PSE ~ ∆TSV. In ∆PSE, PS = 4.4 cm, SE = 5.1 cm, PE = 5.5 cm and
PS /TS=5 /3
Construct APSE and ATSV
Answers
Steps of construction:
1. Construct the A SHR with the given measurements. For this draw SH of length 4.5 cm.
2. Taking S as the centre and radius equal to 5.8 cm draw an arc above SH.
3. Taking H as the centre and radius equal
to 5.2 cm draw an arc to intersect the
previous arc. Name the point of
intersection as R.
4. Join SR and HR. A SHR with the given
measurements is constructed. Extend
SH and SR further on the right side.
5. Draw any ray SX making an acute angle ( i.e; 45°) with SH on the side opposite to
the vertex R.
6. Locate 5 points. (the ratio of old triangle
to new triangle is 3/5 and 5 > 3). Locate A1, A2, A3, A4 and A5 on AX so that SA₁ =
A₁ A2 A2A3 = A3A4 = A4A5.
7. Join A3H and draw a line through As
parallel to A3H, intersecting the extended part of SH at V. 8. Draw a line VU through V parallel to HR. A SVU is the required triangle.
