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Maharashtra State BoardSSC (English Medium) 9th Standard

The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4. Construct the triangle.

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Question

The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4. Construct the triangle.

Sum
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Solution 1

Explanation:

Let ΔABC be the required triangle.

Draw the line XY = 14.4 cm, which is the perimeter of ΔABC.

Divide line XY in the ratio 2:3:4.

Draw ray XP such that, ∠YXP = 30°.

Draw line YQ on the opposite side of ray XP such that, `angle`XYQ = 30°.

Make ray XP into 9 equal parts i.e. `"X"_(x_1)`,x1x2, ... x8x9.

Similarly, according to the figure, make 9 equal parts of ray YQ.

Join x2y7 and x5y4, intersecting line XY at B and C respectively.

Draw an arc considering B as center and BX as radius.

Taking C as center and CY as radius, draw an arc which intersects the first arc at point A.

Draw line AB and line AC.

∴ ΔABC is the required triangle.

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Solution 2

Rough figure:

Explanation:

Let the common multiple be x

∴ In ∆ABC,

AB = 2x cm, AC = 3x cm, BC = 4x cm

Perimeter of triangle = 14.4 cm

∴ AB + BC + AC = 14.4

∴ 9x = 14.4

∴ `x = 14.4/9`

∴ x = 1.6

∴ AB = 2x = 2 × 1.6 = 3.2 cm

∴ AC = 3x = 3 × 1.6 = 4.8 cm

∴ BC = 4x = 4 × 1.6 = 6.4 cm

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Chapter 4: Constructions of Triangles - Problem Set 4 [Page 56]

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Balbharati Mathematics 2 [English] Standard 9 Maharashtra State Board
Chapter 4 Constructions of Triangles
Problem Set 4 | Q 3. | Page 56

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