Question

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose
sides are 7/5
of the corresponding sides of the first triangle.​

Answer 1

Step-by-step explanation:

inabc-

let ab=7 cm

bc=6 cm

can=5cm

now take ab as base and take bc as radius and cut from top

now take ca as radius and cut bc with the help of a rounder

now join ab to bc and bc to ca

Answer 2

Given:

• Triangle sides = 5cm, 6cm, 7cm

• Corresponding Triangle side = 7/5th of 1st triangle

Solution

This is a Different Steps of Construction of Triangle How will we go through.

• Draw a line segment AB =5 cm.

• Take A and B as centre, and draw the arcs of radius 6 cm and 7 cm respectively.

• These arcs will intersect each other at point C, and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.

• Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

• Locate the 7 points such as ${\tt{A_1, A_2, A_3, A_4, A_5, A_6, A_7}}$ , on line AX such that it becomes ${\tt{AA_1 = A_1A_2 = A_2A_3}}$ , ${\tt{ A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7 }}$. Join the points ${\tt{BA_5}}$ and draw a line from ${\tt{A_7}}$ to ${\tt{BA_5}}$ which is parallel to the line ${\tt{BA_5}}$ where it intersects the extended line segment AB at point B’.

• Draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.

Hence,

• ΔAB’C’ is the required triangle.