5 outo
of the given triangle
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm ani
3 cm. Then construct another triangle whose sides are - times the corresponding sides
(class10 math ch-11 ex-11.1 que no. 7 please solve this question)
Answers
Step-by-step explanation:
Given:
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.
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.
