HOLA,
Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Then construct a triangle whose sides are 57 times the corresponding sides of ∆ABC.
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Answered by
13
Answer:
Steps of construction:-
1. Draw a line segment AB=5 cm.
2. From point B, draw ∠ABY=60°.
To measure angle at B:
a. With B as centre and with any radius, draw another arc cutting the line AB at D.
b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.
c. Draw a ray BY passing through E which forms an angle of 60° with the line AB.
3. With B as centre and radius of 6 cm draw an arc intersecting the line BY at C.
4. Join AC, ∆ABC is the required triangle.
5. From A, draw any ray AX downwards making an acute angle.
6. Mark 7 points A1,A2,A3,A4,A5,A6 and A7 on AX such that AA1=A1A2=A2A3=A3A4=A4A5=A5A6=A6A7
7. Join A7B and from A5 draw A5M||A7B intersecting AB at M.
8. From point M draw MN││BC intersecting AC at N. Then, ∆AMN is the required triangle whose sides are equal to 5/7 of the corresponding sides of ∆ABC.
Justification:-
Let AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = x
As per the construction A5M││A7B
AM/MB=AA5/A5A7=5x/2x=5/2
AM/MB=5/2
Now,AB/AM=AM+MB/AM+MB/AM=1+2/5=7/5
Also MN | | BC,
Therefore rAMC is congruent to rAMN
Thus,
AN/AC=NM/BC=AM/AB=5/7
Hence, the new triangle rAMN is similar to the given triangle rABC and its sides are 5/7 times of the corresponding sides of rABC.
Answered by
22
Answer
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.
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