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:

- 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 old triangle to new triangle is 3/5 and 5 > 3). Locate A
_{1}, A_{2}, A_{3}, A_{4}and A_{5}on AX so that SA_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}= A_{4}A_{5}. - Join A
_{3}H and draw a line through A_{5}parallel to A_{3}H, intersecting the extended part of SH at V. - 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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