GSEB Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.2

Question 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Answer:Steps of construction:
1. Draw a circle of radius 6 cm and take O as centre.
2. Take a point P which is 10 cm away from O. Join OP.
3. Bisect OP. Let M be the mid-point of OP.
4. Take M as centre and MO as radius, draw another circle intersecting the previous circle at Q and R.
5. Join PQ and PR.
PQ and PR are the required tangents.
PQ = PR = 8cm. Justification:
Join OQ and OR.
\( \angle OQP = \angle ORP = 90^\circ \) [Angles in semicircles]
Since OQ and OR are radii of the circle and PQ and PR will be the tangents to the circle at Q and R respectively. A circle with diameter OP intersects the given circle in only two points.
Hence, only two tangents can be drawn.
In simple words: First, you draw a circle and mark a point outside it. Then, you use a special method to draw two lines from that outside point that just touch the circle perfectly. Finally, you measure these lines to see how long they are.

Exam Tip: Remember that the tangent at any point of a circle is perpendicular to the radius through the point of contact. This property is crucial for justification.

Question 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer:Steps of Construction:
1. Draw a circle of radius 4 cm with centre O.
2. Taking O as centre, draw another circle of radius 6 cm.
3. Take a point P on the outer circle. Join OP.
4. Bisect OP. Let M be the mid-point of OP.
5. Take M as centre and MO as radius, draw a circle. Let it intersect the given circle at the point Q and R.
6. Join PQ.
PQ is the required tangent. PQ = 4.5 cm. By actual Calculation:
As \( \angle OQP = 90^\circ \) (Angle in a semicircle)
By Pythagoras theorem:
\( OP^2 = PQ^2 + OQ^2 \)
\( PQ = \sqrt{OP^2 - OQ^2} \)
\( = \sqrt{6^2 - 4^2} \)
\( = \sqrt{36 - 16} = \sqrt{20} \)
\( = 4.47 \) cm (approx) Justification:
Since \( \angle OQP = 90^\circ \) (Angle in a semicircle)
\( \implies \) PQ \( \perp \) OQ
Since OQ is the radius of the given circle, PQ has to be a tangent to the circle.
In simple words: Draw two circles from the same center, one smaller and one larger. Pick a point on the larger circle. Now, from this point, draw a line that just touches the smaller circle. Measure this line, and then use math (Pythagoras theorem) to check if your measurement is correct.

Exam Tip: For verification by calculation, always use the Pythagoras theorem, as the radius at the point of tangency forms a right angle with the tangent.

Question 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Answer:Steps of construction:
1. Draw a circle of radius 3 cm of centre O.
2. Take two points P and Q on one of its extended diameter each at a distance 7 cm from its centre O.
3. Bisect PO. Let M be the mid-point of PO.
4. Take M as centre and MO as radius, draw a circle intersecting the given circle at A and B.
5. Join PA and PB.
6. Bisect QO. Let N be the mid-point of QO.
7. Take N as centre and NO as radius, draw a circle intersecting the given circle at C and D.
8. Join QC and QD.
PA, PB, QC and QD are the required tangents. Justification:
Join OA and OB.
\( \implies \angle PAO = 90^\circ \) [Angle in Semicircle]
PA \( \perp \) OA
Since OA is the radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also tangent to the circle.
With the same above explanation, QC and QD are also tangent to the circle.
In simple words: Draw a circle. Extend a line straight through its center. Mark two points on this extended line, both equally far from the center, but on opposite sides. From each of these two points, draw lines that just touch the circle.

Exam Tip: When constructing tangents from two external points on an extended diameter, ensure symmetry in your construction. Remember that the tangent is perpendicular to the radius at the point of contact for both external points.

Question 4. Draw a pair of tangents to a circle to radius 5 cm which are inclined to each other at an angle of 60°.
Answer:Steps of construction:
1. Draw a circle of radius 5 cm and take O as centre.
2. Take a point A on the circle and draw \( \angle AOB = 120^\circ \).
3. At A and B, draw angles of \( 90^\circ \) which other arms meet at C. Then, AC and BC are the required tangents inclining each other at an angle of \( 60^\circ \). Justification:
\( \angle OAC = 90^\circ \) [By construction]
\( \angle OBC = 90^\circ \)
\( \implies \) OA \( \perp \) AC
OB \( \perp \) BC
Also, OA and OB are the radii of the circle.
Therefore, AC and BC are tangents to the circle. In quadrilateral \( \triangle OBC \),
\( \angle AOB + \angle OBC + \angle BCA + \angle CAO = 360^\circ \) [Angle sum property of a quadrilateral]
\( \implies 120^\circ + 90^\circ + \angle BCA + 90^\circ = 360^\circ \)
\( \implies \angle BCA + 300^\circ = 360^\circ \)
\( \implies \angle BCA = 360^\circ - 300^\circ \)
\( \implies \angle BCA = 60^\circ \)
In simple words: Draw a circle. Create a special angle of 120 degrees at the center, touching the circle at two points. Then, from these two points on the circle, draw lines that meet outside, forming an angle of 60 degrees. These meeting lines are your tangents.

Exam Tip: The angle between the two radii to the points of tangency and the angle between the tangents from the external point are supplementary, summing to 180 degrees. If the tangents meet at 60 degrees, the central angle will be 120 degrees.

Question 5. Draw a line segment AB of length 8 cm. Taking A as the centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Answer:Steps of Construction:
1. Draw a line segment AB = 8 cm.
2. Taking A as the centre, draw a circle of radius 4 cm.
3. Now taking B as centre, draw a circle of radius 3 cm.
4. Bisect AB. Let M be the mid-point of AB.
5. Taking M as a centre and AM as radius, draw a circle, intersecting the circle with centre A at P and Q; and intersecting the circle with centre B at R and S.
6. Join BP and BQ.
7. Join AR and AS. BP, BQ, AR and AS are the required tangents. Justification:
\( \angle APB = 90^\circ \) [Angles in the semicircle]
\( \angle ARB = 90^\circ \)
BP \( \perp \) AP
and AR \( \perp \) BR
Since AP and BR are the radii of the circles with A and B as centre respectively. So, BP and AR are the tangents to the circle with centre A and B respectively.
In simple words: Draw a straight line and mark two points, A and B, 8 cm apart. Now, draw a circle using A as the center and a 4 cm radius, and another circle using B as the center and a 3 cm radius. Your task is to draw lines from point A that touch the circle at B, and from point B that touch the circle at A.

Exam Tip: Tangents from a point to a circle are equal in length. The construction involves finding the midpoint of the line joining the center of the first circle to the external point (which is the center of the other circle), and then drawing a third circle.

Question 6. Let ABC be a right triangle in which AB = 6cm, BC = 8cm and \( \angle B = 90^\circ \). BD is perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Answer:Steps of Construction:
1. Draw a \( \triangle ABC \) with AB = 6 cm, BC = 8 cm and \( \angle B = 90^\circ \).
2. Draw BD \( \perp \) AC.
3. Through points B, C and D, draw a circle and mark the centre as O.
4. Join AO and bisect it. Let M be the midpoint of AO.
5. Taking M as centre and MO as radius, draw a circle intersecting the given circle at B and E.
6. Join AB and AE.
Thus, AB and AE are the required tangents. Justification:
By joining OE.
\( \angle AEO = 90^\circ \) [Angle in the semicircle]
\( \implies \) AE \( \perp \) OE
Now since OE is a radius of the given circle, AE will be a tangent to the circle.
In simple words: First, draw a right-angled triangle. Then, draw a line from the right angle corner perpendicular to the opposite side. Now, draw a circle that goes through three special points of your triangle. Finally, from one of the triangle's corners, draw lines that just touch this new circle.

Exam Tip: Recall that the circle passing through B, C, D will have BC as its diameter, as \( \angle BDC = 90^\circ \) (angle in a semicircle). This helps in locating the center O.

Question 7. Draw a circle with the help of a bangle. Take a point outside the circle. Construction the pair of tangents from this point to the circle.
Answer:Steps of Construction:
1. Draw a circle with the help of a bangle.
2. Take two chords AB and CD (non-parallel to each other).
3. Draw the perpendicular bisectors of AB and CD intersecting at O. Then O is the centre of the given circle.
4. Take a point P outside the circle. Join OP.
5. Bisect OP. Let M be the mid-point of OP.
6. Taking M as centre and MO as radius, draw a circle intersecting the given circle at Q and R.
7. Join PQ and PR.
PQ and PR are the required tangents. Justification:
Join OQ and OR.
\( \angle PQO = 90^\circ \) [Angle in the semicircle]
\( \implies \) PQ \( \perp \) OQ
Since OQ is a radius of the given circle, PQ will be a tangent to the circle. Similarly, PR will also be a tangent to the circle.
In simple words: If you have a circle drawn with a bangle, you first need to find its exact center using chords and perpendicular bisectors. Once you find the center, pick a point outside this circle. Then, draw two lines from this outside point that smoothly touch the circle.

Exam Tip: When a circle's center is not given (e.g., drawn with a bangle), the first step is always to locate its center by drawing two non-parallel chords and finding the intersection of their perpendicular bisectors.