# Additional Questions Solutions- Constructions Notes | Study Mathematics (Maths) Class 9 - Class 9

## Class 9: Additional Questions Solutions- Constructions Notes | Study Mathematics (Maths) Class 9 - Class 9

The document Additional Questions Solutions- Constructions Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9 Q1. Construct a triangle ABC in which BC = 7.5 cm, ∠ B = 45º and the difference between the other two sides is 4 cm.

Steps of construction:

• Take BC = 7.5 cm.
• Construct ∠CBX = 45º.
• From cut off BD = 4 cm. • Join D and C.
• Draw the perpendicular bisector of DC which intersects at A.
• Join A and C.

Thus, ABC is the required triangle.

Q2. Construct a ΔABC whose perimeter is 12 cm and sides are in the ratio 2 : 3 : 4.

Steps of construction: • Draw = 12 cm.
• Draw , making an acute angle with XY in the downward direction.
• From X, set off (2 + 3 + 4 = 9) nine equal distances along XZ.
• Mark points P, Q and R on XZ such that XP = 2 units PQ = 3 units QR = 4 units
• Join RY.
• Draw QC || RY and PB || RY.
• With centre B and radius BX, draw an arc.
• With centre C and radius CY, draw an arc that intersects the previous arc at A.
• Join AB and AC.

Thus, ΔABC is the required triangle.

Q3. Construct ΔABC, in which ∠ B = 60º and ∠ C = 45º and the perpendicular from the vertex A to the base BC is 5.2 cm.

Steps of construction: • Draw a line PQ.
• Mark a point L on PQ.
• Draw LM ⊥ PQ.
• From cut off LA = 5.2 cm.
• Draw XY || PQ, through A.
• Construct ∠XAB = 60º and ∠YAC = 45º meeting PQ at B and C respectively.

Thus, ΔABC is the required triangle.

Q4. Construct an equilateral triangle if its altitude is 6 cm. Give justification for your construction.

Steps of construction: • Draw a line XY and mark a point D on it.
• Construct perpendicular PD on XY.
• Cut off AD = 6 cm
• Make ∠CAD = 30° and ∠BAD = 30° at A on both sides of AD, such that B and C lie on XY.

Thus, ABC is the required triangle.
Justification:
∴ In ΔADC, ∠C = 180 ∠ (90° + 30°) = 60°
⇒ ∠A = 60°
Thus, ΔABC is an equilateral triangle in which altitude AD = 6 cm.

Q5. Why we cannot construct a ∆ABC, if ∠A = 60°, AB = 6 cm and AC + BC = 5 cm but construction of ∆ABC is possible if ∠A = 60°, AB = 6 cm and AC – BC = 5 cm?

We know that, by triangle inequality property, construction of triangle is possible if sum of two sides of a triangle is greater than the third side. Here, AC + BC = 5 cm which is less than AB (6 cm) Thus, ∆ABC is not possible.
Also, by triangle inequality property, construction of triangle is possible, if difference of two sides of a triangle is less than the third side
Here, AC – BC = 5 cm, which is less than AB (6 cm)
Thus, ∆ABC is possible.

Q6. Construct an angle of 90° at the initial point of the given ray.

Steps of Construction: • Draw a ray OA.
• With O as centre and any convenient radius, draw an arc, cutting OA at P.
• With P as centre and same radius, draw an arc cutting the arc drawn in step 2 at Q.
• With Q as centre and the same radius as in steps 2 and 3, draw an arc, cutting the arc drawn in step 2 at R.
• With Q and R as centres and same radius, draw two arcs, cutting each other in S.
• Join OS and produce to B.
• Thus, ∠AOB is the required angle of 90°

Q7. Draw a straight angle. Using compass bisect it. Name the angles obtained.

Steps of Construction: • Draw any straight angle (say ∠AOC).
• Bisect ∠AOC and join BO.
• ∠AOB is the required bisector of straight angle AOC.

Q8. Construct a ∆ABC such that BC = 3.2 cm, ∠B = 45° and AC – AB = 2.1 cm.

Steps of Construction: • Draw a line segment BC = 3.2 cm.
• At B, construct an angle ∠CBX = 45° and produce it to point X’.
• Cut-off BD = 2.1 cm and join CD.
• Draw the perpendicular bisector of CD and let it intersect X’BX in A. Join AC.
• Thus, ∆ABC is the required triangle.

Q9. Construct a triangle ABC in which BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°. Given: In ∆ABC, BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°.
Required: To construct ∆ABC.
Steps of Construction:

• Draw BC = 4.7 cm.
• Draw From ray CX, cut off CD = 8.2 cm.
• Join BD. Draw the perpendicular bisector of BD meeting CD at A.
• Join AB to obtain the required triangle ABC.

Justification:
∵ A lies on the perpendicular bisector of BD, therefore, AB = AD
Now, CD = 8.2 cm
⇒ AC + AD = 8.2 cm
⇒ AC + AB = 8.2 cm.

Q10. To construct a triangle, with perimeter 10 cm and base angles 60° and 45°. Given: In ∆ABC,
AB + BC + CA = 10 cm,
∠B = 60° and ∠C = 45°.
Required: To construct ∆ABC.
Steps of Construction:

• Draw DE = 10 cm.
• At D, construct ∠EDP= 5 of 60°= 30° and at E, construct DEQ = 1 of 45° = 22°
• Let DP and EQ meet at A.
• Draw perpendicular bisector of AD to meet DE at B.
• Draw perpendicular bisector of AE to meet DE at C.
• Join AB and AC.
• Thus, ABC is the required triangle.
The document Additional Questions Solutions- Constructions Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9 Use Code STAYHOME200 and get INR 200 additional OFF

## Mathematics (Maths) Class 9

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