Construction of Triangles

# Construction of Triangles

## As you already know that triangle is a 3-sided polygon. It has 3 sides, 3 angles and 3 vertices. Its 3 sides and 3 angles together are called the 6 elements of the triangle. A triangle can be constructed if 3 of its elements are given. Let us see what are the different 3 elements when a triangle is possible to construct.

A triangle can be constructed when:
(i) When All the Three Sides are Given
(ii) When the Lengths of Two Sides and the Included Angle are Given
(iii) When the Length of One Side and Two Angles are Given

## Construction of Triangles

### When All the Three Sides are Given

Example: Construct a PQR where PQ = 4 cm, QR = 5.2 cm and PR = 3.2 cm.

Steps of Construction

1. Draw a rough sketch of the triangle as shown.
2. Construct a line segment PQ measuring 4 cm.
3. At Q, draw an arc of radius 5.2 cm.
4. Taking P as center, draw an arc of radius 3.2 cm. Let the two arcs intersect at a point R.
5. Join PR and QR.
Thus, ∆PQR is the required triangle.

### When the Lengths of Two Sides and the Included Angle are Given

Example: Construct a MAT such that MA = 3.2 cm, AT = 5.3 cm and MAT = 60°.

Steps of Construction

1. Draw a rough sketch of MAT.
2. Draw a line segment MA measuring 3.2 cm.
3. Taking A as center, draw an angle measuring 60° and draw a ray m by extending it.
4. Taking A as center, draw an arc of radius 5.3 cm cutting m at T. Join MT.
Thus, ∆MAT is the required triangle.

### When the Length of One Side and Two Angles are Given

Example: Construct a ABC such that AB = 4.4 cm, CAB = 45° and ABC = 30°.

Steps of Construction

1. Draw a rough sketch of ABC.
2. Draw a line segment AB measuring 4.4 cm.
3. Taking A as center, draw an angle measuring 45° and extend the ray m.
4. Taking B as center, draw an angle measuring 30° and extend the ray n till both the rays m and n intersect. Mark the point of intersection as C.
Thus, ∆ABC is the required triangle.

## Construction of an Equilateral Triangle

Example: Construct an equilateral triangle ABC whose sides measure 4.6 cm.

Steps of Construction

1. Draw a rough sketch of triangle ABC.
2. Construct a line segment AB measuring 4.6 cm
3. At A and B, construct an angle of 60° each.
3. Draw AX and BY and mark their meeting point as C.
Thus, ∆ABC is the required triangle.

## Construction of an Isosceles Triangle

### When its Base and One Base Angle are Given

Example: Construct an isosceles triangle PQR such that PQ = 4.1 cm and P = 45°.

Steps of Construction

1. Draw a rough sketch of triangle PQR.
2. Construct a line segment PQ measuring 4.1 cm.
3. At P and Q, construct an angle measuring 45° and draw rays PX and QY.
4. Mark their meeting point as R.
Thus, ∆PQR is the required triangle.

### When One of the Equal Sides and the Vertical Angle are Given

Example: Construct an isosceles triangle PQR such that PQ = 4.7 cm and P = 45°.

Steps of Construction

1. Draw a rough sketch of PQR.
2. Draw a line segment PQ measuring 4.7 cm. At P, draw an angle measuring 45° and ray PX.
3. Cut off PR = 4.7 cm from PX.
Thus, ∆PQR is the required triangle.

## Construction of a Right-angled Triangle

### When the Length of Sides Forming Right Angle are Given

Example: Construct a CAT such that CA = 4.2 cm, AT = 5.2 cm and the triangle is right-angled at A.

Steps of Construction

1. Draw a rough sketch of the triangle.
2. Draw a line segment CA measuring 4.2 cm.
3. With A as center, draw an angle measuring 90° and cut AT = 5.2 cm. Join CT.
Thus, ∆CAT is the required triangle.

### When One Side and Hypotenuse are Given

Example: Construct a right-angled triangle PQR such that Q = 90°, QR = 5 cm and PR = 6 cm.

Steps of Construction

1. Draw a rough sketch of PQR.
2. Construct a line segment QR measuring 5 cm and construct 90° at Q. Draw line QX.
3. Taking R as center, draw an arc measuring 6 cm, cutting QX at P.
Thus, ∆PQR is the required triangle.