Advertisements
Advertisements
प्रश्न
Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm
उत्तर
Steps of construction:
1. Draw a line segment QR = 6.5 cm.
2. At Q draw QE such that ∠RQE = 60°.
3. At Q, draw QF such that ∠EQF = 90°.
4. Draw the perpendicular of QR which intersects QF at O and QR at G.
5. With O as centre and OQ as radius draw a circle.
6. XY intersects QR at G. On XY, from G mark an arc at M. Such that GM = 4.5 cm.
7. Draw AB through M which is parallel to QR.
8. AB Meets the circle at P and S.
9. Join QP and RP.
10. ∆PQR is the required triangle.
APPEARS IN
संबंधित प्रश्न
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = 3x – 1, find the value of x
ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD
In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
Rhombus PQRB is inscribed in ΔABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
DE || BC and CD || EE Prove that AD2 = AB × AF
Construct a ∆PQR in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 cm.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is
Two circles intersect at A and B. From a point, P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
