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

Steps of construction (i) Draw a line segment BC = 8 cm. (ii) At point B, Using ruler and compass, draw an angle ∠CBX = 45°. (iii) Taking B as centre and radius 3.5 cm, mark an arc, which intersects AX at D. (iv) Jion CD and draw a perpendicular bisector (MN) of CD, which intersects at BD produced aRead more

Steps of construction
(i) Draw a line segment BC = 8 cm.
(ii) At point B, Using ruler and compass, draw an angle ∠CBX = 45°.
(iii) Taking B as centre and radius 3.5 cm, mark an arc, which intersects AX at D.
(iv) Jion CD and draw a perpendicular bisector (MN) of CD, which intersects at BD produced 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]

Steps of construction (i) Draw a line segment QR = 6 cm. (ii) At point Q, Using ruler and compass, draw an angle ∠RQX = 60°. Produce XQ to K. (iii) Taking Q as centre and 2 cm as radius, draw an arc which intersects QK at S. (iv) Jion SR and draw a perpendicular bisector (MN) of SR, which intersectsRead more

Steps of construction
(i) Draw a line segment QR = 6 cm.
(ii) At point Q, Using ruler and compass, draw an angle ∠RQX = 60°. Produce XQ to K.
(iii) Taking Q as centre and 2 cm as radius, draw an arc which intersects QK at S.
(iv) Jion SR and draw a perpendicular bisector (MN) of SR, which intersects QX at point P.
(v) join PR. Triangle PQR is the required triangle.
Justification
Point P lies on the perpendicular bisector of SR. So, PS = PR
Here, QS = PS – PQ
⇒ QS = PR – AC [∵ PS = PR]

Steps of construction (i) Draw a line segment AB = 11 cm. (ii) At A, Using ruler and compass, draw an angle ∠BAX = 15° and at point B, draw an angle ∠ABX = 45°. (iii) Draw the perpendicular bisector (MN) of AX, which intersects AB at Y. (iv) Draw the perpendicular bisector (ST) of BX, which intersecRead more

Steps of construction
(i) Draw a line segment AB = 11 cm.
(ii) At A, Using ruler and compass, draw an angle ∠BAX = 15° and at point B, draw an angle ∠ABX = 45°.
(iii) Draw the perpendicular bisector (MN) of AX, which intersects AB at Y.
(iv) Draw the perpendicular bisector (ST) of BX, which intersects AB at Z
(v) join X to Y and X to Z.
(vi) Triangle XYZ is the required triangle.
Justification
Point Y lies on the perpendicular bisector of AX.
So, AY = XY
Point Z lies on the perpendicular bisector of BX.
So, BZ = ZX
Here, AB = AY + YZ + ZB
⇒ AB = XY + YZ + ZX
[∵ AY = XY and BZ = XZ]
∠XYZ is the exterior angle of triangle AXY. Therefore, ∠XYZ = ∠YXA + ∠YAX = 15° + 15° = 30° Similarly, ∠XYZ is the exterior angle of triangle BXZ.
Hence, ∠XZY = ∠ZXB + ∠ZBX = 45° + 45° = 90°

Steps of construction (i) Draw a line segment AB = 12 cm. (ii) At point A, Using ruler and compass, draw an angle ∠BAX = 90°. (iii) Taking A as centre and 18 cm as radius, draw an arc which intersect AX at D. (iv) Join B to D. Draw a perpendicular bisector (MN) of BD which intersects AD at C. (v) joRead more

Steps of construction
(i) Draw a line segment AB = 12 cm.
(ii) At point A, Using ruler and compass, draw an angle ∠BAX = 90°.
(iii) Taking A as centre and 18 cm as radius, draw an arc which intersect AX at D.
(iv) Join B to D. Draw a perpendicular bisector (MN) of BD which intersects AD at C.
(v) join B to C. Triangle ABC is the required triangle.
Justification
Point C lies on the perpendicular bisector of BD.
So, BC = CD
Here, AD = AC + CD
⇒ AD = AC + BC [∵ BC = CD]

## 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

## Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

Steps of construction (i) Draw a line segment BC = 8 cm. (ii) At point B, Using ruler and compass, draw an angle ∠CBX = 45°. (iii) Taking B as centre and radius 3.5 cm, mark an arc, which intersects AX at D. (iv) Jion CD and draw a perpendicular bisector (MN) of CD, which intersects at BD produced aRead more

Steps of construction

See less(i) Draw a line segment BC = 8 cm.

(ii) At point B, Using ruler and compass, draw an angle ∠CBX = 45°.

(iii) Taking B as centre and radius 3.5 cm, mark an arc, which intersects AX at D.

(iv) Jion CD and draw a perpendicular bisector (MN) of CD, which intersects at BD produced 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]

## Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Steps of construction (i) Draw a line segment QR = 6 cm. (ii) At point Q, Using ruler and compass, draw an angle ∠RQX = 60°. Produce XQ to K. (iii) Taking Q as centre and 2 cm as radius, draw an arc which intersects QK at S. (iv) Jion SR and draw a perpendicular bisector (MN) of SR, which intersectsRead more

Steps of construction

See less(i) Draw a line segment QR = 6 cm.

(ii) At point Q, Using ruler and compass, draw an angle ∠RQX = 60°. Produce XQ to K.

(iii) Taking Q as centre and 2 cm as radius, draw an arc which intersects QK at S.

(iv) Jion SR and draw a perpendicular bisector (MN) of SR, which intersects QX at point P.

(v) join PR. Triangle PQR is the required triangle.

Justification

Point P lies on the perpendicular bisector of SR. So, PS = PR

Here, QS = PS – PQ

⇒ QS = PR – AC [∵ PS = PR]

## Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Steps of construction (i) Draw a line segment AB = 11 cm. (ii) At A, Using ruler and compass, draw an angle ∠BAX = 15° and at point B, draw an angle ∠ABX = 45°. (iii) Draw the perpendicular bisector (MN) of AX, which intersects AB at Y. (iv) Draw the perpendicular bisector (ST) of BX, which intersecRead more

Steps of construction

See less(i) Draw a line segment AB = 11 cm.

(ii) At A, Using ruler and compass, draw an angle ∠BAX = 15° and at point B, draw an angle ∠ABX = 45°.

(iii) Draw the perpendicular bisector (MN) of AX, which intersects AB at Y.

(iv) Draw the perpendicular bisector (ST) of BX, which intersects AB at Z

(v) join X to Y and X to Z.

(vi) Triangle XYZ is the required triangle.

Justification

Point Y lies on the perpendicular bisector of AX.

So, AY = XY

Point Z lies on the perpendicular bisector of BX.

So, BZ = ZX

Here, AB = AY + YZ + ZB

⇒ AB = XY + YZ + ZX

[∵ AY = XY and BZ = XZ]

∠XYZ is the exterior angle of triangle AXY. Therefore, ∠XYZ = ∠YXA + ∠YAX = 15° + 15° = 30° Similarly, ∠XYZ is the exterior angle of triangle BXZ.

Hence, ∠XZY = ∠ZXB + ∠ZBX = 45° + 45° = 90°

## Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Steps of construction (i) Draw a line segment AB = 12 cm. (ii) At point A, Using ruler and compass, draw an angle ∠BAX = 90°. (iii) Taking A as centre and 18 cm as radius, draw an arc which intersect AX at D. (iv) Join B to D. Draw a perpendicular bisector (MN) of BD which intersects AD at C. (v) joRead more

Steps of construction

See less(i) Draw a line segment AB = 12 cm.

(ii) At point A, Using ruler and compass, draw an angle ∠BAX = 90°.

(iii) Taking A as centre and 18 cm as radius, draw an arc which intersect AX at D.

(iv) Join B to D. Draw a perpendicular bisector (MN) of BD which intersects AD at C.

(v) join B to C. Triangle ABC is the required triangle.

Justification

Point C lies on the perpendicular bisector of BD.

So, BC = CD

Here, AD = AC + CD

⇒ AD = AC + BC [∵ BC = CD]

## Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

In each pair either 0 or 1 or 2 points are common. The maximum number of common points is 2.

In each pair either 0 or 1 or 2 points are common. The maximum number of common points is 2.

See less