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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. Q.7

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Sir give me the best and simple way to solve the problem of question from class 10th ncert of constructions chapter of exercise 11.1 of question no.7, how i solve this problem 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.

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  1. The sides other than hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other

    Construction Procedure:

    The required triangle can be drawn as follows.

    1. Draw a line segment BC =3 cm.

    2. Now measure and draw ∠= 90°

    3. Take B as centre and draw an arc with the radius of 4 cm and intersects the ray at the point B.

    4. Now, join the lines AC and the triangle ABC is the required triangle.

    5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

    6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB1 = B1B2 = B2B3= B3B4 = B4B5

    7. Join the points B3C.

    8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.

    9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.

    10. Therefore, ΔA’BC’ is the required triangle.

    Ncert solutions class 10 Chapter 11-8

    Justification:

    The construction of the given problem can be justified by proving that

    Since the scale factor is 5/3, we need to prove

    A’B = (5/3)AB

    BC’ = (5/3)BC

    A’C’= (5/3)AC

    From the construction, we get A’C’ || AC

    In ΔA’BC’ and ΔABC,

    ∴ ∠ A’C’B = ∠ACB (Corresponding angles)

    ∠B = ∠B (common)

    ∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

    Since the corresponding sides of the similar triangle are in the same ratio, it becomes

    Therefore, A’B/AB = BC’/BC= A’C’/AC

    So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3

    Hence, justified.

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