Gujarat Board Solutions Class 10 Maths Chapter 11 Constructions
Gujarat Board Textbook Solutions Class 10 Maths Chapter 11 Constructions
Ex 11.1
Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction
1. Draw AB = 7.6 cm
2. At A, draw an acute angle ∠BAX below BA.
3. On AX mark 13 (5 + 8) arcs A1, A2, …, A13.
4. Join B to A13.
5. From A5, draw A5C || A13B.
AC : CO = 5 : 8.
Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
Solution:
Steps of Construction:
1. Draw ΔABC with AC = 6cm, AB = 5 cm, BC = 4 cm.
2. At A1 draw an acute angle∠CAX below the base AC.
3. Mark 3 arcs on AX such that AA1 = A1A2 = A2A3.
4. Join A3C.
5. Draw A2C’ || A3C
6. From C’ draw B’C’ || BC
Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 5/8 of the corresponding sides of the first triangle.
Solution:
1. Construct a ΔABC with AB = 7 cm, AC = 5 cm, BC = 6 cm
2. At A draw an acute ∠BAX below AB.
3. On AX, mark 7 arcs A1, A2, …, A7 such that
AA1 = A1A2 =…=A6A7.
4. Join A5B.
5. From A7 draw A7B’ || A5B meeting AB at B’ (extent AB to B’)
6. From B’ draw B’C’ || BC meeting AC at C’ [extend AC to C’]
AB’C’ is the required triangle each of whose sides is 5/8 of the corresponding sides of ΔABC.
Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm and then another triangle whose sides are 1 times the corresponding sides of the isosceles triangle. (CBSE 2017)
Solution:
Steps of Construction:
1. Draw a line segment BC = 8 cm.
2. Draw its perpendicular bisector AD (4 cm).
3. Joining AB and AC, we get isosceles triangle ABC
4. Construct an acute ∠CBX.
5. Along BX cut 3 arcs B1, B2 and B3 such that
BB1 = B1B2 = B2B3.
6. Join C to B2 and draw a line B3C’ || B2C [extend BC to C’]
7. From C’ draw C’A’ || CA [extend BA to A’]
∴ ΔA’BC’ is a required triangle.
Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the ABC.
Solution:
Steps of Construction:
1. Draw a line segment BC = 6 cm.
2. At B draw ∠CBY = 60° on which take AB = 5 cm.
3. Join AC. ΔABC is required triangle.
4. From B, draw an acute ∠CBX downwards.
5. Mark four arcs B1, B2, B3, B4 on BX such that BB1 = B1B2 = B2B3 = B3B4.
6. Join B4C and from B3 draw B3C’ B4C.
7. From C’, draw A’C’ || AC.
When ΔA’BC’ is the required triangle whose sides are 3/4 of the corresponding sides of ΔABC.
Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct a triangle whose sides are 3/4 times the corresponding sides of the ΔABC.
Solution:
Steps of Construction:
1. Draw a ABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°
[∠C = 180° – 45° – 105° = 180° – 150° = 30°]
2. At B, draw an acute ∠CBX below BC.
3. On BX, mark the arcs B1, B2, B3, B4 such
that BB1 = B1B2 = B2B3 = B1B4.
4. Join B3C.
5. Draw B4C’|| B3C [Extend BC to C’]
6. Draw C’A’ || CA [Extend BA to A’]
∴ ΔA’BC’ is the required triangle.
Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are times the corresponding sides of the given triangle. (CBSE 2010)
Solution:
1. Draw a line segment AB = 4 cm.
2. By making right angle at B. Draw BC = 3 cm.
3. Join AC. ΔABC is the required right-angled triangle.
4. At A, Draw acute angle ∠BAX downwards.
5. On AX, mark 5 arcs A1, A2, A3, A4, A5 such
that AA1 = A1A2 = … = A4A5
6. Join A3B.
7. Draw A5B’|| A3B [Extend AB to B’]
8. Draw B1C’ || BC [Extend AC to C’]
Hence ΔAB’C’ is the required triangle
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)
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.