Thinking process
(i)Here, for making two similar triangles with one vertex is base We assume that In
ΔABC and
ΔPQR, vertex B = vertex Q.
(ii) In
ΔABC and
ΔPQR, vertex B = vertex Q.
So, we get the required scale factor.
Now, construct a
ΔABC and then a
ΔPBR, similar to
ΔABC whose sides are
PQAB of the corresponding sides of the
ΔABC.
Let
ΔPQR and
ΔABC are similar triangle, then its scale factor between the corresponding sides is
PQAB=86=43
Steps of construction
1. Draw a line segment BC = 5 cm
2. Construct OQ the perpendicular bisector of line segment BC meeting BC at
P′
3. Taking B and C as centers draw two arcs of equal radius 6cm intersecting each other at A
4. Join BA and CA. So.
ΔABC is the required isosceles triangle.
5. From B, draw a ray BX making an acute
∠CBX
6. Locate four points
B1,B2,B3,B4 on BX Such that
BB1=B1B2=B3B4
7. Join
B3C and from
B4 draw a line
B4R∥B3C intersecting the extended line segment BC at R.
8. From point R draw
RP∥CA meeting BA produced at p
Then,
ΔPBR is the required triangle.
Justification
∵B4R∥B3C
∴BCCR=31
Now
BRBC=BC+CRBC
=1+CRBC=1+13=43
Also
RP∥CA
∴ΔABC≅ΔPBR
And
PBAB=RPCA=BRBC=43
Hence the new triangle is similar to the given triangle whose sides are
43 times of the corresponding sides of the isosceles
ΔAC.