Draw a right triangle in which the sides are of lengths 4cm and 3cm. Then construct another triangle whose sides are 5:3 About the author Daisy

Step-by-step explanation: The ΔA ′ BC ′ whose sides are 4 3 of the corresponding sides of ΔABC can be drawn as follows: Step 1: Draw a ΔABC with side BC=6cm,AB=5cm,∠ABC=60 ∘ Step 2: Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Step 3: Locate 4 points, B 1 ,B 2 ,B 3 ,B 4 on line segment BX. Step 4: Join B 4 C and draw a line through B 3 , parallel to B 4 C intersecting BC at C ′ . Step 5: Draw a line through C ′ parallel to AC intersecting AB at A ′ . The triangle A ′ BC ′ is the required triangle. Reply

Steps of construction:- (i) Draw a line segment AB having a length of 4 cm. (ii) Now, construct a right angle at point A and make a line of 3 cm. (iii) Name this point C. Thus ￼ ABC is the required triangle. (iv) Draw a line AX which makes an acute angle with AB and is opposite of vertex C. (v) Cut four equal parts of line AX namely AA1, AA2, AA3, AA4, AA5. (vi) Now join A3 to B. Draw a line A5B’ parallel to A3B. (vii) And then draw a line B’C’ parallel to BC. Hence ￼ AB’C’ is the required triangle. Reply

Step-by-step explanation:The ΔA

′

BC

′

whose sides are

4

3

of the corresponding sides of ΔABC can be drawn as follows:

Step 1: Draw a ΔABC with side BC=6cm,AB=5cm,∠ABC=60

∘

Step 2: Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3: Locate 4 points, B

1

,B

2

,B

3

,B

4

on line segment BX.

Step 4: Join B

4

C and draw a line through B

3

, parallel to B

4

C intersecting BC at C

′

.

Step 5: Draw a line through C

′

parallel to AC intersecting AB at A

′

.

The triangle A

′

BC

′

is the required triangle.

Steps of construction:-

(i) Draw a line segment AB having a length of 4 cm.

(ii) Now, construct a right angle at point A and make a line of 3 cm.

(iii) Name this point C. Thus ￼ ABC is the required triangle.

(iv) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(v) Cut four equal parts of line AX namely AA1, AA2, AA3, AA4, AA5.

(vi) Now join A3 to B. Draw a line A5B’ parallel to A3B.

(vii) And then draw a line B’C’ parallel to BC.

Hence ￼ AB’C’ is the required triangle.