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
Solution:

Draw the triangle with the given conditions.

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

Two triangles are called similar if their corresponding angles are equal. They are said to satisfy AngleAngleAngle (AAA)

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 constructions:
 Draw BC = 4. At B, make an angle ∠CBY = 90° and mark A on BY such that BA = 3 cm. Join A to C. Thus ΔABC is constructed.
 Draw the ray BX so that ∠CBX is acute.
 Mark 5 (5 > 3 in 5/3) points B_{1}, B_{2}, B_{3}, B_{4}, B_{5} on BX so that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}
 Join B (3^{rd} point on BX as 3 < 5) to C and draw B_{5}C' parallel to B_{3}C so that C' lies on the extension of BC.
 Draw C'A' parallel to CA to intersect of the extension of BA at A’. Now ΔBA'C' is the required triangle similar to ΔBAC where BA'/BA = BC'/BC = C'A'/CA = 5/3
Proof:
In ΔBB_{3}C', B_{3}C  B_{3}C'
Hence by Basic proportionality theorem,
B_{3}B_{5}/BB_{3} = CC'/BC = 2/3
CC'/BC + 1 = 2/3 + 1 (Adding 1)
(CC' + BC)/BC = 5/3
BC'/BC = 5/3
Consider ΔBAC and ΔBA'C'
∠ABC = ∠A'BC' = 90°
∠BCA = ∠BC'A' (Corresponding angles as CA  C'A')
∠BAC = ∠BA'C'
By AAA axiom, ΔBAC ~ ΔBA'C'
Therefore, corresponding sides are proportional,
Hence,
BA'/BA = BC'/BC = C'A'/CA = 5/3
Video Solution:
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
NCERT Solutions Class 10 Maths  Chapter 11 Exercise 11.1 Question 7:
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
A right triangle ABC in which the sides other than the hypotenuse having lengths 4 cm and 3 cm and another triangle A'BC' whose sides are 5/3 times the corresponding sides of triangle ABC have been constructed