Division of a line segment in the given ratio. Draw a line segment
AB=10cm .Divide it in the ratio 3:4
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Answers
Answer:
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Step-by-step explanation:
We follow the following step of construction.
Steps of construction
Step I
Drawn a line segment AB=10 cm by using a ruler.
Step II
Drawn any ray making an acute angle ∠BAX with AB.
Step III
Along AX, mark-off 5(=3+2) points A
1
,A
2
,A
3
,A
4
and A
5
such that
AA
1
=A
1
A
2
=A
3
A
4
=A
4
A
5
.
Step Iv
Join BA
5
Step v
Through A
3
draw a line A
3
P parallel to A
5
B by making an angle equal to ∠AA
5
B at A
3
intersecting AB at a point P.
The point P so obtained is the required point, which divides AB internally in the ration 3:2.
ALTERNATIVE METHOD FOR DIVISION OF A LINE SEGMENT INTERNALLY IN A GIVEN RATIO
We may use the following steps to divide a given line segment AB internally in a given ratio m:n, where m and n are natural numbers.
Steps of construction
Step I
Draw line segment AB of given length.
Step II
Draw any ray AX making an acute angle ∠BAX with AB.
Step III
Draw a ray BY, on opposite side of AX, parallel to AX by making an angle ∠BAY equal to ∠BAX.
Step IV
Mark off m points A
1
,A
2
,,A
m
, on AX and n points B
1
,B
2
,,B
n
on BY such that
AA
1
=A
1
A
2
=.=A
m−1
A
m
=BB
1
=B
1
B
2
=.=Bn−1B
n
.
Step V
Join A
m
B
n
. Suppose it intersects AB at P.
The point P is the required point dividing AB in the ratio m:n.
In triangles AA
m
P and BB
n
P, we have
∠A
m
AP=∠PBB
n
and, APA
m
=∠BPB
n
So, by AA similarly criterion, we have
△A A
m
P−△BB
n
P
⇒
BB
n
AA
m
=
BP
AP
⇒
BP
AP
=
n
m