NCERT Solutions Class 9 Mathematics Chapter 11 Construction

Exercise 11.1

Q.1) Construct an angle of 90° at the initial point of a given ray and justify the construction.
Sol.1) Steps of construction:

i) Let us take a ray AB with initial point A.
ii) Taking A as a centre & some radius, draw an arc of a circle, which intersects AB at C.
iii) With C as centre & the same radius as before, draw an arc, intersecting the previous arc at E.
iv) With E as a centre as a same radius, as before, draw an arc, which intersects the arc drawn in step (ii) at F.
v) With E as a centre & some radius draw an arc.
vi) With F as centre & the same radius as before, draw another arc, intersecting the previous arc at G.
vii) Draw the ray AG. Then ∠BAG is the required angle of 90°.
Justification:
Join 𝐴𝐸, 𝐶𝐸, 𝐸𝐹, 𝐹𝐺 AND 𝐺𝐸.
𝐴𝐶 = 𝐶𝐸 = 𝐴𝐸 (By construction)
⇒ ΔACE is an equilateral triangle.
⇒ ∠CAE = 60° (i)
Similarly, ∠AEF = 60° (ii)
From (i) & (ii), 𝐹𝐸||AC (iii) (Alternate angles are equal)
Also, 𝐹𝐺 = 𝐸𝐺 (By construction)
⇒ G lies on a perpendicular bisector of EF.
⇒ ∠GIE = 90° (iv)
∴ ∠GAB = ∠GIE = 90° (corresponding angles)
𝐺𝐹 = 𝐺𝐸 (arcs of equal radii)

Q.2) Construct an angle of 45°at the initial point of a given ray and justify the construction.
Sol.2) Steps of construction:
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc cutting OA at B.
3. With B as centre and same radius cut the previous drawn arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°
7. Draw the bisector 𝑂𝐹 of ∠AOE. Then ∠AOF = 45°

Justification :
By construction ∠AOE = 90° and OF is the bisector of ∠AOE.
Therefore, ∠AOF = ∠AOE = 1/2 × 90° = 45°.

Q.3) Construct the angles of the following measurements:
i) 30° (ii) 22(1/2°)
(iii) 15°
Sol.3) (i) Steps of construction:
1. Draw a ray OA.
2. With its initial point O as centre and any radius, draw an arc, cutting OA at C.
3. With centre C and Same radius (as in step 2). Draw an arc, cutting the arc of step 2 in D.
4. With C and D as centers, and any convenient radius
(more than 1/2 𝐶𝐷), draw two arcs intersecting at B.
5. Join OB. Then ∠AOB = 30°.

(ii) Steps of Construction :
1. Draw an angle 𝐴𝑂𝐵 = 90°
2. Draw the bisector OC of ∠AOB, then ∠AOC = 45°.
3. Bisect ∠AOD, such that ∠AOD = ∠COD = 22.5°.
Thus ∠AOD = 22.5°.

(iii) Steps of Construction :
1. Construct an ∠AOB = 60°.
2. Bisect ∠AOB so that ∠AOC = ∠BOC = 30°.
3. Bisect ∠AOD, so that ∠AOD = ∠COD = 15°.
Thus ∠AOD = 15°.

Q.4) Construct the following angles and verify by measuring them by a protractor :
(i) 75° (ii) 105° (iii) 135°
Sol.4) i) 75°
Steps of constructions:

Step 1: A ray OY is drawn.
Step 2: An arc BAE is drawn with O as a center.
Step 3: With E as a center, two arcs are A and C are made on the arc BAE.
Step 4: With A and B as center, arcs are made to intersect at
X and ∠XOY = 90° is made.
Step 5: With A and C as center, arcs are made to intersect at D
Step 6: OD is joined and ∠DOY = 75° is constructed.
Thus, ∠DOY is the required angle making 75° with OY.

ii) 105°
Steps of constructions:

Step 1: A ray OY is drawn.
Step 2: An arc ABC is drawn with O as a center.
Step 3: With A as a center, two arcs are B and C
are made on the arc ABC.
Step 4: With B and C as center, arcs are made to
intersect at E and ∠𝐸𝑂𝑌 = 90° is made.
Step 5: With B and C as center, arcs are made to intersect at X
Step 6: OX is joined and ∠𝑋𝑂𝑌 = 105° is constructed.
Thus, ∠XOY is the required angle making 105° with OY.

iii) 135°
Steps of constructions:

Step 1: A ray DY is drawn.
Step 2: An arc ACD is drawn with O as a center.
Step 3: With A as a center, two arcs are B and C
are made on the arc ACD.
Step 4: With B and C as center, arcs are made to
intersect at E and ∠𝐸𝑂𝑌 = 90° is made.
Step 5: With F and D as center, arcs are made to intersect at X or bisector of ∠𝐸𝑂𝐷 is constructed.
Step 6: OX is joined and ∠𝑋𝑂𝑌 = 135° is constructed.
Thus, ∠𝑋𝑂𝑌 is the required angle making 135° with DY

Q.5) Construct an equilateral triangle, given its side and justify the construction.
Sol.5) Steps of constructions:
Step 1: A line segment 𝐴𝐵 = 4 𝑐𝑚 is drawn.
Step 2: With A and B as centres, two arcs are made.
Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60° each.
Step 5: Lines from A and B are extended to meet each other at C.
Thus, ABC is the required triangle formed.
Justification:
By construction,
𝐴𝐵 = 4 𝑐𝑚, ∠𝐴 = 60° and ∠𝐵 = 60°
We know that,
∠𝐴 + ∠𝐵 + ∠𝐶 = 180° (Sum of the angles of a triangle)
⇒ 60° + 60° + ∠𝐶 = 180°
⇒ 120° + ∠𝐶 = 180°
⇒ ∠𝐶 = 60°
𝐵𝐶 = 𝐶𝐴 = 4 𝑐𝑚 (Sides opposite to equal angles are equal)
𝐴𝐵 = 𝐵𝐶 = 𝐶𝐴 = 4 𝑐𝑚
∠𝐴 = ∠𝐵 = ∠𝐶 = 60°

Exercise 11.2

Q.1) Construct a triangle ABC in which 𝐵𝐶 = 7𝑐𝑚, ∠𝐵 = 75° and 𝐴𝐵 + 𝐴𝐶 = 13 𝑐𝑚.
Sol.1) Steps of Construction:
Step 1: A line segment BC of 7 cm is drawn.
Step 2: At point B, an angle ∠𝑋𝐵𝐶 is constructed
such that it is equal to 75°.
Step 3: A line segment 𝐵𝐷 = 13 𝑐𝑚
is cut on BX (which is equal to AB+AC).
Step 3: DC is joined and ∠𝐷𝐶𝑌 = ∠𝐵𝐷𝐶 is made.
Step 4: Let CY intersect BX at A.
Thus, 𝛥𝐴𝐵𝐶 is the required triangle.

Q.2) Construct a triangle ABC in which 𝐵𝐶 = 8𝑐𝑚, ∠𝐵 = 45° and 𝐴𝐵 – 𝐴𝐶 = 3.5 𝑐𝑚.
Sol.2) Steps of Construction:
Step 1: A line segment 𝐵𝐶 = 8 𝑐𝑚 is drawn and at point B,
make an angle of 45° i.e. ∠𝑋𝐵𝐶.
Step 2: Cut the line segment 𝐵𝐷 = 3.5 𝑐𝑚
(equal to AB – AC) on ray BX.
Step 3: Join DC and draw the perpendicular bisector PQ of DC.
Step 4: Let it intersect BX at point A. Join AC.
Thus, 𝛥𝐴𝐵𝐶 is the required triangle.

Q.3) Construct a triangle PQR in which 𝑄𝑅 = 6𝑐𝑚, ∠𝑄 = 60° and 𝑃𝑅 – 𝑃𝑄 = 2𝑐𝑚.
Sol.3) Steps of Construction:
Step 1: A ray 𝑄𝑋 is drawn and cut off a line
segment 𝑄𝑅 = 6 𝑐𝑚 from it.
Step 2:. A ray QY is constructed making an angle
of 60° with QR and YQ is produced to form a line 𝑌𝑄𝑌’
Step 3: Cut off a line segment 𝑄𝑆 = 2𝑐𝑚
from 𝑄𝑌’. 𝑅𝑆 is joined.
Step 5: Draw perpendicular bisector of RS intersecting
QY at a point 𝑃. 𝑃𝑅 is joined.
Thus, 𝛥𝑃𝑄𝑅 is the required triangle.

Q.4) Construct a triangle 𝑋𝑌𝑍 in which ∠𝑌 = 30°, ∠𝑍 = 90° and 𝑋𝑌 + 𝑌𝑍 + 𝑍𝑋 = 11 𝑐𝑚.
Sol.4) Steps of Construction:
Step 1: A line segment 𝑃𝑄 = 11 𝑐𝑚 is
drawn. (𝑋𝑌 + 𝑌𝑍 + 𝑍𝑋 = 11 𝑐𝑚)
Step 2: An angle, ∠𝑅𝑃𝑄 = 30° is constructed at point A
and an angle ∠𝑆𝑄𝑃 = 90° at point B.
Step 3: ∠𝑅𝑃𝑄 and ∠𝑆𝑄𝑃 are bisected . The bisectors of these
angles intersect each other at point 𝑋.
Step 4: Perpendicular bisectors TU of 𝑃𝑋 and WV of 𝑄𝑋 are constructed.
Step 5: Let TU intersect PQ at Y and 𝑊𝑉 intersect PQ at Z.
Step 6: XY and XZ are joined.
Thus, 𝛥𝑋𝑌𝑍 is the required triangle.

Q.5) Construct a right triangle whose base is 12𝑐𝑚 and sum of its hypotenuse and other side is 18 cm
Sol.5) Steps of Construction:
Step 1: A ray 𝐵𝑋 is drawn and a cut off a line segment 𝐵𝐶 = 12 𝑐𝑚 is made on it.
Step 2: ∠𝑋𝐵𝑌 = 90° is constructed.

Step 3: Cut off a line segment 𝐵𝐷 = 18 𝑐𝑚 is made on 𝐵𝑌. 𝐶𝐷 is joined.
Step 4: Perpendicular bisector of CD is constructed intersecting BD at 𝐴.
Step 5: 𝐴𝐶 is joined.
Thus, 𝛥𝐴𝐵𝐶 is the required triangle.