Advertisements
Advertisements
प्रश्न
Construct a ∆PQR in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 cm.
Advertisements
उत्तर


Steps of construction:
1. Draw a line segment PQ = 4.5 cm
2. At P, draw PE such that ∠QPE = 60°
3. At P, draw PF such that ∠EPF = 90°
4. Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
5. With O as centre and OP as radius draw a circle.
6. From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
7. Join PR and RQ.
8. 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"/"DB" = 3/4` and AC = 15 cm find AE
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
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`

If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`

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
ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F, respectively. Prove that EF || BD.
Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm
Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is

ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle
