Question

Construct a triangle , having its perimeter = 55 mm and the ratio of its angles 2:1:3 . I want the steps along with the drawing . this is ENGINEERING GRAPHICS grade 11 question .​

Answer 1

see attached

Step-by-step explanation:

You want to construct a triangle with a perimeter of 55 mm and angles in the ratio 2:1:3.

Angles

The 2+1+3 = 6 ratio units represent 180°, the sum of angles in a triangle. Hence each unit in the ratio is 180°/6 = 30°.

The angles are 60° : 30° : 90°.

Perimeter

From your knowledge of "special triangles," you know the side lengths in a 30°-60°-90° triangle have the ratios 1 : √3 : 2.

That is, the perimeter of the triangle is (1+√3 +2) = 3+√3 times the length of the shortest side.

Side lengths

The shortest side of the triangle will have the length ...

`\frac{55\text{ mm}}{3+√(3)}=\frac{(3-√(3))(55\text{ mm})}{(3-√(3))(3+√(3))}= \frac{(165-55√(3))\text{ mm}}{3^2-(√(3))^2}\\\\\\=(27.5-(55)/(6)√(3))\text{ mm}\approx 11.6229\text{ mm}`

The longest side will be double this length, about 23.2457 mm. The side of intermediate length will be (55√3 -55)/2 ≈ 20.1314 mm.

Construction

If you're doing the drawing using only a compass and straightedge, you will start by constructing perpendicular lines (the right angle). Mark off 1 and two lengths of the shortest side on one of these perpendicular segments. Then draw the hypotenuse as 2 times the short length from the end of the short length. It will intersect the perpendicular segment at √3 times the short length.

The tricky part here is constructing a line segment of the desired irrational length. One way to do this is to construct a 30-60-90 triangle with a unit-length short side, then "unfold" it to a line segment with parts that are 1, √3, and 2 units long. A "proportioning" construction can divide the 55 mm length of the perimeter into these proportions, (10 mm might be a good "unit length" for this.)

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Additional comment

The desired dimensions are sufficiently small that an accurate compass/straightedge construction will be difficult. It is unlikely your pencil line width is much less than about 0.2 mm, so already the error is about 2% in any length.

The attached drawing was made by telling the software what lengths we wanted in "exact" terms. Its calculation is generally good to 10 sf or better. To scale the drawing properly, you'd have to make the longest side about 23.25 mm, or 0.9152 inches.