Chapter-11 Construction

Explanation:

Construction steps:
(i) Draw the ray OA. With point O as the center, draw an arc whose radius intersects OA at B.
(ii) With B as the center and the same radius as above, draw an arc that intersects the previously drawn arc at C.
(iii) With C as the center and the same radius as above, draw an arc that intersects the arc at D.
(iv) With C and D as centers, draw an arc of equal radius which intersects at E.
(v) Join OE, i.e. the desired radius at 90° to the given radius OA.

Proof:

∠AOC = 30°
∠AOC + ∠COB = 30° + 60°
∠AOB = 90°

Explanation:

Construction steps:
(i) Draw an OA ray. With O as the center, draw a radius arc and intersect OA at P.
(ii) With P as the center and the same radius, draw an arc that intersects the arc previously drawn at Q.
(iii) Draw an arc with Q as the center and with the same radius, and intersect the arc at R
( iv) Take Q and R as median arc of same radius and cut in S.
(v) Add OS. Let it cut the arc at point T.
(vi) With T and P as centers, draw 2 arcs that intersect in U and draw the median of angle BOA.

                    
 



Proof:
Bisector of ∠BOA = ½ × ∠BOA
= ½ × 90°
= 45°
∠BOA = 45°

Explanation:

Explanation:

Construction steps:
(i) Draw a radius PQ.
(ii) Now take P as the center of the circle, draw an arc of any radius, then mark a point B on the arc.
(iii) Take B as the center of the circle with the same radius and mark a point C on the arc ABC.
(iv) Now take B and C as centers, and from each center draw an arc which intersects at point D.
Then connect P and D to form a 90° angle. (v) Now, from centers A and E, mark 2 intersecting arcs at F, and we get the median at an angle of 90°.
(vi) Similarly, centering on G and A, and marking 2 intersecting arcs at H, we obtain the median at an angle of 45°.

                                        

Explanation:

Construction steps:
(i) Draw a radius OA.
(ii) Now take O as the center of the circle, take any radius, draw the arc, then mark point P on the arc, and add OP
(iii) Center P and Q, then draw the arc from each midpoint, each arc intersects other arcs at point R. Then by connecting R and O we get the midline at the angle of 60°.

(iv) Similarly, with S and Q as centers, mark 2 arcs that intersect at T, and we get the midline at an angle of 30°

Explanation:

Construction steps:
(i) Draw a radius OA.
(ii) Now draw an arc with O as the center.
(iii) With P as center, draw two arcs Q and R.
(iv) Now, with Q and R as centers, draw two arcs so that they meet at S, forming an angle of 90°.
(v) Centered in P and Q, the arc intersects in U.
(vi) Now join OR and thus construct the ∠UOA = 75°.

Explanation:

Construction steps:
(i) Draw a ray OA.
(ii) Now draw an arc with O as the center.
(iii) With P as center, draw two arcs Q and R.
(iv) Now, with Q and R as centers, draw two arcs so that they meet at S, forming an angle of 90°.
(v) Centered on R and T, make the arc intersect in U.
(vi) Connecting OR gives ∠COA =105°

Explanation:

Construction steps:
(i) Draw radius OA.
(ii) Now draw an arc with O as the center.
(iii) Now take P as the center, draw two arcs Q and R on the arc.
(iv) With Q and R as the center, draw an arc to intersect at B, and the angle formed is BOA=90°.
(v) With S and T as the center, draw an arc that intersects at point U or the centerline of angle BOD.
(vi) Join OR and form an ∠COA = 135°.

Explanation:

Construction steps:
(i) Draw a line segment AB = 6 cm.
(ii) With A and B as centers, draw two circular arcs on segment AB. Label them D and E respectively.
(iii) Now, centering on D and E, make two arcs intersect the previous arc respectively, forming an angle of 60°.
(iv) Extend the straight lines of A and B until they intersect at point C.

Prove:
In ∆ABC: 
∠A = ∠B=∠C
Use angles and properties:
∠A + ∠B + ∠C = 180°
60°+ 60°

∠ C = 180°
∠C= 180°− 120°
∠C= 60°

Explanation:

Construction steps:
(i) Draw a line segment 7 cm BC. Draw an angle of 75° at point B.
(ii) Starting from B, cut the arc BO = 13 cm (equal to AB + AC) on radius BX.
(iii) Connecting OC, draw the perpendicular bisector of OC.
(iv) Mark point A where the center line intersects XB.
(iv) Joining AC, ABC is the desired triangle.

Explanation:

Construction steps:
(i) Draw line segment BC = 8 cm and form a 45° angle at point B.
(ii) Section line segment BO = 3.5 cm (equal to AB - AC) on the BY radius.
(iii) Join OC and draw the bisector of OC.
(iv) Let it intersect BY at point A. Join AC. ABC is the desired triangle.

Explanation:

Construction steps:
(i) Draw a line segment QR of 6 cm. Draw an angle of 60° at point Q.
(ii) From point Q, cut an arc QO 2 cm from the line segment QT, which extends to the other side of the line segment XQ. (if PR>PQ and PR-PQ=2cm). join or.
(iii) Draw the perpendicular bisector of the OR line.
(iv) Let it cut QX into P. Join PQ, PR. PQR is the desired triangle