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प्रश्न
ΔSHR ~ ΔSVU. In ΔSHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and `"SH"/("SV")=3/5`. Construct ΔSVU.
उत्तर
Steps of construction:
- Construct the Δ SHR with the given measurements. For this draw SH of length 4.5 cm.
- Taking S as the centre and radius equal to 5.8 cm draw an arc above SH.
- 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.
- Join SR and HR. Δ SHR with the given measurements is constructed. Extend SH and SR further on the right side.
- Draw any ray SX making an acute angle (i.e. 45°) with SH on the side opposite to the vertex R.
- 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.
- Join A3H and draw a line through A5 parallel to A3H, intersecting the extended part of SH at V.
- Draw a line VU through V parallel to HR.
Δ SVU is the required triangle.
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