Mathematics Part ii Solutions Solutions for Class 9 Maths Chapter 4 Constructions Of Triangles are provided here with simple step-by-step explanations. These solutions for Constructions Of Triangles are extremely popular among class 9 students for Maths Constructions Of Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 9 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 53:
Question 1:
Construct PQR, in which QR = 4.2cm Q = and PQ + PR = 8.5cm
Answer:
Steps of constructions:
(1) Draw seg QR of length 4.2 cm.
(2) Draw ray QT such that ∠RQT = 40°.
(3) Mark point S on ray QT such that QS = 8.5 cm
(4) Draw seg RS.
(5) Construct the perpendicular bisector of seg RS .
(6) Name the point of intersection of ray QS and the perpendicular bisector of RS as P.
(7) Draw seg PR.
Therefore, â–³ABC is the required triangle
Page No 53:
Question 2:
Construct XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. XYZ =
Answer:
Steps of constructions:
(1) Draw seg YZ of length 6 cm.
(2) Draw ray YV such that ∠ZYV = 50°.
(3) Mark point W on ray YV such that YW = 9 cm
(4) Draw seg WZ.
(5) Construct the perpendicular bisector of seg WZ .
(6) Name the point of intersection of ray YW and the perpendicular bisector of WZ as X.
(7) Draw seg XZ.
Therefore, â–³XYZ is the required triangle.
Page No 53:
Question 3:
Construct ABC, in which BC = 6.2 cm, ACB= , AB + AC = 9.8 cm
Answer:
Steps of constructions:
(1) Draw seg BC of length 6.2 cm.
(2) Draw ray CD such that ∠BCD = 50°.
(3) Mark point E on ray CD such that CE = 9.8 cm
(4) Draw seg BE.
(5) Construct the perpendicular bisector of seg BE .
(6) Name the point of intersection of ray CE and the perpendicular bisector of BE as A.
(7) Draw seg AB.
Therefore, â–³ABC is the required triangle.
Page No 53:
Question 4:
Construct ABC, in which BC =5.2 cm, ACB = and perimeter of ABC is 10 cm.
Answer:
Perimeter of â–³ABC = 10 cm
⇒ AB + BC + AC = 10 cm
⇒ AB + 5.2 + AC = 10 cm
⇒ AB + AC = 4.8 cm
A triangle can be only possible if the sum of the sides of a traingle is greater than the third side.
But here, AB + AC < BC
Hence, the construction of â–³ABC is not possible.
Page No 54:
Question 1:
Construct XYZ, such that YZ = 7.4 cm, XYZ = and XY XZ = 2.7 cm.
Answer:
Steps of construction:
(1) Draw seg YZ of length 7.4 cm.
(2) Draw ray YP such that ∠ZYP = 45°
(3) Take point Q on ray YP such that YQ = 2.7 cm.
(4) Construct the perpendicular bisector of seg QZ.
(5) Name the point of intersection of ray YP and the perpendicular bisector of seg QZ as X.
(6) Draw seg XZ.
Therefore, â–³XYZ is required triangle.
Page No 54:
Question 2:
Construct PQR, such that QR = 7.4 cm, PQR= and PQ PR = 2.5 cm.
Answer:
​Steps of construction:
(1) Draw seg QR of length 6.5 cm.
(2) Draw ray QA such that ∠RQA = 60°
(3) Take point B on ray QA such that QB = 2.5 cm.
(4) Construct the perpendicular bisector of seg BR.
(5) Name the point of intersection of ray QA and the perpendicular bisector of seg BR as P.
(6) Draw seg PR.
Therefore, â–³PQR is required triangle.
Page No 54:
Question 3:
Construct ABC, such that BC = 6 cm, ABC = and AC AB = 2.5cm.
Answer:
​Steps of construction:
(1) Draw seg BC of length 6 cm.
(2) Draw ray BE such that ∠CBE = 100°
(3) Take point D on the opposite of ray BE such that BD = 2.5 cm.
(4) Construct perpendicular bisector of seg DC.
(5) Name the point of intersection of ray BE and the perpendicular bisector of DC as A.
(6) Draw seg AC.
Therefore, â–³ABC is required triangle.
Page No 56:
Question 1:
Construct PQR, in which Q = , R = and PQ + QR + PR = 9.5 cm.
Answer:
Steps of construction:
(1) Draw seg AB of 9.5 cm length.
(2) Draw a ray making angle of 35° at point A.
(3) Draw another ray making an angle of 40° at point B.
(4) Name the point of intersection of the two rays as P.
(5) Draw the perpendicular bisector of seg AP and seg BP.
Name the points as Q and R respectively where the perpendicular bisectors intersect line AB.
(6) Draw seg PQ and seg PR.
Therefore, â–³PQR is the required triangle.
Page No 56:
Question 2:
Construct XYZ, in which Y = , X = and perimeter of triangle is 10.5 cm.
Answer:
Steps of construction:
​
(1) Draw seg PQ of 10.5 cm length.
(2) Draw a ray making angle of 29° at point P.
(3) Draw another ray making an angle of 23° at point Q.
(4) Name the point of intersection of the two rays as Z.
(5) Draw the perpendicular bisector of seg PZ and seg QZ.
Name the points as Y and X respectively where the perpendicular bisectors intersect line PQ.
(6) Draw seg XZ and seg YZ.
Therefore, â–³XYZ is the required triangle.
Page No 56:
Question 3:
Construct LMN, in which M = , N = and LM + MN + NL = 11 cm.
Answer:
Steps of construction:
​
(1) Draw seg AB of 11 cm length.
(2) Draw a ray making angle of 30° at point A.
(3) Draw another ray making an angle of 40° at point B.
(4) Name the point of intersection of the two rays as L.
(5) Draw the perpendicular bisector of seg AL and seg BL.
Name the points as M and N respectively where the perpendicular bisectors intersect line AB.
(6) Draw seg LM and seg LN.
Therefore, â–³LMN is the required triangle.
Page No 56:
Question 1:
Construct XYZ, such that XY + YZ =10.3 cm, YZ = 4.9 cm, XYZ = .
Answer:
Steps of constructions:
(1) Draw seg YZ of length 4.9 cm.
(2) Draw ray YV such that ∠ZYA = 45°.
(3) Mark point W on ray YA such that YB = 10.3 cm
(4) Draw seg BZ.
(5) Construct the perpendicular bisector of seg BZ .
(6) Name the point of intersection of ray YA and the perpendicular bisector of BZ as X.
(7) Draw seg XZ.
Therefore, â–³XYZ is the required triangle.
Page No 56:
Question 2:
Construct ABC, in which B = , C = , AB + BC + AC = 11.2 cm.
Answer:
Steps of construction:
​
(1) Draw seg XY of 11.2 cm length.
(2) Draw a ray making angle of 35° at point X.
(3) Draw another ray making an angle of 30° at point Y.
(4) Name the point of intersection of the two rays as A.
(5) Draw the perpendicular bisector of seg AX and seg AY.
Name the points as B and C respectively where the perpendicular bisectors intersect line XY.
(6) Draw seg AB and seg AC.
Therefore, â–³ABC is the required triangle.
Page No 56:
Question 3:
The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4. Construct the triangle.
Answer:
Steps of construction:
1: A line AB = 14.4 cm is drawn.
2: Darw a ray AC making an acute angle at A in the downward direction. Then, divid it into (2 + 3 + 4) = 9 parts.
3: Join B and A9.
4: Draw a line parallel to BA9 at A2 and A5 intersecting AB at X and Y respectively..
5: With X as centre and XA as radius, draw an arc.
6: With Y as centre and YB as radius, draw an arc cutting the previous arc at Z.
7: Join XZ and YZ.
Hence, â–³XYZ is the required triangle.
Page No 56:
Question 4:
Construct PQR, in which PQ PR = 2.4 cm, QR = 6.4 cm and PQR = .
Answer:
​Steps of construction:
(1) Draw seg QR of length 6.4 cm.
(2) Draw ray QA such that ∠RQA = 55°
(3) Take point B on ray QA such that QB = 2.4 cm.
(4) Construct the perpendicular bisector of seg BR.
(5) Name the point of intersection of ray QA and the perpendicular bisector of seg BR as P.
(6) Draw seg PR.
Therefore, â–³PQR is required triangle.
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