Given : Circle C (P,r) and circle C (Q, r') intersects each other at the points A and B. To prove: points p and Q lies on the perpendicular bisector of common chord AB. Construction: Join point P and Q to mid-point M of chord AB. Proof: AB is chord of circle C (P, r) and PM is bisector of chord AB.Read more
Given : Circle C (P,r) and circle C (Q, r’) intersects each other at the points A and B.
To prove: points p and Q lies on the perpendicular bisector of common chord AB.
Construction: Join point P and Q to mid-point M of chord AB.
Proof: AB is chord of circle C (P, r) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠PMA = 90°
Similarly, AB is chord of circle C (Q, r) and QM is bisector of chord AB.
Therefore, QM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠QMA = 90
Now, ∠PMA + ∠QMA = 90° + 90° = 180°
Since, ∠PMA and ∠QMA are forming liner pair. So PMQ is a straight line.
Hence, points P and Q lies on the perpendicular bisector of common chord AB.
I/Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D. (iv) Taking C and D as centre, draw arcs with equal raRead more
I/Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D.
(iv) Taking C and D as centre, draw arcs with equal radius( more than half of CD), which intersect at E.
(v)Draw the ray AE ∠EAD is the required angle of 30°.
II/Steps of construction.
(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, Draw an arc which intersent the previous arc at M.
(iv) Similarly, taking M as centre and with the same radius, draw an arc which intersect at N.
(v) Taking M and N as centre, draw arcs with equal radius (more than half of MN), which intersect at P.
(vi) Draw a ray AP which intersects the main arc at D.
(vii) Taking C and D as centre, draw arcs with equal radius (more than half of CD), which intersect at G.
(viii) Draw a ray AG which intersects the main arc at E.
(iX) Taking C and E as centre, draw arcs with equal radius (more than half of CE), which intersect at F.
(x) draw an arc AF. ∠FAB is the required angle of 22(1/2)°.
III/Steps of construction.
(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(III) Taking A as centre and with the same radius, draw an arc which intersect the previous arc at D.
(iv) Taking C and D as centre, draw arcs with equal radius (more than half of CD), which intersect at G.
(v) Draw a ray AG which intersects the main arc at F.
(vi) Taking C and F as centre, draw arcs with equal radius (more than half of CF), which intersect at H.
(vii) Draw a ray AH.
(viii) ∠HAB is the required angle of 15°.
I/Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at F. (iv) Similarly, taking E as centre and with the same rRead more
I/Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at F.
(iv) Similarly, taking E as centre and with the same radius, draw an arc which intersect at E.
(v)Taking E and F as as center, draw arcs with equal radius (more than half of EF). which intersect at H.
(vi) Draw a ray AH which intersects the main arc at D.
(vii) Taking F and D as centre, draw arcs with equal radius (more than half of FD), which intersect at G.
(viii) Draw a ray AG. ∠GAB is the required angle of 75°.
II/Steps of construction.
(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, Draw an arc which intersect the previous arc at D.
(iv) Similarly, taking E as centre and with the equal radius, draw an arc which intersect at G.
(v) Taking D and G as centre, draw arcs with same radius (more than half of DG), which intersect at F.
(vi) Draw a ray AF which intersects the main arc at E.
(vii) Taking E and G as centre, draw arcs with equal radius (more than half of EG), which intersect at H.
(viii) Draw a ray AH. ∠HAB is the required angle of 105°.
III/Steps of construction.
(i) Draw a ray AD at the point A.
(II) Taking B as centre and a convenient radius, draw an arc which intersect AD at C.
(III) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at P.
(iv) Similarly, Taking P as centre and with the equal radius, draw an arc which intersect at F.
(v) Taking P and F as centre, draw arcs with same radius (more than half of PF), which intersect at H.
(vi) Draw a ray BH from the point B.
(vii) Taking B as center, draw an arc taking some radius, which intersects AB at E and BH at Q.
(viii) Taking E and Q as centre, draw arcs with equal radius (more than half of EQ), which intersect at G.
(ix) Draw a ray BG. ∠GBD is the required angle of 135°.
Steps of construction (i) Draw a line segment AB of given measurement. (ii) Taking A and B as centre, draw arcs with same radius (equal to AB), which intersect at C. (iii) Join A to C and B to C. (iv) ΔABC is the required equilateral triangle. Justification In ΔABC, AB = BC [∵ By construction] AC =Read more
Steps of construction
(i) Draw a line segment AB of given measurement.
(ii) Taking A and B as centre, draw arcs with same radius (equal to AB), which intersect at C.
(iii) Join A to C and B to C.
(iv) ΔABC is the required equilateral triangle.
Justification
In ΔABC,
AB = BC [∵ By construction]
AC = BC [∵ By construction]
Hence, AB = BC = AC
⇒ Triangle ABC is an equilateral triangle.
Steps of construction (i) Draw a line segment BC = 7 cm. (ii) Using ruler and compass, draw an angle ∠CBX = 75°. (iii) Taking B as centre and 13 cm as radius, mark an arc on BX, which intersects at D. (iv) Jion CD and draw a perpendicular bisector (PQ) of CD, which intersects BD at A. (v) join AC. (Read more
Steps of construction
(i) Draw a line segment BC = 7 cm.
(ii) Using ruler and compass, draw an angle ∠CBX = 75°.
(iii) Taking B as centre and 13 cm as radius, mark an arc on BX, which intersects at D.
(iv) Jion CD and draw a perpendicular bisector (PQ) of CD, which intersects BD at A.
(v) join AC.
(vi) Triangle ABC is the required triangle.
Justification
Point A lies on the perpendicular bisector of DC. So, AD = AC
Here, AB = BD – AD
⇒ AB = BD – AC [∵ AD = AC]
⇒ AB + AC = BD
If two circles intersect at two points, prove that their centers lie on the perpendicular bisector of the common chord.
Given : Circle C (P,r) and circle C (Q, r') intersects each other at the points A and B. To prove: points p and Q lies on the perpendicular bisector of common chord AB. Construction: Join point P and Q to mid-point M of chord AB. Proof: AB is chord of circle C (P, r) and PM is bisector of chord AB.Read more
Given : Circle C (P,r) and circle C (Q, r’) intersects each other at the points A and B.
See lessTo prove: points p and Q lies on the perpendicular bisector of common chord AB.
Construction: Join point P and Q to mid-point M of chord AB.
Proof: AB is chord of circle C (P, r) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠PMA = 90°
Similarly, AB is chord of circle C (Q, r) and QM is bisector of chord AB.
Therefore, QM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠QMA = 90
Now, ∠PMA + ∠QMA = 90° + 90° = 180°
Since, ∠PMA and ∠QMA are forming liner pair. So PMQ is a straight line.
Hence, points P and Q lies on the perpendicular bisector of common chord AB.
Construct the angles of the following measurements: (i) 30° (ii) 22/5° (iii) 15°
I/Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D. (iv) Taking C and D as centre, draw arcs with equal raRead more
I/Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D.
(iv) Taking C and D as centre, draw arcs with equal radius( more than half of CD), which intersect at E.
(v)Draw the ray AE ∠EAD is the required angle of 30°.
II/Steps of construction.
See less(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, Draw an arc which intersent the previous arc at M.
(iv) Similarly, taking M as centre and with the same radius, draw an arc which intersect at N.
(v) Taking M and N as centre, draw arcs with equal radius (more than half of MN), which intersect at P.
(vi) Draw a ray AP which intersects the main arc at D.
(vii) Taking C and D as centre, draw arcs with equal radius (more than half of CD), which intersect at G.
(viii) Draw a ray AG which intersects the main arc at E.
(iX) Taking C and E as centre, draw arcs with equal radius (more than half of CE), which intersect at F.
(x) draw an arc AF. ∠FAB is the required angle of 22(1/2)°.
III/Steps of construction.
(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(III) Taking A as centre and with the same radius, draw an arc which intersect the previous arc at D.
(iv) Taking C and D as centre, draw arcs with equal radius (more than half of CD), which intersect at G.
(v) Draw a ray AG which intersects the main arc at F.
(vi) Taking C and F as centre, draw arcs with equal radius (more than half of CF), which intersect at H.
(vii) Draw a ray AH.
(viii) ∠HAB is the required angle of 15°.
Construct the following angles and verify by measuring them by a protractor: (i) 75° (ii) 105° (iii) 135°
I/Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at F. (iv) Similarly, taking E as centre and with the same rRead more
I/Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at F.
(iv) Similarly, taking E as centre and with the same radius, draw an arc which intersect at E.
(v)Taking E and F as as center, draw arcs with equal radius (more than half of EF). which intersect at H.
(vi) Draw a ray AH which intersects the main arc at D.
(vii) Taking F and D as centre, draw arcs with equal radius (more than half of FD), which intersect at G.
(viii) Draw a ray AG. ∠GAB is the required angle of 75°.
II/Steps of construction.
(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, Draw an arc which intersect the previous arc at D.
(iv) Similarly, taking E as centre and with the equal radius, draw an arc which intersect at G.
(v) Taking D and G as centre, draw arcs with same radius (more than half of DG), which intersect at F.
(vi) Draw a ray AF which intersects the main arc at E.
(vii) Taking E and G as centre, draw arcs with equal radius (more than half of EG), which intersect at H.
(viii) Draw a ray AH. ∠HAB is the required angle of 105°.
III/Steps of construction.
See less(i) Draw a ray AD at the point A.
(II) Taking B as centre and a convenient radius, draw an arc which intersect AD at C.
(III) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at P.
(iv) Similarly, Taking P as centre and with the equal radius, draw an arc which intersect at F.
(v) Taking P and F as centre, draw arcs with same radius (more than half of PF), which intersect at H.
(vi) Draw a ray BH from the point B.
(vii) Taking B as center, draw an arc taking some radius, which intersects AB at E and BH at Q.
(viii) Taking E and Q as centre, draw arcs with equal radius (more than half of EQ), which intersect at G.
(ix) Draw a ray BG. ∠GBD is the required angle of 135°.
Construct an equilateral triangle, given its side and justify the construction.
Steps of construction (i) Draw a line segment AB of given measurement. (ii) Taking A and B as centre, draw arcs with same radius (equal to AB), which intersect at C. (iii) Join A to C and B to C. (iv) ΔABC is the required equilateral triangle. Justification In ΔABC, AB = BC [∵ By construction] AC =Read more
Steps of construction
See less(i) Draw a line segment AB of given measurement.
(ii) Taking A and B as centre, draw arcs with same radius (equal to AB), which intersect at C.
(iii) Join A to C and B to C.
(iv) ΔABC is the required equilateral triangle.
Justification
In ΔABC,
AB = BC [∵ By construction]
AC = BC [∵ By construction]
Hence, AB = BC = AC
⇒ Triangle ABC is an equilateral triangle.
Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.
Steps of construction (i) Draw a line segment BC = 7 cm. (ii) Using ruler and compass, draw an angle ∠CBX = 75°. (iii) Taking B as centre and 13 cm as radius, mark an arc on BX, which intersects at D. (iv) Jion CD and draw a perpendicular bisector (PQ) of CD, which intersects BD at A. (v) join AC. (Read more
Steps of construction
See less(i) Draw a line segment BC = 7 cm.
(ii) Using ruler and compass, draw an angle ∠CBX = 75°.
(iii) Taking B as centre and 13 cm as radius, mark an arc on BX, which intersects at D.
(iv) Jion CD and draw a perpendicular bisector (PQ) of CD, which intersects BD at A.
(v) join AC.
(vi) Triangle ABC is the required triangle.
Justification
Point A lies on the perpendicular bisector of DC. So, AD = AC
Here, AB = BD – AD
⇒ AB = BD – AC [∵ AD = AC]
⇒ AB + AC = BD