KC Sinha Mathematics Solution Class 10 Chapter 12 Constructions Exercise 12.1

Question 1

Divide a line segment 4.2 cm long internally in the ratio 5:3 Also, write the steps of construction.

Sol :

Step1: Draw a line segment AB of length 4.2 cm

Step2: Draw a line AC at any angle below AB

Step3: take any distance in compass and keep the needle of the compass on point A and draw an arc intersecting the line AC. Name the intersection point as X₁. Keeping the distance same in compass keep the needle on point X₁ and mark an arc intersecting AC at X₂. Draw 8 (5:3 given ratio 5 + 3 = 8) such parts i.e. upto X₈

Thus we have made 8 equal parts on line AC
Step4: Join points X₈ and B

Now we have to divide the segment AB in ratio 5:3, i.e. 5 parts and 3 parts.
Step5: from point, X₅ draw a line parallel to BX₈ intersecting AB at D, and we have divided the segment AB in ratio 5:3

Question 2

Divide a line segment of length 3.2 cm in the ratio of 3:5 internally.

Sol :
Step1: Draw a line segment AB of length 3.2 cm

Step2: Draw a line AC at any angle below AB

Step3: take any distance in compass and keep the needle of the compass on point A and draw an arc intersecting the line AC. Name the intersection point as X₁. Keeping the distance same in compass keep the needle on point X₁ and mark an arc intersecting AC at X₂. Draw 8 (3:5 given ratio 3 + 5 = 8) such parts i.e. upto X₈

Step4: Join points X₈ and B

Now we have to divide the segment AB in ratio 3:5, i.e. 3 parts and 5 parts.

Step5: from point, X₃ draw a line parallel to BX₈ intersecting AB at D, and we have divided the segment AB in ratio 3:5

Question 3

Draw a line segment of length 5.6 cm and divide it internally in the ratio 5:8. Measure the two parts.

Sol :
Step1: Draw a line segment AB of length 5.6 cm

Step2: Draw a line AC at any angle below AB

Step3: take any distance in compass and keep the needle of the compass on point A and draw an arc intersecting the line AC. Name the intersection point as X₁. Keeping the distance same in compass keep the needle on point X₁ and mark an arc intersecting AC at X₂. Draw 13 (5:8 given ratio 5 + 8 = 13) such parts i.e. upto X₁₃

Step4: Join points X₁₃ and B

Now we have to divide the segment AB in ratio 5:8, i.e. 5 parts and 8 parts.

Step5: from point, X₅ draw a line parallel to BX₁₃ intersecting AB at D, and we have divided the segment AB in ratio 5:8

And measure the length of parts, i.e. AD and DB which are 2.2 cm and 3.4 cm respectively

Question 4

Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Sol :
Step1: Draw a line segment AB of length 7.6 cm

Step2: Draw a line AC at any angle below AB

Step3: Take any distance in compass and keep the needle of the compass on point A and draw an arc intersecting the line AC. Name the intersection point as X₁. Keeping the distance same in compass keep the needle on point X₁ and mark an arc intersecting AC at X₂. Draw 13 (5:8 given ratio 5 + 8 = 13) such parts i.e. upto X₁₃

Step4: Join points X₁₃ and B

Question 5 A

Draw a line segment AB = 2 cm. Divide it externally in the ratio of 5:3

Sol :
5:3

Step1: Draw a line segment AB of length 2 cm

Step2: draw a ray at any angle below AB from A

Now we have to divide the segment AB in ratio 5:3 externally which is greater than 1. So 5 parts will be the segment AB and 3 parts will be externally added, so we have to divide the ray into 8 (5 + 3 = 8) parts

Step3: Take any distance in compass and keep the needle of the compass on point A and draw an arc intersecting the ray drawn in step2. Name the intersection point as X₁. Keeping the distance same in compass keep the needle on point X₁ and mark an arc intersecting ray at X₂. Draw 8 such parts, i.e. upto X₈

Step4: Join X₅B. Extend AB and draw a line parallel to X₅B which intersects the extended AB at C. Hence C divides externally the segment AB

Question 5 B

Draw a line segment AB = 2 cm. Divide it externally in the ratio of 3:5

Sol :
3:5

Step1: Draw a line segment AB of length 2 cm

Step2: draw a ray at any angle below AB from A

Step3: Take any distance in compass and keep the needle of the compass on point A and draw an arc intersecting the ray drawn in step2. Name the intersection point as X₁. Keeping the distance same in compass keep the needle on point X₁ and mark an arc intersecting ray at X₂. Draw 5 such parts, i.e. upto X₅

Step4: Join X₃B(X₃ because the segment AB should be 3 parts). Extend AB and draw a line parallel to X₃B which intersects the extended AB at C. Hence C divides externally the segment AB

Question 6

Construct a triangle of scale AB = 2.3 cm, BC = 4 cm and AC = 2.9 cm and then construct a triangle similar to a given ΔABC whose sides are of the corresponding sides of the triangle. Also, write the steps of construction.

Sol :
Let the triangle with sides 2.3 cm, 4 cm and 2.9 cm be ΔABC

Step1: construct segment AC of 2.9 cm

Step2: take distance 2.3 cm in compass keep the needle of the compass on point A and mark an arc above AC

Step3: take distance 4 cm in compass keep the needle of the compass on point C and mark an arc intersecting the arc drawn in step2. Mark intersection point as B join AB and AC

Step4: draw a ray from point A below AC at any angle

Step5: take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step4 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 3 such parts (greater of 2 and 3 in 2/3), i.e. by repeating this process mark points upto X₃

Step6: join X₃ and C and from X₂ (because X₂ is the second point 2 being smaller in 2/3) construct line parallel to X₃C and mark the intersection point with AC as D

Step7: construct line parallel to BC from point D and mark the intersection point with AB as E thus ΔADE ~ ΔACB is ready

Question 7

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Sol :

Let the triangle with sides 4 cm,5 cm and 6 cm be ΔABC
Step1: construct segment AC of 6 cm

Step2: take distance 4 cm in compass keep the needle of the compass on point A and mark an arc above AC

Step3: take distance 5 cm in compass keep the needle of the compass on point C and mark an arc intersecting the arc drawn in step2. Mark intersection point as B join AB and AC

Step4: draw a ray from point A below AC at any angle

Step6: join X₃ and C and from X₂ (because X₂ is the second point 2 being smaller in 2/3) construct line parallel to X₃C and mark the intersection point with AC as D

Step7: construct line parallel to BC from point D and mark the intersection point with AB as E thus ΔADE ~ ΔACB is ready

Question 8

Draw a triangle ABC with side BC = 6 cm, AB = 5cm, and ∠ABC = 60°, then construct a triangle whose sides are of the corresponding sides of ABC.

Sol :
Step1: Construct segment BC of 6 cm

Step2: construct a ray at angle 60° above BC from point B

Step3: take 5 cm in compass because of AB = 5 cm, keep the needle of the compass on point B and mark an arc intersecting the ray drawn in step 2. Mark the intersection point as A and join A and C hence ΔABC is ready

Step4: draw a ray at any angle from point B below BC

Step5: take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step4 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 4 such parts (greater of 3 and 4 in 3/4), i.e. by repeating this process mark points upto X₄

Step6: join X₄ and C and from X₃ (because X₃ is the third point 3 being smaller in 3/4) construct line parallel to X₃C and mark the intersection point with BC as D

BD is three forth of BC

Step7: construct line parallel to AC from point D and mark the intersection point with AB as E thus ΔBDE which have sides three forth of ΔABC is ready.

Question 9

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 2.2 cm and 2.2 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Sol :
Step1: Construct a segment AB of 2.2 cm

Step2: Construct AC of 2.2 cm at 90°. Join B and C to get right-angled triangle ABC

Step3: Draw a ray at any angle from point A below AB

Step4: Take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step3 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 5 such parts (greater of 5 and 3 in 5/3), i.e. by repeating this process mark points upto X₅

Step5: Join X₃ and B (3 being smaller of 5 and 3 in

and not X₅ because the ratio

is greater than 1)

Step6: Now extend AB and draw a line parallel to X₃B from X₅ intersecting AB at D

Step7: Extend AC and draw a line parallel to BC from point D intersecting AC at E and ΔADE whose sides are

times ΔABC is ready

Question 10

Construct an isosceles triangle whose base is 3.2 cm and altitude 1.7 cm and then construct another triangle whose sides are times the corresponding sides of the isosceles triangle.

Sol :
Step1: Draw the base of triangle AB = 3.2 cm

Step2: Using scale mark the centre of AB as M and from M using protractor draw a line perpendicular to AB

Step3: Take distance 1.7 cm in compass keep the needle of the compass on point M and mark an arc intersecting ray drawn in step2. Mark the intersection point as C and join AC and BC.

Step4: Draw a ray at any angle below AB from A

The scaling factor is

Step5: Take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step4 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 3 such parts (greater of 3 and 2 in 3/2), i.e. by repeating this process mark points upto X₃

Step6: Join X₂ and B (2 being smaller of 3 and 2 in

and not X₃ because the ratio

is greater than 1)

Step7: Now extend AB and draw a line parallel to X₂B from X₃ intersecting AB at D

Step8: Extend AC and draw a line parallel to BC from point D intersecting AC at E and ΔADE whose sides are

i.e. 1 1/2 times ΔABC is ready

Question 11

Draw triangle ABC with side BC = 4 cm, ∠B = 45°, ∠C = 30°. Then construct a triangle whose sides are times the corresponding sides ABC.

Sol :
Step1: Draw segment BC of 4 cm

Step2: using protractor draw a ray at angle 45° from point B and a ray at angle 30° from point C. mark intersection of both these rays as point A

Step3: Draw a ray at any angle below BC from B

Step4: Take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step3 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 4 such parts (greater of 4 and 3 in 4/3), i.e. by repeating this process mark points upto X₄

Step5: Join X₃ and C (3 being smaller of 4 and 3 in

and not X₄ because the ratio

is greater than 1)

Step6: Now extend BC and draw a line parallel to X₃C from X₄ intersecting BC at D

Step7: Draw a line parallel to AC from point D intersecting BA at E and ΔEBD whose sides are

times ΔABC is ready

Question 12

Construct a triangle ABC, similar to a given isosceles triangle PQR, with QR = 2.8 cm, PQ = 2.5 cm, such that each of its side th of the corresponding sides of the ΔPQR. Also draw the circumcircle of ΔPBC.

Sol :
Note: we have to construct triangle PBC and not ABC

Step1: Draw PQ = 2.5 cm

As ΔPQR is isosceles QR = PR = 2.8 cm

Step2: Take distance 2.8 cm in compass, keep the needle on point P and mark an arc above PQ. Keeping the distance in the compass same keep the needle on point Q and mark an arc intersecting the previous arc. Mark intersection point as R

Step3: Join PR and QR and draw a ray from point P below PQ

Step4: Take any distance in compass and keeping the needle of the compass on point P cut an arc on ray constructed in step3 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 7 such parts (greater of 6 and 7 in 6/7), i.e. by repeating this process mark points upto X₇

Step5: join X₇ and Q and from X₆ (6 being smaller in

) draw a line parallel to X₇Q intersecting PQ at B

Step6: Draw a line parallel to QR from point B intersecting PR at C and ΔPBC is ready

Now to construct the circumcircle of ΔPBC. The centre of circumcircle is the intersection of perpendicular bisectors, and the radius is the distance from the centre to any vertex of the triangle. We will draw perpendicular bisector of PC and BC

Step7: Take any distance approximately greater than half of BC in compass. Keep the needle of the compass on point B and mark arcs to both sides of BC.

Step8: Keeping the distance in the compass same keep the needle on point C and mark arcs intersecting arcs drawn in step7. Draw a line between these intersecting arcs

Step9: Repeat step7 and step8 to draw perpendicular bisector for PC and mark the intersection point of both perpendicular bisectors as O

Step10: Keep the needle of the compass on point O and draw circle taking radius OC. Circumcentre of ΔPBC is ready

Question 13

Construct a ΔABC in which AB = 2.6 cm ∠B = 60° and altitude CD = 1.8 cm. Construct a ΔAQR similar to ΔABC, such that each side of ΔAQR is 1.5 times that of the corresponding side of ΔABC.

Sol :
Step1: Construct AB = 2.6 cm

First, we have to make a line parallel to AB at 1.8 cm

Step2: Mark a point P at 1.8 cm from segment AB above it. Draw a line passing through point P and intersecting AB at T as shown

Step3: Take any distance in compass keep the needle on point T and mark an arc intersecting AB and PT at K and L respectively. Keeping the distance in compass same keep the needle on point P and draw an arc which intersects line TP at J

Step4: Take the distance of arc LK in compass and keep the needle on point J and draw an arc intersecting the arc passing from J at point M. Draw a line through point M and N and is parallel to AB

Step5: Draw the line at 60° from point B intersecting the line drawn in step4 at point C. Join AC and BC

Step6: draw a ray at any angle below BA from point A

Now we have to construct the triangle AQR which is 1.5 times that of the corresponding side of ΔABC

The scaling factor is

Step7: Take any distance in compass and keeping the needle of the compass on point A and cut an arc on ray constructed in step6 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 3 such parts (greater of 3 and 2 in 3/2), i.e. by repeating this process mark points upto X₃

Step8: Join X₂ and B (2 being smaller of 3 and 2 in

and not X₃ because the ratio

is greater than 1)

Step9: Extend AB and draw a line parallel to X₂B from X₃ intersecting AB at R

Step10: Extend AC and draw a line parallel to BC from R intersecting AC at Q and Δ AQR is ready

Question 14

Construct an isosceles triangle whose base is 8cm and altitude 4cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.

Sol :
Step1: Draw the base of triangle AB = 8 cm

Step2: Using scale mark the centre of AB as M and from M using protractor draw a line perpendicular to AB

Step3: Take distance 4 cm in compass keep the needle of the compass on point M and mark an arc intersecting ray drawn in step2. Mark the intersection point as C and join AC and BC.

Step4: Draw a ray at any angle below AB from A

The scaling factor is

Step6: Join X₂ and B (2 being smaller of 3 and 2 in

and not X₃ because the ratio

is greater than 1)

Step7: Now extend AB and draw a line parallel to X₂B from X₃ intersecting AB at D

Step8: Extend AC and draw a line parallel to BC from point D intersecting AC at E and ΔADE whose sides are

i.e. 1 1/2 times ΔABC is ready

Question 15

Draw a right triangle in which the sides (other than hypotenuse) are the length 4cm and 3cm. Then construct another triangle whose sides are times the corresponding sides of the given triangle.

Sol :
Step1: Construct a segment AB of 4 cm

Step2: Construct AC of 3 cm at 90°. Join B and C to get right-angled triangle ABC

Step3: Draw a ray at any angle from point A below AB

Step5: Join X₃ and B (3 being smaller of 5 and 3 in

and not X₅ because the ratio

is greater than 1)

Step6: Now extend AB and draw a line parallel to X₃B from X₅ intersecting AB at D

Step7: Extend AC and draw a line parallel to BC from point D intersecting AC at E and ΔADE whose sides are

times ΔABC is ready

Question 16

Draw a triangle ABC with side BC = 7cm, ∠A = 105°. Then construct a triangle whose sides are times the corresponding sides of ΔABC.

Sol :
Note: I think one more angle should have been given I am taking ∠B = 45°

Sum of angles of a triangle is 180°

⇒ ∠A + ∠B + ∠C = 180°

⇒ 105° + 45° + ∠C = 180°

⇒ ∠C = 180° - 150°

⇒ ∠C = 30°

Step1: Draw segment BC of 7 cm

Step2: using protractor draw a ray at angle 45° from point B and a ray at angle 30° from point C. mark intersection of both these rays as point A

Step3: Draw a ray at any angle below BC from B

Step4: Take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step3 and name that point X₁. Keeping the distance in compass same keep the needle of the compass on point X₁ and cut an arc on the same ray and mark that point as X₂. Draw 4 such parts (greater of 4 and 3 in 4/3), i.e. by repeating this process mark points upto X₄

Step5: Join X₃ and C (3 being smaller of 4 and 3 in

and not X₄ because the ratio

is greater than 1)

Step6: Now extend BC and draw a line parallel to X₃C from X₄ intersecting BC at D

Step7: Draw a line parallel to AC from point D intersecting BA at E and ΔEBD whose sides are

times ΔABC is ready

Question 17

Draw a circle with radius 4cm. Mark a point on it. Draw a tangent at P to the circle.

Sol :
Step1: Take distance 4 cm in compass and draw a circle with centre O

Step2: Take a point P on the circle and draw a line segment OA passing through P as shown

Step3: take any distance in compass keep the needle of the compass on point P and mark arcs to the left and right of P intersecting OA at J and K respectively

Step4: Take any distance in compass greater than JP, keep the needle on point J and mark arcs above and below OA.

Step6: Keeping the distance in the compass same as in step4, keep the needle on point K and mark arcs intersecting the arcs drawn in step4 at points R and T. Draw a line passing through R and T which is the tangent to circle at point P

Question 18

Draw a circle of radius 3cm. Draw any diameter of the circle. At the end points of the diameter of the circle, draw tangents to the circle. Any they parallel?

Sol :
Step1: Take distance 3 cm in compass and draw a circle with centre O

Step2: Draw diameter AB. We know that the radius is perpendicular to the tangent. Using protractor draw lines at 90° from point A and B

Take points C and D on tangents as shown

⇒ ∠BAC = ∠ABD …both 90° as the radius is perpendicular to the tangent ∠BAC and ∠ABD are alternate angles for the two tangents with transversal as AB.

As alternate angles are equal the tangents are parallel.

Question 19

Draw a circle of radius 5cm. Take a point P on the circle. Draw the tangent of the circle at point P without using the centre of the circle.

Sol :
Step1: Take distance 5 cm in compass and draw a circle and take a point P on circle

Step2: Draw a chord PQ and subtend an angle ∠PRQ on the major arc of the circle

Using the alternate segment theorem, we will draw an ∠QPT congruent to ∠PRQ so that the line passing through PT will be tangent to circle at point P
Step3: Take any distance in compass keep the needle on point R and mark an arc intersecting PR and QR at J and K respectively

Step4: Keeping the distance in the compass same as that in step3 keep the needle on P and mark an arc intersecting PQ at S

Step5: Measure the distance of arc JK in compass, keep the needle on point S and mark an arc intersecting the arc drawn in step4 at point T. Draw line passing through point P and T and it is the tangent.

Question 20

Draw a circle of radius 6cm. From a point 10cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Sol :
Step1: Draw circle of radius 6 cm with centre O

Step2: Draw segment OP of 10 cm

Step3: Using scale take midpoint of OP as M

Step4: take distance MP in compass and draw arcs intersecting the circle at points T and Q as shown

Step5: Draw lines passing through PT and PQ which are the required tangents and measure length PT and PQ them with scale

Question 21

Draw two concentric circles with centre O and radii 2 cm and 4 cm. From a point on the outer circle draw a tangent to the inner circle.

Sol :
Step1: Draw a circle of radius 2 cm with centre O by taking 2 cm in compass. This is the inner circle

Step2: Now take 4 cm in compass keep the needle on point O and draw a circle. This is the outer circle. Take any point P on the outer circle

Now we have to draw a tangent from point P to the inner circle. This is the same as drawing tangents to circle from an external point.

Step4: Join OP and using scale mark the midpoint of OP which will lie on the inner circle

Step5: Take the distance MO in compass keep the needle on point M and mark arcs cutting the inner circle at point Q and T as shown

Step6: Construct a line passing through PQ and PT which are the required tangents

Question 22

Draw a circle of radius 4cm form a point on the concentric circle of radius 6cm and measure its length. Also, verify the measurement by actual calculation.

Sol :
Step1: Draw a circle of radius 4 cm with centre O by taking 4 cm in compass. This is the inner circle

Step2: Now take 6 cm in compass keep the needle on point O and draw a circle. This is the outer circle. Take any point P on the outer circle

Now we have to draw a tangent from point P to the inner circle. This is the same as drawing tangents to circle from an external point.

Step4: Join OP and using scale mark the midpoint of OP as M

Step5: Take the distance MO in compass keep the needle on point M and mark arcs cutting the inner circle at point Q and T as shown

Step6: Construct a line passing through PQ and PT which are the required tangents and measure the lengths PQ and PT using a scale

For verification

Let's join OQ

⇒ ∠OQP = 90° …radius OQ is perpendicular to tangent PQ at the point of contact Q

Consider ΔOQP

⇒ OQ = 4 cm …radius of inner circle

⇒ OP = 6 cm …radius of outer circle

Using Pythagoras

⇒ OP² = OQ² + PQ²

⇒ 6² = 4² + PQ²

⇒ 36 = 16 + PQ²

⇒ PQ² = 20

⇒ PQ = √20

⇒ PQ = √(5 × 4)

⇒ PQ = 2√5

⇒ PQ = 4.5 cm

Hence verified

Question 23

Draw a circle of radius 3cm. Take two points P and Q on one of its extended diameter each at a distance of 7cm from its centre. Draw tangents to the circle from these two points P and Q.

Sol :
Step1: Take 3 cm in compass and draw a circle with centre O

Step2: Draw a straight line passing through the centre and mark points P and Q on both sides of O at 7 cm each.

Step3: Using scale mark the midpoint of OP as M. Keep the needle on M take distance MO in compass and mark arcs intersecting the circle at point Q and T as shown

Step4: Join points PT and PQ thus PT and PQ are required tangents from point P

Step5: Similarly repeat steps step3 and step4 for tangents from point Q. Take midpoint as N then centre as N and radius NO cut arcs on the circle at K, and L. QK and QL will be the required tangents from point Q

Question 24

Draw a circle of radius 5cm. Take a point P outside the circle. Construct a pair of tangents from P to the circle without using its centre.

Sol :
Step1: Take distance 5 cm in compass and draw a circle. Take any point P outside the circle and draw a straight line which cuts the circle at A and B where AB is the chord

Step2: Take distance PA in compass and keep the needle on point P and mark arc to the left of P intersecting the line at C. Hence we have PA = PC

Now we have to draw perpendicular bisector of BC

Step3: Take any distance in compass approximately greater than half of CB and keeping the needle on point C mark arcs above and below CB

Step4: Keeping the distance in the compass same as that of in step3 keep the needle on B and mark arcs intersecting the arcs drawn in step3. Join these intersection points we get the perpendicular bisector of CB at M

Step5: Take distance MB in compass keep the needle on point M and draw a semicircle as shown

Step6: Using protractor draw a line from P perpendicular to CB intersecting the semicircle drawn in step5 at L

Step7: Take distance PL in compass, keep the needle on P and mark arc intersecting the circle at Q and T. Join PT and PQ thus PT and PQ are the required triangles

Question 25

Draw a circle of radius 4 cm. Draw two tangents to the circle such that they include an angle of 135°.

Sol :
Consider a rough figure as shown DB and DC are tangents centre of circle is A

In quadrilateral ABDC

∠BDC = 135° …given

∠DBA = 90° …radius is perpendicular to tangent at point of contact

∠DCA = 90° …radius is perpendicular to tangent at point of contact

As the sum of angles of a quadrilateral is 360°

⇒ ∠BDC + ∠DBA + ∠DCA + ∠BAC = 360°

⇒ 135° + 90° + 90° + ∠BAC = 360°

⇒ 315° + ∠BAC = 360°

⇒ ∠BAC = 45°

Now let us construct

Step1: Construct a circle of radius 4 cm mark the centre as A and draw radius AB

Step2: Using protractor draw the line at 45° to AB from point A and mark its intersection point with a circle as C join AC

Step3: Using protractor draw a line perpendicular to AB from point B because tangent is perpendicular to the radius. Thus this line is tangent to circle at point B

Step4: Using protractor draw a line perpendicular to AC from point C and mark the intersection point with a line drawn in step3 as D

Hence tangents DB and DC are ready at angle 135°

Question 26

Draw a circle of radius 5cm. Draw any line through the centre of the circle. Draw a tangent to the circle making an angle of 45° with the line. What is the length of the tangent?

Sol :
Consider a rough figure as shown CB is tangent and centre of the circle is A. CA is a line passing through the centre

∠BCA = 45° …given

∠CBA = 90° …radius is perpendicular to the tangent

Consider ΔABC

⇒ ∠ABC + ∠ACB + ∠BAC = 180° …sum of angles of triangle

⇒ 90° + 45° + ∠BAC = 180°

⇒ 135° + ∠BAC = 180°

⇒ ∠BAC = 45°

Now let us construct

Step1: Take distance 5 cm in compass and draw a circle with centre A and draw a line passing through A

Step2: Using protractor draw the line at 45° to the line drawn in step1 from point A intersecting circle at B

Step3: Using protractor draw a line perpendicular to AB from point B because the radius is perpendicular to the tangent. Mark the intersection of this line with a line passing through the centre as C hence CB is the required tangent at 45°

Measure the length CB with a scale which is the length of the tangent

CB = 5 cm length of tangent is 5 cm

Question 27

Draw a pair of tangents to a circle of radius 2.3 cm which is inclined to each other at an angle of 60°.

Sol :
Consider a rough figure as shown DB and DC are tangents centre of circle is A

In quadrilateral ABDC

∠BDC = 60° …given

∠DBA = 90° …radius is perpendicular to tangent at point of contact

∠DCA = 90° …radius is perpendicular to tangent at point of contact

As the sum of angles of a quadrilateral is 360°

⇒ ∠BDC + ∠DBA + ∠DCA + ∠BAC = 360°

⇒ 60° + 90° + 90° + ∠BAC = 360°

⇒ 240° + ∠BAC = 360°

⇒ ∠BAC = 120°

Now let us construct

Step1: Construct a circle of radius 2.3 cm mark the centre as A and draw radius AB

Step2: Using protractor draw the line at 120° to AB from point A and mark its intersection point with a circle as C join AC

Step3: Using protractor draw a line perpendicular to AB from point B because tangent is perpendicular to the radius. Thus this line is tangent to circle at point B

Step4: Using protractor draw a line perpendicular to AC from point C and mark the intersection point with a line drawn in step3 as D

Hence tangents DB and DC are ready at angle 60°

S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1