4+5+6=15
a:b:c=4:5:6
a=\(\frac { 4 }{ 15 } \times 150\)=40 cm
b=\(\frac { 5 }{ 15 } \times 150\)=50 cm
c=\(\frac { 6 }{ 15 } \times 150\)=60 cm

\(\angle OAB=\angle OBA=20°\)  
\(\angle OAC=\angle OCA=30°\) 
\(x=2\angle BAC\)

\(\angle OAB=\angle OBA=45^o\)  
\(\angle OAC=\angle OCA=35°\) 
\(\angle BOC=2\angle BAC\)

OB=OC=BC
∴ \(\angle BOC=60°\)
∴ \(\angle BAC=\frac { 1 }{ 2 } 60°\angle BOC=30°\)

A median of a triangle divides it into two triangles of equal areas.

A rectangle is essentially a parallelogram. If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

Steps of construction:
i) Construct a triangle ABC.
ii) Mark an exterior angle outside the triangle ABC,and name the point as E.

iii) Now, ACE is the exterior angle.
iv) Draw a bisecter of ㄥACE

\(\Delta \)PSR and \(\Delta \)PSQ are on the same base PS and between the same parallels PS and QR.
ar(\(\Delta \)PSR) = ar(\(\Delta \)PSQ)
ar(\(\Delta \)PSR) - ar(\(\Delta \)PSO) = ar(\(\Delta \)PSQ) - ar(\(\Delta \)PSO)
ar(\(\Delta \)ROS) = ar(\(\Delta \)POQ).

84 cm²

\(18\sqrt { 2 } \)cm²

50°

Construction: Join OA and Oc.
Since perpendicular from centre of the circle to the chord bisects the chord

AP = PB = \(\frac { 1 }{ 2 } \) AB = 4 cm
CQ = QD = \(\frac { 1 }{ 2 } \) CD = 3 cm
In \(\Delta\)OAP,
OP² = OA²-AP²     (Pythagoras theorem)
OP² = 5²-4²
= 25-16
=9
\(\therefore\)  OP =3
In \(\Delta\)OCQ,
OQ² = OC² - CQ²   (Pythagoras theorem)
= 5²-3²
= 25-9
= 16
\(\therefore\)   OQ =4
\(\therefore\)   PQ =OP+OQ
=3+4
= 7cm.

Steps of construction:
i) Draw a line segment AB = 11 cm(As XY + YZ + ZX = 11cm)
ii) Construct an angle ㄥPAB of 30^o at point A and an angle ㄥQAB = 90^o at point B.

iii) Bisect ㄥPAB of 30^o and ㄥQBA.These bisectors intersect each other at point X.
iv) Draw perpendicular bisectors ST of AX and UV of BX
v) ⊥ bisector ST intersects AB at Y and UV intersects AB at Z. Join XY,XZ. ΔXYZ is the required triangle.

Steps of construction:
i) Draw a line segment BC = 7 cm. At point B draw ㄥXBC=  75^o
ii) Cut a line segment BD = 13 cm (i.e., equal to AB + AC) from BX
iii) Join DC and make an angle DCY equal to ㄥBDC
iv) Line CY intersects BX at A. ΔABC is the required triangle.

Rs.5376

In figure, AD is the median of \(\Delta\)ABC.

\(\therefore \ ar(\triangle ABD)=\frac { 1 }{ 2 } ar(\triangle ABC)\)      ...........(i)
Now, BE is the median of \(\Delta\)ABD
\(\therefore \ ar(\triangle BED)=\frac { 1 }{ 2 } ar(\Delta ABD)\)       ..............(ii)
From (i) and (ii), we get
ar(\(\Delta\)BED) = \(\frac { 1 }{ 2 } ar(\triangle ABC)\)

Let ABC be the right angled triangle right angled at B. Let O be the centre of the circumcircle.
Then, O is the midpoint of the hypotenuse AC. I by geometry
OA = OB =OC
= Radius of the circumcircle = 3 cm
 
  \(\therefore \) Hypotenuse AC = Diameter of the circle
= 2 \(\times \) Radius of the circumcircle
= 2 \(\times \) 3 = 6 cm
Let BM be the perpendicular from B on AC.
\(\therefore \) BM =2 cm
\(\therefore \)  Area of the right angled triangle ABC
= \(\frac { 1 }{ 2 } \) \(\times \)Base \(\times \) Altitude
= \(\frac { 1 }{ 2 } \)\(\times \) AC \(\times \) BM = \(\frac { 1 }{ 2 } \)\(\times \) 6\(\times \) 2 = 6 cm².