
Construct a cyclic quadrilateral KLMN with $KL=4$ cm, $LM=4.8$ cm, $KM=6.8$cm and $KN=4.3$ cm
Answer
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Hint: We first draw the side $KL=4$ cm and then construct the triangle KLM. We find the circumcenter O of triangle KLM by finding the point of intersection of perpendicular bisectors of sides KL and LM. We draw the circum-circle with O as a center and OK as the radius. We take an arc of 4.3 cm at the point K and denote its point of intersection with the circle as N.
Complete step-by-step solution:
Step-1: We construct a line segment KL whose length is 4.3cm with scale and ruler. KL will act as a side of the quadrilateral.
Step-2: We take an arc of 6.8cm and then take an arc of 4.8 cm and denote their point of intersection as M. We join the line segments $KM=6.8$cm and $LM=4.8$cm. Now we have constructed the triangle KLM.
Step-3: We construct PQ the perpendicular bisector of KL by taking an arc of $KL=4$ cm at both the points K and L with help of a compass.
Step-4: We similarly construct RS the perpendicular bisector of KL by taking an arc of $LM=4.8$ cm at both the points L and M with help of a compass. We denote the point of intersection of perpendicular bisectors PQ and RS as O.
Step-5: We draw the circumcircle of triangle KLM taking OM as the radius at the center O.
Step-6: We take an arc of 4.3 cm at the point K and denote its point of intersection with the circle as N. We join KN and MN.
We have the required cyclic quadrilateral KLMN.
Note: We note that a cyclic quadrilateral is a quadrilateral whose all vertices lie on a circle. The circle is called the circumcircle of the quadrilateral and the vertices are called concyclic. The circle passing through vertices of a triangle is called circum-circle and the center of circum-circle is called circum-center. The circum-center is the point of intersection of the perpendicular bisector of the circle.
Complete step-by-step solution:
Step-1: We construct a line segment KL whose length is 4.3cm with scale and ruler. KL will act as a side of the quadrilateral.
Step-2: We take an arc of 6.8cm and then take an arc of 4.8 cm and denote their point of intersection as M. We join the line segments $KM=6.8$cm and $LM=4.8$cm. Now we have constructed the triangle KLM.
Step-3: We construct PQ the perpendicular bisector of KL by taking an arc of $KL=4$ cm at both the points K and L with help of a compass.
Step-4: We similarly construct RS the perpendicular bisector of KL by taking an arc of $LM=4.8$ cm at both the points L and M with help of a compass. We denote the point of intersection of perpendicular bisectors PQ and RS as O.
Step-5: We draw the circumcircle of triangle KLM taking OM as the radius at the center O.
Step-6: We take an arc of 4.3 cm at the point K and denote its point of intersection with the circle as N. We join KN and MN.
We have the required cyclic quadrilateral KLMN.
Note: We note that a cyclic quadrilateral is a quadrilateral whose all vertices lie on a circle. The circle is called the circumcircle of the quadrilateral and the vertices are called concyclic. The circle passing through vertices of a triangle is called circum-circle and the center of circum-circle is called circum-center. The circum-center is the point of intersection of the perpendicular bisector of the circle.
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