Haryana State Board HBSE 9th Class Maths Important Questions Chapter 11 Constructions Important Questions and Answers.

## Haryana Board 9th Class Maths Important Questions Chapter 11 Constructions

Very Short Answer Type Questions

Question 1.

With the help of protractor, draw an angle of 110°. Bisect it to get an angle of measure 55°. [NCERT Exemplar Problems]

Solution:

Steps of construction:

Step – I: Draw a ray BC.

Step – II: Construct ∠ABC = 110°, with the help of protractor.

Step – III: Taking B as centre and a suitable radius, draw an arc intersecting ray AB and BC at Q and P respectively.

Step – IV: Taking P and Q as the centres and radius more than \(\frac{1}{2}\)PQ, draw two arcs intersecting each other at D.

Step – V: Draw ray BD. Ray BD is required bisector of ∠ABC. On measuring with the help of the protractor, we get ∠CBD = ∠ABD = 55°.

Question 2.

Draw adjacent supplementary angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.

Solution:

Steps of construction :

Step – I: Draw ∠AOB = 180°.

Step – II: Draw ray OC such that it divides ∠AOB into two adjacent supplementary angles.

i.e., ∠AOC + ∠BOC = 180°.

Step – III: Draw the bisector OD of ∠AOC as followed by the steps in 1.

Step – IV: Draw the bisector OE of ∠BOC as followed by the steps in 1.

Verification: Since ray OD is the bisector of ∠AOC.

∴ ∠AOD = ∠COD = \(\frac{1}{2}\)∠AOC…(i)

Similarly, QE is the bisector of ∠BOC.

∴ ∠BOE = ∠COE = \(\frac{1}{2}\)∠BOC …..(ii)

Adding (i) and (ii), we get

∠COD + ∠COE = \(\frac{1}{2}\)∠AOC + \(\frac{1}{2}\)∠BOC

⇒ ∠COD + ∠COE = \(\frac{1}{2}\)(∠AOC + ∠BOC)

⇒ ∠DOE = \(\frac{1}{2}\) × 180°

[∵ ∠AOC and ∠BOC are supplementary angles]

⇒ ∠DOE = 90°

DO ⊥ OE

Yes, bisecting rays are perpendicular to each other.

Question 3.

Draw a circle whose centre is O. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle.

Solution:

Steps of construction:

Step – I: With the help of compass and scale draw a circle of radius 4.5 cm (say) whose centre is O.

Step – II: Draw chord AB of suitable length.

Step – III: Taking A as the centre and radius more than \(\frac{1}{2}\)AB draw two arcs one on each side of AB.

Step – IV: Taking B as the centre and same radius as before, draw two arcs intersecting previous arc at C and D respectively.

Step – V: Join CD, intersecting AB at M. So, CD is the perpendicular bisector of AB. Yes, perpendicular bisector CD passes through the centre of the circle.

Question 4.

Draw a circle whose centre is O. Draw its two chords PQ and QR such that PQ ≠ RS. Do they intersect at point O?

Solution:

Steps of construction:

Step – I: Draw a circle of suitable radius whose centre is O.

Step – II: Draw two chords PQ and QR of suitable lengths.

Step – III: Draw the perpendicular bisector AB of chord PQ as followed by the steps in 3.

Step – IV: Similarly draw the perpendicular bisector CD of chord QR.

Step – V: AB and CD intersect each other at point O.

Yes, perpendicular bisectors of chord PQ and chord QR intersect each other at point O.

Question 5.

Draw a line segment of length 10 cm with the help of ruler and compass to obtain a length of 7.5 cm.

Solution:

Steps of construction :

Step – I: Draw a line segment of length 10 cm.

Step – II: Draw the perpendicular bisector PQ of line AB as followed by the steps in 3 intersecting AB at M.

∴ AM = MB = \(\frac{10}{2}\) = 5 cm.

Step – III: Similarly, draw the perpendicular bisector of AM intersecting AM at N.

∴ AN = NM = \(\frac{5}{2}\) = 2.5 cm

Now, BN = BM + NM

⇒ BN = 5 cm + 2.5 cm

⇒ BN = 7.5 cm

Hence, we obtain BN of length 7.5 cm.

Question 6.

Using ruler and compass only, construct the following angles :

(a) 120° (b) 150° (c) 37\(\frac{1^{\circ}}{2}\)

Solution:

(a) Steps of construction:

Step – I: Draw a ray AB with initial point A.

Step – II: Taking A as centre and a suitable radius, draw an arc intersecting AB at P.

Step – III: Taking P as centre and same radius as before, draw an arc intersecting the previous arc at Q.

Step – IV: Taking Q as centre and same radius as before, draw an arc intersecting the previous arc at R.

Step – V: Join AR and produced it to C. Then ∠CAB = 120° is required angle.

(b) Steps of construction:

Step – I: Draw a ray AB with initial point A.

Step – II: Taking A as centre and a suitable radius, draw an arc intersecting AB at P.

Step – III: Taking P as centre and same radius as before, draw an arc intersecting previous arc at Q.

Step – IV: Taking Q as centre and same radius as before, draw an arc intersecting previous arc at R. Join AR and produced it to C. Then ZCAB = 120°.

Step – V: Taking R as centre and same radius as before, draw an arc intersecting previous arc at S. Join AS. Then ∠CAS = 180° – 120° = 60°.

Step – VI: Draw bisector AD of ∠CAS.

∴ ∠CAD = \(\frac{1}{2}\) × ∠CAS

⇒ ∠CAD = \(\frac{1}{2}\) × 60° = 30°

∠DAB = 120° + 30° = 150°.

Hence, ∠DAB = 150° is the required angle.

(c) Steps of construction:

Step – I: Draw a ray AB with initial point A.

Step – II: Draw ∠CAB = 60° as followed by the steps in Construction 11.3.

Step – III: Draw ∠DAB = 90° as followed by the steps in 1. of Ex. 11:1. Then

∠DAC = 90° – 60° = 30°.

Step – IV: Draw bisector AE of ∠DAC.

∴ ∠DAE = ∠CAE = \(\frac{30^{\circ}}{2}\) = 15°

∠EAB = 90° – ∠DAE

⇒ ∠EAB = 90° – 15° = 75°.

Step – V: Draw the bisector AF of ∠EAB.

∴ ∠FAB = \(\frac{1}{2}\)∠EAB

⇒ ∠FAB = \(\frac{75^{\circ}}{2}\) = 37\(\frac{1^{\circ}}{2}\)

Hence, ∠FAB = 37\(\frac{1^{\circ}}{2}\) is the required angle.

Question 7.

Construct an equilateral triangle, if one of its altitude is 3.2 cm. [NCERT Exemplar Problems]

Solution:

Steps of construction:

Step – I: Draw a line l.

Step – II: Mark any point D on the line l.

Step – III: At point D, draw DX ⊥ l and cut off AD = 3.2 cm.

Step – IV: Construct ∠DAB = 30° and ∠DAC = 30° intersecting l at B and C respectively.

Then ABC is the required triangle.

Question 8.

Construct a triangle ABC such that AB = 6 cm, BC = 6 cm and median CM = 4.3 cm.

Solution:

Steps of construction:

Step – I: Draw a line segment AB = 6 cm.

Step – II: Draw the bisector of AB intersecting AB at M.

Step – III: Taking M as the centre and radius 4.3 cm draw an arc.

Step – IV: Taking B as the centre and radius 6 cm, draw another arc which intersects the previous arc at point C.

Step – V: Join AC and BC, then ABC is the required triangle.

Question 9.

Construct an isosceles triangle whose base 6 cm and vertical angle is 60°.

Solution:

Steps of construction:

Step – I: Draw a line segment BC = 6 cm.

Step – II: Construct ∠CBP = 60°, below the line segment BC.

Step – III: Construct ∠PBQ = 90°.

Step – IV: Draw the perpendicular bisector XY of BC, intersecting BQ at O.

Step – V: Taking O as the centre and radius OB, draw a circle intersecting XY at A.

Step – VI: Join AB and AC, then ABC is the required triangle.

Question 10.

Construct right angled triangle whose hypotenuse measures 7 cm and the length of one of whose sides containing the right angle is 5 m.

Solution:

Steps of construction:

Step – I: Draw a line segment BC = 7 cm.

Step – II: Draw the perpendicular bisector of BC intersecting BC at O.

Step – III: Taking O as centre and radius OB, draw a semi-circle on BC.

Step – IV: Taking B as the centre and radius equal to 5 cm, draw an arc, interscting the semicircle at A.

Step – V: Join AB and AC. Then ABC is the required triangle.

Question 11.

Construct a right angled triangle in which ∠B = 90°, hypotenuse AC = 13 cm and the sum of the remaining sides AB and BC is 17 cm.

Solution:

Sum of the remaining two sides = 17 cm (given)

Let one of its side be x, then other its side is (17 – x) cm

hypotenuse = 13 cm

In right triangle, we have

(hypotenuse)^{2} = (one side)^{2} + (other side)^{2}

⇒ 132 = x^{2} + (17 – x)^{2}

⇒ 169 = x^{2} + 289 + 32 – 34x

⇒ 0 = 2x^{2} – 34x + 289 – 169

⇒ 2x^{2} – 34x + 120 = 0

⇒ x^{2} – 17x + 60 = 0

⇒ x^{2} – 12x – 5x + 60 = 0

⇒ x(x – 12) -5 (x – 12) = 0

⇒ (x – 12) (x – 5) = 0

⇒ x – 12 = 0 or x – 5 = 0

x = 12 or x = 5

So, sides are 12 cm and 17 – 12 = 5 cm.

So, we construct a right triangle in which ∠B = 90°, hypotenuse (AC) = 13 cm and other sides AB = 12 cm and BC = 5 cm.

Steps of construction:

Step – I: Draw a line segment AC = 13 cm.

Step – II: Draw the perpendicular bisector of AC intersecting AC at O.

Step – III: Taking O as the centre and radius AO, draw a semicircle on AC.

Step – IV: Taking A as the centre and radius equal to 12 cm, draw an arc interscting the semicircle at B.

Step – V: Join AB and BC. The ABC is the required triangle.

Question 12.

Construct a triangle ABC whose perimeter is 12 cm and sides are in the ratio 3 : 4 : 5.

Solution:

Steps of construction:

Step – I: Draw a line segment PQ = 12 cm (= perimeter).

Step – II : At the point Q, draw an acute ∠PQX with PQ.

Step – III: Divide QX into 3 + 4 + 5 = 12 equal parts.

Step – IV: Join Q_{12} to P.

Step – V: From Q_{3}, draw BQ_{3} || PQ_{12} intersecting PQ at B.

Step – VI: From Q_{7}, draw CQ_{7} || PQ_{12} intersecting PQ at C.

Step – VII: Taking B as centre and radius as QB, draw an arc.

Step – VIII: Taking C as the centre and radius as CP, draw another arc, which intersects the previous arc at A.

Step – IX: Join AB and AC, then ABC is the required triangle.

Question 13.

Construct a triangle having its perimeter 11.5 cm and ratio of the angles as 3 : 4 : 5.

Solution:

Ratio of angles = 3 : 4 : 5

Sum of ratios = 3 + 4 + 5 = 12

Ist angle = \(\frac{13}{2}\) × 180°

[∵ Sum of angles of a triangle is 180°]

⇒ I^{st} angle = 45°

⇒ II^{nd} angle = \(\frac{4}{12}\) × 180°

⇒ II^{nd} angle = 60°

III^{rd} angle = \(\frac{5}{12}\) × 180°

⇒ III^{rd} angle = 75°.

Let base angles be 45° and 60° and perimeter of triangle is 11.5 cm (given).

Now we construct a ΔABC as follows:

Steps of construction:

Step – I: Draw a line segment PQ = 11.5 cm.

Step – II: At the point Q construct ∠PQX = 45° and at the point P construct ∠QPY = 60°.

Step – III: Draw the bisectors of ∠PQX and ∠QPY intersecting at A.

Stop – IV: Draw the perpendicular bisectors of AQ and AP intersecting PQ at B and C respectively.

Step – V: Join AB and AC, then ABC is the required triangle.

Question 14.

Construct a triangle ABC, the lengths of whose medians are 5 cm, 5 cm and 6 cm. Measure the length of sides of ΔABC.

Solution:

In a ΔABC, let medians AD = 5 cm, BE = 6 cm and CF = 5 cm. We need to construct a ΔABC.

Steps of construction:

Step – I: Construct a triangle AXD with sides AX = 5 cm, AD = 5 cm and XD = 6 cm.

Step – II: Draw the medians AP and XR intersecting each other at point G.

Step – III: Extend GP to B such that GP = PB.

Step – IV: Join BD and produced to it C such that BD = CD.

Step – V: Join A to C, then ABC is the required triangle and AD its median.

On measuring the sides of ΔABC, we get AB = 5.95 cm, AC = 6.5 cm, BC = 6.5 cm.

Question 15.

Construct a rhombus whose diagonals are 4 cm and 6 cm in lengths. [NCERT Exemplar Problems]

Solution:

We know that, diagonals of a rhombusbisect each other at right angles.

Let AC and BD are diagonals of a rhombus and bisect each other at O.

Steps of construction:

Step – I: Draw diagonal AC of length 6 cm.

Step – II: Draw perpendicular bisector XY of AC intersecting AC at O.

Step – III: From O, cuts OB = OD = 2 cm.

Step – IV: Join AB, BC, CD and DA. Then ABCD is the required rhombus.

Multiple Choice Questions

Choose the correct option in each of the following:

Question 1.

Which of the following angle cannot be constructed using ruler and compass only: [NCERT Exemplar Problems]

(a) 22°\(\frac{1^{\circ}}{2}\)

(b) 67\(\frac{1^{\circ}}{2}\)

(c) 40°

(d) 37\(\frac{1^{\circ}}{2}\)

Answer:

(c) 40°

Question 2.

Which of the following angle cannot be constructed using ruler and compass only:

(a) 60°

(b) 150°

(c) 65°

(d) 22\(\frac{1^{\circ}}{2}\)

Answer:

(c) 65°

Question 3.

Which of the following angle can be constructed using ruler and compass only:

(a) 85°

(b) 115°

(c) 47\(\frac{1^{\circ}}{2}\)

(d) 75°

Answer:

(d) 75°

Question 4.

Which of the following angle can be constructed using ruler and compass only:

(a) 82\(\frac{1^{\circ}}{2}\)

(b) 92°

(c) 100°

(d) 122\(\frac{1^{\circ}}{2}\)

Answer:

(a) 82\(\frac{1^{\circ}}{2}\)

Question 5.

It is not possible to construct a triangle when its sides are :

(a) 4 cm, 2 cm and 6 cm

(b) 7 cm, 4 cm and 5 cm

(c) 4 cm, 5 cm and 8 cm

(d) 5 cm, 12 cm and 13 cm

Answer:

(a) 4 cm, 2 cm and 6 cm

Question 6.

It is possible to construct a triangle when its sides are :

(a) 5 cm, 6 cm and 13 cm

(b) 5 cm, 7 cm and 12 cm

(c) 6 cm, 4 cm and 9 cm

(d) 3 cm, 5 cm and 10 cm

Answer:

(c) 6 cm, 4 cm and 9 cm

Question 7.

It is possible to construct a triangle when its angles are :

(a) 40°, 55°, 60°

(b) 55°, 60°, 65°

(c) 50°, 70°, 80°

(d) 67°, 63°, 51°

Answer:

(b) 55°, 60°, 65°

Question 8.

The construction of ΔABC in which BC = 8 cm, ∠C = 50° is possible when (AC + BC) is :

(a) 10 cm

(b) 8 cm

(c) 7 cm

(d) 6 cm

Answer:

(a) 10 cm

Question 9.

The construction of ΔABC in which AB = 5 cm, ∠B = 70° is not possible when (BC + AC) is :

(a) 7 cm

(b) 8.5 cm

(c) 4.5 cm

(d) 9 cm

Answer:

(c) 4.5 cm

Question 10.

The construction of a triangle ABC in which AB = 4 cm, ∠A = 60° is not possible when difference of BC and AC is equal to: [NCERT Exemplar Problems]

(a) 3.5 cm

(b) 4.5 cm

(c) 3 cm

(d) 2.5 cm

Answer:

(b) 4.5 cm

Question 11.

The construction of a triangle ABC, given that BC = 6 cm, ∠B = 45° is not possible when difference of AB and AC is equal to : [NCERT Exemplar Problems]

(a) 6.9 cm

(b) 5.2 cm

(c) 5.0 cm

(d) 4.0 cm

Answer:

(a) 6.9 cm

Question 12.

The construction of triangle ABC, given that BC = 3 cm, ∠C = 60° is possible when difference of AB and AC is equal to : [NCERT Exemplar Problems]

(a) 3.2 cm

(b) 31 cm

(c) 3 cm

(d) 2.8 cm

Answer:

(b) 31 cm