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Question

Draw an isosceles triangle ABC in which AB=AC = 6 cm and BC=5 cm Construct a triangle PQR similar to ΔABC in which PQ = 8 cm, Also justify the construction.

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Solution

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 B4RB3C intersecting the extended line segment BC at R.
8. From point R draw RPCA meeting BA produced at p
Then, ΔPBR is the required triangle.
Justification
B4RB3C
BCCR=31
Now BRBC=BC+CRBC
=1+CRBC=1+13=43
Also RPCA
ΔABCΔPBR
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.

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