# Mathematics Part II Solutions for Class 9 Math

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#### Question 4:

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Problem Set No. 4

Question 1:

Construct ∆ XYZ, such that XY + YZ =10.3 cm, YZ = 4.9 cm, ∠XYZ = 450 .

Steps of constructions:

(1) Draw seg YZ of length 4.9 cm.

(2) Draw ray YV such that ∠ZYA = 450.

(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, ∆ XYZ is the required triangle.

Construct ∆ ABC, in which ∠B = 700, ∠C = 600 , AB + BC + AC = 11.2 cm.

Steps of construction:

(1) Draw seg XY of 11.2 cm length.

(2) Draw a ray making angle of 350 at point X.

(3) Draw another ray making an angle of 300 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, ∆ ABC is the required triangle.

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.

Steps of construction:

1: A line AB = 14.4 cm is drawn.

2: Draw a ray AC making an acute angle at A in the downward direction. Then, divide 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, ∆ XYZ is the required triangle.

Construct ∆ PQR, in which PQ − PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 550 .

Steps of construction:

(1) Draw seg QR of length 6.4 cm.

(2) Draw ray QA such that ∠RQA = 550

(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, ∆ PQR is required triangle.

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