NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Class 9 maths chapter 11 question answer - Exercise: 11.1

Q1 Construct an angle of 90 ^o at the initial point of a given ray and justify the construction.

Answer:

The steps of construction to follow:

Step 1: Draw a ray OP.

Then, take O as the centre and any radius draw an arc cutting OP at Q.

Step 2: Now, taking Q as the centre and with the same radius as before draw an arc cutting the previous arc at R. Repeat the process with R to cut the previous arc at S.

Step 3: Take R and S as centre draw the arc of radius more than the half of RS and draw two arcs intersecting at A. Then, join OA.

Hence, $\mathrm{\angle }POA={90}^{\circ }$ .

Justification:

We need to justify, $\mathrm{\angle }POA={90}^{\circ }$

So, join OR and OS and RQ. we obtain

By construction OQ = OS = QR.

So, $\mathrm{△}ROQ$ is an equilateral triangle. Similarly $\mathrm{△}SOR$ is an equilateral triangle.

So, $\mathrm{\angle }SOR={60}^{\circ }$

Now, $\mathrm{\angle }ROQ={60}^{\circ }$ that means $\mathrm{\angle }ROP={60}^{\circ }$ .

Then, join AS and AR:

Now, in triangles OSA and ORA:

$SR=SR$ (common)

$AS=AR$ (Radii of same arcs)

$OS=OR$ (radii of the same arcs)

So, $\mathrm{\angle }SOA=\mathrm{\angle }ROA=\frac{1}{2}\mathrm{\angle }SOR$

Therefore, $\mathrm{\angle }ROA={30}^{\circ }$

and $\mathrm{\angle }POA=\mathrm{\angle }ROA+\mathrm{\angle }POR={30}^{\circ }+{60}^{\circ }={90}^{\circ }$

Hence, justified.

Q2 Construct an angle of 45 ^o at the initial point of a given ray and justify the construction.

Answer:

The steps of construction to follow:

Step 1: Draw a ray OY.

Then, take O as the centre and any radius, mark a point A on the arc ABC.

Step 2: Now, taking A as the centre and the same radius, mark a point B on the arc ABC.

Step 3: Take B as a centre and the same radius, mark a point C on the arc ABC.

Step 4: Now, taking C and B as centre one by one, draw an arc from each centre intersecting each other at a point X.

Step 5: X and O are joined and a ray making an angle ${90}^{\circ }$ with OY is formed.

Let the arc AC touches OX at E

Step 6: With A and E as centres, 2 arcs are marked intersecting each other at D and the bisector of angle XOY is drawn.

Justification:

By construction we have,

$\mathrm{\angle }XOY={90}^{\circ }$

We constructed the bisector of $\mathrm{\angle }XOY$ as $\mathrm{\angle }DOY$

Thus,

$\mathrm{\angle }DOY=\frac{1}{2}\mathrm{\angle }XOY=\frac{1}{2}\times {90}^{\circ }={45}^{\circ }$

Q3 (i) Construct the angles of the following measurements: 30 ^o

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, take A as a center and take any radius, then draw an arc AB cutting OY at A.

Step 3: Take A and B as centres, draw 2 arcs are marked intersecting each other at X and hence, the bisector of ${30}^{\circ }$ is constructed.

Thus, $\mathrm{\angle }XOY$ is the required angle.

Q3.Construct the angles of the following measurements: (i) 30o
Edit Q

Q3 (ii) Construct the angles of the following measurements: $22{\frac{1}{2}}^{\circ }$

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, take A as a centre and take any radius, then mark a point B on the arc

Step 3: Take B as a centre with the same radius, mark a point C on the arc ABC.

Step 4: Now, taking B and C as centres simultaneously, then draw an arc from each centre intersecting each other at a point X. Then join X and O and a ray making an angle with OY is formed.

Let the arc AC touches OX at E

Step 5: Now, with A and E as centres, mark 2 arcs which intersect each other at D and we obtain the bisector of the angle $XOY$ .

Q3 (iii) Construct the angles of the following measurements: 15 ^o

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, take A as a centre and take any radius, draw an arc AB which cuts OY at A.

Step 3: Take A and B as a centre, then mark 2 arcs which intersect each other at X and Hence, the bisector is constructed of ${30}^{\circ }$ .

Step 4: Now, with A and E as centres, mark 2 arcs which intersect each other at D and we obtain the bisector of the angle $XOY$ .

Thus, the angle of ${15}^{\circ }$ is obtained which is $\mathrm{\angle }EOY.$

Q4 (i) Construct the following angles and verify by measuring them by a protractor: 75 ^o

Answer:

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, taking O as the centre draw an arc ABC.

Step 3: On taking A as a centre, draw two arcs B and C on the arc ABC.

Step 4: Now, taking B and C as centres, arcs are made to intersect at point E and the $\mathrm{\angle }EOY={90}^{\circ }$ is constructed.

Step 5: Taking A and C as centres, arcs are made to intersect at D.

Step 6: Now, join OD and hence, $\mathrm{\angle }DOY={75}^{\circ }$ is constructed.

Hence, $\mathrm{\angle }DOY$ is the required angle.

Q4 (ii) Construct the following angles and verify by measuring them by a protractor: 105 ^o

Answer:

The steps of construction to be followed:

Step 1: Draw a ray OY.

Step 2: Then, taking O as a centre, draw an arc ABC.

Step 3: Now, with A as a centre, draw two arcs B and C which are made on the arc ABC.

Step 4: Taking B and C as centres simultaneously, arcs are made to intersect at E and $\mathrm{\angle }EOY={90}^{\circ }$ is constructed.

Step 5: With B and C as centres, arcs are made to intersect at X.

Step 6: Join the OX and we get $\mathrm{\angle }XOY={105}^{\circ }$ is constructed.

Thus, the angle $XOY$ is ${105}^{\circ }.$

Q4 (iii) Construct the following angles and verify by measuring them by a protractor: 135 ^o

Answer:

The steps of construction to be followed:

Step 1: Draw a ray DY.

Step 2: Draw an arc ACD with O as a center.

Step 3: Now, with A as a centre, draw two arcs B and C on the arc ACD.

Step 4: Taking B and C as centres, arcs are made to intersect at E and the angle formed is $\mathrm{\angle }EOY={90}^{\circ }$ .

Step 5: Take F and D as centres, draw arcs to intersect at point X or the bisector of angle EOD is made.

Step 6: Join OX and the $\mathrm{\angle }XOY={135}^{\circ }$ is made.

Hence, the angle required $\mathrm{\angle }XOY$ is ${135}^{\circ }$ .

Q5 Construct an equilateral triangle, given its side and justify the construction.

Answer:

The following steps to make an equilateral triangle:

Step 1: Draw a line segment AB = 4 cm.

Step 2: With A and B as centres, make two arcs in the line segment AB. Mark it as D and E respectively.

Step 3: Now, with D and E as centres, make the two arcs cutting the previous arcs respectively, and forming an angle of ${60}^{\circ }$ each.

Step 4: Extend the lines of A and B until they intersect each other at point C.

Hence, triangle constructed is ABC which is equilateral.

Now, Justification ;

Since the angles constructed are of ${60}^{\circ }$ each, so the third angle will also be ${60}^{\circ }$ .

$[\because Thesumofallanglesoftriangle={180}^{\circ }]$

Class 9 maths chapter 11 NCERT solutions - Excercise: 11.2

Q1 Construct a triangle ABC in which $BC=7cm$ , $\mathrm{\angle }B={75}^{\circ }$ and $AB+AC=13cm$ .

Answer:

The steps of construction are as follows:

Step 1: Draw a line segment BC of 7cm length. Taking the help of protractor make an $\mathrm{\angle }XBC={75}^{\circ }$ .

Step 2: Now, cut a line segment BD having 13 cm on BX which is $(AB+AC).$

Step 3: Now, join CD.

Step 4: Draw a perpendicular bisector of CD to intersect BD at a point A. Join AC. Then ABC is the required triangle

Hence, the required triangle is ABC.

Q2 Construct a triangle ABC in which $BC=8cm$ , $\mathrm{\angle }B={45}^{\circ }$ ° and $AB-AC=3.5cm$

Answer:

The steps of construction to be followed:

Step 1: Draw a line segment BC = 8cm and make an angle of ${45}^{\circ }$ at point B i.e., $\mathrm{\angle }XBC.$

Step 2: Now, cut the line segment BD = 3.5 cm on ray BX. i.e. $(AB-AC)$ .

Step 3: Join CD and draw a perpendicular bisector of CD i.e., PQ.

Step 4: Let the perpendicular bisector of CD intersects BX at point A. Then,

Step 5: Join AC, to get the required triangle $\mathrm{△}ABC.$

Q3 Construct a triangle PQR in which $QR=6cm$ , $\mathrm{\angle }Q={60}^{\circ }$ and $PR-PQ=2cm$ .

Answer:

The steps of construction to be followed:

Step 1: Draw a ray QX and cut off a line segment QR which is equal to 6 cm in length.

Step 2: With an angle of ${60}^{\circ }$ with QR, construct a ray QY and extend it to form a line YQY'.

Step 3: Now, cut off a line segment QS equal to 2 cm from QY' and join RS.

Y

Y'

Step 4: Draw a perpendicular bisector of RS which intersects QY at a point P.

Step 5: Join PR to get the required triangle $\mathrm{△}PQR.$

Q4 Construct a triangle XYZ in which $\mathrm{\angle }Y={30}^{\circ }$ , $\mathrm{\angle }Z={90}^{\circ }$ ° and $XY+YZ+ZX=11cm$ .

Answer:

The steps of construction to be followed:

Step 1: For given $XY+YZ+ZX=11cm$ , a line segment $PQ=11cm$ is drawn.

Step 2: At points, P and Q angles of $\mathrm{\angle }RPQ={30}^{\circ }$ and $\mathrm{\angle }SQP={90}^{\circ }$ are constructed respectively.

Step 3: Now, bisects the angle RPQ and SQP. The bisectors of these angles intersect each other at a point X.

Step 4: Construct the perpendicular bisector of PX and QX, name them as TU and WV respectively.

Step 5: Let the bisector TU intersect PQ at Y and bisector WV intersect PQ at Z. Then XY and ZY are joined.

Therefore, $\mathrm{△}XYZ$ is the required triangle.

Q5 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Answer:

The steps of construction to follow:

Step 1: Draw a ray BX and Cut off a line segment $BC=12cm$ from the ray.

Step 2: Now, construct an angle $\mathrm{\angle }XBY={90}^{\circ }$ .

Step 3: Cut off a line segment BD of length 18 cm on BY. Then join the CD.

Step 4: Now, construct a perpendicular bisector of CD which intersects BD at A and AC is joined.

Thus, the constructed triangle is ABC.