Advertisements
Advertisements
प्रश्न
Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Advertisements
उत्तर
Steps of Construction:
- Draw a line segment QR = 4.5 cm
- At Q, draw a ray QX making an angle of 90°
- With centre R and radius 8 cm, draw an arc which intersects QX at P.
- Join RP. ΔPQR is the required triangle.
- Draw the bisector of ∠PQR that meets PR in T.
- From T, draw perpendicular PL and PM respectively on PQ and QR.
Proof: In ΔLTQ and ΔMTQ
∠TLQ = ∠TMQ ...(Each = 90°)
∠LQT = ∠TQM ...(QT is angle bisector)
QT = QT ...(Common)
∴ By Angle – Angle – side criterion of congruence,
∴ ΔLTQ ≅ ΔMTQ ...(AAS postulate)
The corresponding parts of the congruent triangles are congruent
∴ TL = TM ...(C.P.C.T.)
Hence, T is equidistant from PQ and QR.
APPEARS IN
संबंधित प्रश्न
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:
F is equidistant from AB and AC.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
In the figure given below, find a point P on CD equidistant from points A and B.
Describe the locus of the moving end of the minute hand of a clock.
Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
Describe the locus of the centres of all circles passing through two fixed points.
Describe the locus of points at distances greater than 4 cm from a given point.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
Show that the locus of the centres of all circles passing through two given points A and B, is the perpendicular bisector of the line segment AB.
