.

  • Answer)

    a)  

        

    b)

        

    c) 


Qn 2.

The numbers 1,4,8 on a clock’s face are joined to make a triangle.

Calculate the angles of this triangle.
How many equilateral triangles can we make by joining numbers on the clock’s face?

  • Answer)

    Angle between 1 and 4 is 30 x 3 = 90

    The angles of these triangles are 45o,60o,75o. We can make 4 equilateral by joining the numbers (1, 5, 9) (2, 6, 10) (3, 7, 11) and (4, 8, 12)


Qn 3.

In each problem below, draw a circle and a chord to divide it into two parts such that the parts are as specified.
i. All angles on one part 80°.
ii. All angles on one part 110°.
iii. All angles on one part half of all angles on the other.
iv. All angles on one part, one and a half times the angles on the other.

  • Answer)

     


Qn 4.

A rod bent into an angle is placed with its corner at the center of a circle and it is found that  of the circle lies within it. it is placed with its corner on another circle, what part of the circle would be within it?

  • Answer)


Qn 5.

  • Answer)


Qn 6.

  • Answer)


Qn 7.

In the picture, AB and CD are mutually perpendicular chords of the circle. Prove that the arcs APC and BQD joined together would make half the circle.

  • Answer)

        


Qn 8.

In the picture, A, B, C, D are points on a circle centred at O. The lines AC and BD are extended to meet at P. The lines AD and BC intersect at Q. Prove that the angle which the small arc AB makes at O is the sum of the angles it makes at P and Q.

  • Answer)


Additional Exercise 1.2

Qn 1.

Calculate  the angles of the quadrilateral in the picture and also the angles between their diagonals:

  • Answer)


Qn 2.

Prove that any outer angle of a cyclic quadrilateral is equal to the inner angle at the opposite vertex.

  • Answer)


Qn 3.

Prove that a parallelogram which is not a rectangle is not cyclic.

  • Answer)


Qn 4.

Prove that any non-isosceles trapezium is not cyclic.

  • Answer)


Qn 5.

In the picture, bisectors of adjacent angles of the quadrilateral ABCD intersect at P, Q, R, S. Prove that PQRS is a cyclic quadrilateral.

 

  • Answer)


Qn 6.

i)  The two circles below intersect at P, Q and lines through these points meet the circles at A, B, C, D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral.

ii) In the picture, the circle on the left and right intersect the middle circle at P, Q, R, S; the lines joining them meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.

  • Answer)

    i)

      

     

    ii) 


Qn 7.

In the picture, points P, Q, R are marked on the sides BC, CA, AB of ΔABC and the circumcircles of ΔAQR and ΔBRP are drawn. M is a point where these circles intersect. Prove that the circumcircle of ΔCPQ also passes through M.

  • Answer)

     


Additional Exercise 1.3

Qn 1.

In the picture, chords AB and CD of the circle are extended to meet at P.

i)Prove that the angles of Δ APC and Δ PBD, formed by joining AC and BD, are the same.
ii. Prove that PA × PB = PC × PD.
iii. Prove that if PB = PD, then ABDC is an isosceles trapezium.

  • Answer)


Qn 2.

Draw a rectangle of width 5 centimeters and height 3 centimeters.
i. Draw a rectangle of the same area with width 6 centimeters.
ii. Draw a square of the same area
.

  • Answer)

    i) 

    Draw a rectangle of length 5 cm and width 3cm.

    Extend AB up to 6cm.
    (AE = 6cm) Draw an arc having radius as AE and A as centre. Extend DA and mark the point F.

    Extend BA towards left. Mark G as AD = AG
    Draw Δ GFB.

    Circum circle of Δ GFB meets AD at D.
    ∴ AG × AB = AF × AH.
    That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.

    ii)  Draw a rectangle of width 5 cm and height 3 cm. 

    Area = 5 x 3 = 15 sq cm.

    Draw a semicircle of diameter AH.
    Extend BC, and mark the point F.
    AB × BH = 5 × 3 = 15
    AB × BH = BF2 ; BF = √5 cm
    Area of BEGF = √15 × √15 = 15 cm2


Qn 3.

Draw a square of area 15 square centimeters.

  • Answer)

    Draw a rectangle of length 5cm and width 3cm. Area = 3 × 5 = 15 cm2. Side of the square is √15.

    Draw a semicircle of diameter AH.
    BC can touch the point F.
    AB × BH = BF2  = 15 cm2.

    BF = √15
    Area of BEGF = 15 cm2.


Qn 4.

Draw a square of area 5 square centimeters in three different ways. (Recall Pythagoras theorem).

  • Answer)

    Method 1

    Draw a rectangle of length 5cm and width 1 cm.

    Draw a semicircle of diameter AE. Extend BC up to F. √5 is the side of the square BGHF.

    BGHF is the required square.

     

    Method 2

    Draw a right-angled triangle of perpendicular sides 2 cm and 1cm.
     
    Hypotenuse will be √5  cm. The area of the square ACDE is 5 cm2, because here we take the hypotenuse as side of the square.

     

    Method 3

    If the hypotenuse of right triangle is 3cm and base is 2 cm then its altitude is √5  cm .

    BEDC is the required square.


Qn 5.

In the picture, a line through the centre of a circle cuts a chord into two parts:

What is the radius of the circle?

  • Answer)

    The intersection may be with in the circle.
    Chords AB & CD intersect at P i. e.,
    PA × PB = PC × PD
    (The intersection will be inside the circle)
    4 × 6 = PC × (OP + OD)
    24 = PC × (OP + OC)
    24 = PC × (OP + OP + CP)
    24 = PC × (5 + 5 + CP)
    24 = PC × (10 + CP)
    PC =2
    Radius = PC + OP = 2 + 5 = 7cm


Qn 6.

In the picture, a line through the center of a circle meets a chord of the circle:

What are the lengths of the two pieces of the chord?

  • Answer)


 SSLC Maths Textbook Solution


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