The DAV Class 7 Maths Solution and **DAV Class 7 Maths Chapter 10 Worksheet 1** Solutions of Construction of Triangles offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 10 WS 1 Solutions

Question 1.

Construct a ΔLMN in which LM = 5 cm, MN = 5.6 cm and NL = 4.2 cm.

Answer:

Steps of Construction:

- Draw MN = 5.6 cm.
- Take M and N as centre and draw two arcs with radius 5 cm and 4.2 cm respectively.
- L is the point of intersection of the two arcs.
- Join ML and NL.
- ALMN is the required triangle.

Question 2.

Draw ΔDEF such that DE = DF = 4 cm and EF = 6 cm.

Answer:

Steps of Construction:

- Draw EF = 6cm.
- Draw two arcs with centre E and F and radius 4 cm.
- Let the two arcs be cut at D.
- Join ED and FD.
- ΔDEF is the required triangle.

Question 3.

Draw an equilateral triangle one of whose sides is of length 5 cm.

Answer:

Steps of Construction:

- Draw BC = 5 cm.
- Draw two arcs with centre B and C and with radius 5 cm.
- Let the two arcs be cut at A.
- Join AB and AC.
- ΔABC is the required equilateral triangle.

Question 4.

Draw A PQR in which QR = 4 cm, PQ = 3 cm, RP = 5 cm. Also draw perpendicular bisector of side PQ.

Answer:

Steps of Construction:

- Draw PQ = 3 cm.
- Draw the arcs with centre P and Q and with radius 5 cm and 4 cm respectively.
- Let the two arcs be cut at R.
- Join PR and QR.
- ΔPQR is the required triangle.
- Draw a perpendicular bisector ST at PQ.

Question 5:

Is it possible to construct a triangle whose sides are 3 cm, 6.1 cm and 2.6 cm?

Answer:

To construct a triangle, the necessary condition is that the sum of any two sides of the triangle must be greater than the third side.

Here, 3 cm + 2.6 cm < 6.1 cm

⇒ 5.6 cm < 6.1 cm

Hence, triangle is not possible to construct.

Question 6.

Draw a ΔPQR in which PQ = 7 cm, QR = 3 cm, RP = 4 cm. Is the construction of triangle possible?

Answer:

Here, PQ = 7 cm, QR = 3cm and RP = 4 cm

QR + RP = PQ

3 + 4 = 7

7 = 7

∴ P, Q and R are collinear

Hence, the triangle is not possible to construct.

### DAV Class 8 Maths Chapter 10 Worksheet 1 Notes

If A, B, C are three non-collinear points, the figure made by three line segments AB, BC and CA is called the triangle with vertex A, B, C.

The three line segments forming a triangle are called the sides of the triangle.

While constructing a triangle, the following facts are to be kept in mind.

- The sum of the three angles of a triangle is 180°.
- The sum of any two sides of a triangle is greater than the third side.
- There are six components of a triangle, three angles and three sides.
- Triangle is possible to construct if any three components are known independently.

A triangle can be constructed when

- All the three sides are given. (SSS triangle construction)
- Two sides and an included angle are given. (SAS triangle construction)
- Two angles and an included side are given. (ASA triangle construction)
- Right angle with hypotenuse and one side are given. (RHS triangle construction)

Example 1:

Construct a Δ PQR in which PQ = 3cm, PR = 4 cm and QR = 5 cm. What type of this triangle is?

Solution:

Steps of construction:

- Draw QR = 5 cm.
- Draw two arcs with radius 3 cm and 4 cm taking centres Q and R respectively to meet each other at P.
- Join PQ and PR.
- Δ PQR is the required triangle.
- Δ PQR is a right angled triangle.

Example 2:

Construct a triangle whose any two sides are 3.5 cm and 4.5 cm and the included angle is 60°.

Solution:

Steps of Construction:

- Draw BC = 3.5 cm
- Draw ∠B = 600
- Cut BA = 4.5 cm B
- Join AC
- Δ ABC is the required triangle.

Example 3:

Construct Δ ABC in which ∠B = 60°, ∠C = 40° and BC = 4 cm.

Solution:

Steps of Construction:

- Draw BC = 4 cm.
- Draw ∠B = 60° and ∠C = 40° to meet each other at A.
- Δ ABC is the required triangle.

Example 4:

Draw a right angle triangle whose hypotenuse is 5.5 cm and a side is 3.5 cm.

Answer:

Steps of Construction:

- Draw QR = 3.5 cm
- Draw ∠RQX = 90°
- Taking R as centre and radius 5.5 cm, draw an arc to meet QX at P
- Join PR
- Δ PQR is the required triangle.