Question

Construct a triangle similar to AABC with its sides equal to of the
corresponding sides of AABC. Justify the construction,?


solve it fast and get 50 point

only one person put wrong answer because see need 50 points​

Answer 1

Let ΔABC be a triangle having BC=6cm;∠B=50°;∠C=60°.

Steps of construction of triangle (3/4)th of △ABC:

(1). Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

(2). Mark four points on BX as B

1

,B

2

,B

3

,B

4

such that BB

1

=B

1

B

2

=B

2

B

3

=B

3

B

4

.

(3). Join B

4

C and draw a line through B

3

parallel to B

4

C intersect BC at C

′

.

(4). Draw a line C

′

parallel to AC intersect AB at A

′

.

Thus, ΔA

′

BC

′

is required triangle.

Answer 2

Answer:

where

n = no. of ribbons

Now, by substituting the given values in the formula, we get

39.2 = n \times 1.439.2=n×1.4

⇒ n = \frac{39.2\:cm}{1.4\:cm}n=1.4cm39.2cm

⇒ \bold{n = 28}n=28

Thus, the number of ribbons each of length 1.4 cm long that can be cut from the ribbon roll, 39.2 cm long is 28.

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Also View:

Length of ribbon is 1/3 meter.Ribbons are joined together to get 2 2/3 meter length.Then how many such ribbons were joined together.

thanks