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Step 1: Draw a line segment AB=13 cm.

Step 2: At A, construct an angle of 45∘ and at B construct an angle of 90∘.

Step 3: Bisect these angles. Let bisector of these angles intersect at point X. Join AX and BX.

Step 4: Draw perpendicular bisectors of AX and BX to intersect AB at Y and Z respectively. Join AB and AC.

Given: In ∆ABC
AB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.
Required: To construct ∆ABC.

Steps of Construction :
1. Draw DE = 10 cm
2. At D, construct ∠EDP = 12 of 60°= 30° and at E, construct ∠DEQ =12 of 45° = 22 12∘
3. Let DP and EQ meet at A.
4. Draw perpendicular bisector of AD to meet DE at B.
5. Draw perpendicular bisector of AE to meet DE at C.
6. Join AB and AC. Thus, ABC is the required triangle.

Steps of construction:
1. Draw a line segment PQ = 11.6 cm.
2. Construct an angle of 45° and bisect it to get ∠QPX.
3. Construct an angle of 60° and bisect it to get ∠PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, △ABC is the required triangle.

It is given that AB: BC: CA = 3: 4: 5

3x + 4x + 5x = 12

12x = 12

x = 1

AB = 3 cm, BC = 4 cm and CA = 5 cm

Steps of construction:

(1) Draw a sufficiently long line segment using a ruler.

(2) Locate points A and B on it such that AB = 3 cm.

(3) With A as the centre and radius 5 cm, draw an arc.

(4) With B as the centre and radius 4 cm, draw another arc that cuts the previous arc at C.

(5) Join AC and BC.

Then, ABC is the required triangle.