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ΔSHR ~ ΔSVU. In ΔSHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and SH/SV=3/5. Construct ΔSVU. - Geometry

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Sum

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

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Solution

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 old triangle to 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.

Concept: Division of a Line Segment
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