Advertisements
Advertisements
प्रश्न
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
उत्तर
Steps of construction:
1. Draw a line segment AB = 5.5 cm.
2. At A draw AE such that ∠BAE = 25°.
3. At A draw AF such that ∠EAF = 90°.
4. Draw the perpendicular bisector of AB which intersects AF at O and AB at G.
5. With O as centre and OB as radius draw a circle.
6. XY intersects AB at G. On XY, from G mark an arc at M. Such that GM = 4 cm.
7. Through M draw a line parallel to AB intersect the circle at C and D.
8. Join AC and BC.
9. ∆ABC 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
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
In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
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
∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR
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 in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 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
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.