Gujarat Board GSEB Solutions Class 10 Maths Chapter 11 Constructions Ex 11.2 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 10 Maths Chapter 11 Constructions Ex 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.
Solution:
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.
∠OQP = ∠ORP = 90° [Angles in semicircies]
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. Circle with diameter OP intersects the given circle in only two points.
Hence, only two tangents can be drawn.
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.
Solution:
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 an 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 ∠OQP = 90° (Angle in a semicircle)
By Pythagoras theorem:
OP2 = PQ2 + OQ2
PQ = \(\sqrt{\mathrm{OP}^{2}-\mathrm{OQ}^{2}}\)
= \(\sqrt{6^{2}-4^{2}}\)
= \(\sqrt{36-16}\) = \(\sqrt{20}\)
= 4.47 cm (approx)
Justification:
Since ∠OQP = 90° (Angle in a semicircle)
⇒ PQ ⊥ OQ
Since OQ is the radius of the given circle, PQ has to be a tangent to the circle.
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.
Solution:
Steps of construction:
1. Draw a circle of radius 3 cm of centre O.
2. Take two points P arid 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 of C and D.
8. Join QC and QD.
PA, PB, QC and QD are the required tangents.
Justification:
Join OA and OB.
⇒ ∠PAO = 90° [Angle in Semicircle]
PA ⊥ OA
Since OA is the radius of the given circle PA has to be a tangent to the circle. Similarly, PB in also tangent to the circle. ,.
With the same above explanation, QC and QD are also tangent to the circle.
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°.
Solution:
Steps of construction
1. Draw a circle of radius 5 cm and take O as centre.
2. Take a point A on a circle and draw ∠AOB = 120°.
3. At A and B draw angles of 90° which other arms meet at C. Then, AC and BC are the required tangents inclining each other at an angle of 60°.
Justification:
∠OAC = 90° [By construction]
∠OBC = 90°
⇒ OA ⊥ AC
OB ⊥ BC
Also, OA and OB are the radii of the circle.
∴ AC and BC is a tangent to the circle. In quadrilateral ΔOBC,
∠AOB + ∠OBC + ∠BCA + ∠CAO = 360°
[Angle sum property of a quadrilaterall
⇒ 120° + 90° + ∠BCA + 90° = 360°
⇒ ∠BCA + 300° = 360°
⇒ ∠BCA = 360° – 300° = 60°
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.
Solution:
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:
∠APB = 90° [Angles in the semicircle]
∠ARB = 90°
BP ⊥ AP
and AR ⊥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.
Question 6.
Let ABC be a right triangle in which AB = 6cm, BC = 8cm and ∠B = 90°. BD is perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of Construction:
1. Draw a ABC with AB 6 cm, BC = 8 cm and ∠B = 90°.
2. Draw BD ⊥ 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.
∠AEO = 90° [Angle in the semicircie]
⇒ AE ⊥ OE
Now since OE is a radius of the given circle, AE will be a tangent to the circle.
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.
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
∠PQO = 90° [Angle in the semicircle]
⇒ PQ ⊥ OQ
Since OQ in a radius of the given circle. PQ will be a tangent to the circle. Similarly, PR will also be a tangent to the circle.