Question

ΔSHR ~ ΔSVU. In ΔSHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and `"SH"/("SV")=3/5`. Construct ΔSVU.

Answer 1

Steps of construction:

1. Construct the Δ 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. Δ 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 the old triangle to the new triangle is 3/5 and 5 > 3). Locate A₁, A₂, A₃, A₄ and A₅ on AX so that SA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
7. Join A₃H and draw a line through A₅ parallel to A₃H, intersecting the extended part of SH at V.
8. Draw a line VU through V parallel to HR.

Δ SVU is the required triangle.