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प्रश्न
Draw an acute angled Δ PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
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उत्तर
Steps of construction:
- Draw any acute angled ∆ PQR.
- With P as centre, draw an arc that cut the side QR at X and Y.
- With X as centre and radius more than half of XY, draw an arc below QR. With Y as centre and same radius draw another arc that cut the previous arc at A.
- Join PA that intersects QR at L. So, PL is the altitude on side QR.
In the same manner, draw QM⊥PR and RN⊥PQ.
Hence, ∆PQR is the required triangle with altitudes PL, QM and RN on sides QR, RP and PQ respectively, with O as the point of concurrence of all the three altitudes.
संबंधित प्रश्न
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Draw rough sketch for the following:
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Median is also called ______ in an equilateral triangle.
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How many altitudes can be drawn in a triangle?
Which of the following is NOT true about the altitude of a triangle?
What do we call the segment from a vertex to the opposite side that forms a 90° angle with that side?
