Steps for construction:
Step 1: Draw the ΔABC with the given measurement.
Step 2: Construct the perpendicular bisector of any two sides (AB and AC) and let them meet at S which is the circumcentre.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, Band C.

\(\angle POS+\angle SOR=180^0\) (angle ofa straight line)
105° + x°= 180°
x°=180°-105°
=75°
The value of x = 75°.

\(\angle AOB =90^0\)
\(\angle BOC +\angle COA=90^0\)
3x⁰+2x⁰ = 90⁰
5x⁰= 90⁰
\(x^0=\frac{90}{5}=18^0\)
The value of x = 18°

Supplement of Right angle (90°) = 180° - 90°
= 90°

Supplement of 140° = 180⁰- 140°
= 40°

Complement of 70° = 90° - 70°
=20°

ㄥACD = ㄥA + ㄥB
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x⁰
x⁰ = 120° - 50°
= 70°
In the the triangle ABC
ㄥA + ㄥB + ㄥACB = 180° (Sum of the angles of a Δ)
50° +x + ㄥACB = 180°
50° + 70° + ㄥACB = 180°
ㄥACB = 180° - 120°
y = 60°    (OR)
ㄥACD + ㄥACB = 180⁰(Angles of a linear pair)
ㄥACB = 180° - 120°
= 60°
The value ofx = 70° andy = 60°.

ABCD is a parallelogram.
OA = OC and
OB = OD (∴ Diagonals bisect each other)
now OB + BQ = OD + DP
OQ = OP and OA = OC
APCQ is a parallelogram.

Through T draw TE|| AB.
\(\therefore \angle BAT+\angle ATE=180^0(AB||TE)\)
\(140^0+\angle ATE=180^0\)
\(\angle ATE=180^0-140^0=40^0\)
Similarly \(\angle ETC+\angle TCD=180^0(TE||CD)\)
\(\angle ETC+150^0=180^0\)
\(\angle ETC=180^0-150^0=30^0\)
\(x=\angle ATE+\angle ETC\)
= 40° + 30° = 70°
x = 70°

Step 1: Draw the ΔPQR with the given measurements


Step 2:
Construct altitudes from any two vertices (say) R and P, to their opposite sides PQ and QR respectively.
The point of intersection of the altitude H is the Orthocentre of the given ΔPQR.

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (AB and BC) and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, Band C.


Circum radius = 4.3 cm.

Since the diagonals of a rhombus bisect each other at right angles
\(AO=\frac{1}{2}AC=\frac{1}{2}\times12=6\ cm\)
\(BO=\frac{1}{2}BD=\frac{1}{2}\times16=8\ cm\)
In the right triangle AOD
AD² = AO² + DO²
= 6² + 8²
= 36 + 64
= 100
\(\therefore\ AD=\sqrt {100}\)
= 10
\(\therefore\) AB = BC = CD = AD = 10 cm.