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प्रश्न
Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?
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उत्तर
A ΔP'Q'R' whose sides are `3/5`of the corresponding sides of ΔPQR can be drawn as follows:
Step 1
Draw a ΔPQR with side QR = 6 cm, PQ = 5 cm and ∠PQR = 60°.
Step 2
Draw a ray QX making an acute angle with QR on the opposite side of vertex P.
Step 3
Locate 5 points (as 5 is greater in 3 and 5), Q1, Q2, Q3, Q4 and Q5 on line segment QX.
Step 4
Join Q5R and draw a line through Q3, parallel to Q5R intersecting QR at R'.
Step 5
Draw a line through R' parallel to PR intersecting PQ at P'.
Δ P'Q'R' is the required triangle.
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Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
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