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ICSE Class 10 Answered

using ruler and compass construct a triangle ABC in which AB=5.5cm BC=3.4cm and CA=4.9cm (1)locus of points equidistant from A and C (2)the  circle touching AB at A and passing through C (please explain each step of construction in sentences)
Asked by ushanihar12 | 05 Feb, 2019, 10:07: PM
answered-by-expert Expert Answer
Step 1.  Draw a line segment AB of length 5.5 cm by ruler. Using A as centre, draw an arc of radius 4.9 cm as shown in figure.
               Similarly draw an arc of radius 3.4 cm using B as center. let the arcs intersect at a point C.
 
Step 2. Join AC and BC. ΔABC is the required triangle of given sides
 
Locus of points equidistant from A and C is perpendicular bisector of side AC.
Hence we need to draw a perpendicular bisector to side A
 
Step-3 :  Using the points A and C as centre,  draw arcs of equal length on both side of AC as shown in figure.
               Length of arc is arbitrary but it should be greater than half of length of AC. Let the arcs intersect at E and F.
               Join EF. EF is perpendicular to AC. Any point on the line EF is equidistant from A and C.
 
A Circle touching at AB at A and passing through C means, AB is tangent to that circle at A and AC is the chord of the circle.
Hence to draw the circle so that AB is tangent, we draw a perpendicular line at A.
Let the perpendicular line meet the perpendicular bisector at G. since G is at equidistant from A and C,
if we draw a circle using G as centre and GA as radius, we get the required circle that is touching AB at A and passing through C.
Figure shown below shows the construction
 
Answered by Thiyagarajan K | 06 Feb, 2019, 01:07: PM

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