Advertisements
Advertisements
Question
Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm
Advertisements
Solution
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
RELATED QUESTIONS
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
If `"AD"/"DB" = 3/4` and AC = 15 cm find AE
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
DE || BC and CD || EE Prove that AD2 = AB × AF
Check whether AD is bisector of ∠A of ∆ABC of the following
AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
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 ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm
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.
