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.