from a handpicked tutor in LIVE 1to1 classes
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
Solution:

Draw the line segment of the largest length 7 cm. Measure 5 cm and 6 cm separately and cut arcs from both the ends of the line segment such that they cross each other at one point. Connect this point from both the ends

Then draw another line that makes an acute angle with the given line (7 cm). Divide the line into m + n parts where m and n are the ratios given.

Two triangles are said to be similar if their corresponding angles are equal and are said to satisfy the AA criteria.

The basic proportionality theorem states that “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally".
Steps of construction:
 Draw BC = 7cm with B and C as centres and radii 5 cm and 6 cm respectively. Draw arcs to intersect at A. ΔABC is obtained.
 Draw ray BX making ∠CBX an acute angle.
 Mark 7 points (greater of 7 and 5 in 7/5 ) B₁, B₂,………B₇ on BX such that BB₁ = B₁B₂ =............... = B₆B₇
 Join B₅ (smaller of 7 and 5 in 7/5 which is the 5^{th} point) to C and draw B₇C' parallel to B₅C intersecting the extension of BC at C'.
 Through C' draw C'A' parallel to CA to meet the extension of BA at A’. Now, ΔA' B'C' is the required triangle similar to ΔABC where BA'/BA = C'A'/CA = BC'/BC = 7/5
Proof:
In ΔBB₇C', B₃C is parallel to B₇C'
Hence by Basic proportionality theorem,
B₅B₇/BB₅ = CC'/BC = 2/5
Adding 1 to both the sides of CC'/BC = 2/5,
CC'/BC + 1 = 2/5 + 1
(BC + CC')/BC = 7/5
BC'/BC = 7/5
Consider ΔBAC and ΔBA'C'
∠ABC = ∠A'BC' (Common)
∠BCA = ∠BC'A' (Corresponding angles ∵ CA  C'A')
By AA criteria, ΔBAC ∼ ΔBA'C'
∴ Corresponding sides are proportional
Hence,
BA'/BA = C'A'/A = BC'/BC = 7/5
☛ Check: NCERT Solutions for Class 10 Maths Chapter 11
Video Solution:
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.
NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.1 Question 3
Summary:
A triangle BAC of sides 5 cm, 6 cm, and 7 cm and another triangle B'A'C whose sides are 7/5 of the corresponding sides of the first triangle BAC have been constructed
☛ Related Questions:
 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1(1/2) times the corresponding sides of the isosceles triangle.
 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/2 of the corresponding sides of the triangle ABC.
 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/2 of the corresponding sides of the triangle ABC.
 Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
visual curriculum