WBBSE Solutions For Class 10 Maths Chapter 11 Construction Of Circumcircle And Incircle Of A Triangle Exercise 11.1
Question 1. Let us draw the following triangles. By drawing the circumcircle in each case, let us write the position of the circumstances and the length of the circumradius by measuring it.
1. An equilateral triangle having each side of length 6 cm.
Solution: ABC is an equilateral triangle whose each side is 6 cm. Circumcentre O is inside the triangle.
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Question 2. An isosceles triangle whose length of the base is 5.2 cm and each of the equal sides is 7 cm.
Solution: O is the circumstances of the circumcircle of the PQR, whose length of circumradius is 4 cm. Centre O is inside the triangle.
Question 3. A right-angled triangle has two sides 4 cm and 8 cm in length, containing the right angle.
Solution: We draw the circumcircle of the right-angled triangle XYZ, whose centre is at the midpoint of the hypotenuse length of the circumradius is 4.6 cm.
Question 4. A right-angled triangle has a length of hypotenuse 12 cm and another side of 5 cm in length.
Solution: ABC is a right-angled triangle whose length of the hypotenuse is 12 cm & length of the other side is 5 cm. Centre O is on the midpoint of the hypotenuse.
Question 5. A triangle whose length of one side is 6.7 cm and the two angles adjacent to this side are 75° and 55°.
Solution:
Question 6. ABC is a triangle whose BC = 5 cm, ∠ABC= 100°, and AB = 4 cm.
Solution: Draw the circumcircle of ABC, whose center is outside the triangle & the length of the circumradius is 3.7 cm.
Question 7. Given PQ = 7.5 cm. QPR = 45°, PQR = 75°; PQ = 7.5 cm. ZQPS = 60°, PQS = 60°. Let us draw APQR and APQS in such a way that the points R and S lie on the same side of PQ, let us draw the circumcircle of APQR, and let us observe and write the position of the point S within, on, and outside the circumcircle. Let us find out its explanation.
Solution:
Given PQ = 7.5 cm. QPR = 45°, PQR = 75°; PQ = 7.5 cm. ZQPS = 60°, PQS = 60°. Let us draw APQR and APQS in such a way that the points R and S lie on the same side of PQ
We draw the circumcircle of the APQR. Point S is on that circumcircle.
Question 8. Given AB = 5 cm, <BAC = 30°, ∠ABC = 60°; AB = 5 cm, ∠BAD = 45°, ∠ABD = 45°. Let us draw AABC and AABD in such a way that points C and D lie on opposite sides of AB. Let us draw the circle circumscribing AABC. Let us write the position of point D with respect to the circumcircle. Let us write by understanding what other characteristics we are observing here.
Solution:
Given AB = 5 cm, <BAC = 30°, ZABC = 60°; AB = 5 cm, BAD = 45°, ZABD = 45°. Let us draw AABC and AABD in such a way that points C and D lie on opposite sides of AB.
We draw the circumcircle of AABC. Point D is on the circumcircle. The centre is on the midpoint of AB.
Question 9. We draw a quadrilateral ABCD having AB = 4 cm, BC = 7 cm CD = 4 cm, ZABC = 60°, BCD = 60°. Let us draw the circle circumscribing ABC and write what other characteristics we observe.
Solution: We draw the circumcircle of AABC whose center O is inside the AABC & inside the quadrilateral ABCD.
Question 10. Let us draw the rectangle PQRS having PQ = 4 cm, and QR = 6 cm. Let us draw the diagonals of the rectangle. Let us write by calculating the position of the center of the circumcircle of △PQR and the length of the circumradius without drawing. By drawing the circumcircle of △PQR, let us verify.
Solution: The diagonals PQ & RS of the rectangle PQRS are drawn. The centre of the △PQR is at the intersecting point of the diagonals. The length of the circumradius is 3.8 cm.
Question 11. If any circular picture is given, then how shall we find its center? Let us find the center of the circle in the adjoining.
Solution: By drawing, the centre & incentre of an equilateral triangle are the same.
