Given:
BE = 7 cm, BA = 7.5 cm and ㄥBEA = 80⁰


Steps:
1. Draw a line segment BE = 7 cm.
2. Make an angle ㄥBEX =80° at E on \(\overset { \_ \_ }{ BE } \).
3. With B as centre, draw an arc of radius 7.5 cm cutting \(\overset { \_ \_ }{ EX } \)  at A and Join BA.
4. With B as centre, draw an arc of radius equal to the length of \(\overset { \_ \_ }{ AE } \).
5. With A as centre, draw an arc of radius 7 cm. Let both arcs cut at R.
6. Join BR and AR.
7. BEAR is the required parallelogram.
Calculation of area:
Area of the parallelogram BEAR = bh sq. units
= 7 x 4.1 = 28.7 sq. cm

Given:
CA = 7 cm, CF = 6 cm and AF = 10 cm


Steps:
1. Draw a line segment CA = 7 cm.
2. With C and A as centres, draw arcs of radii 7 cm and 6 cm respectively. Let them cut at F.
3. Join CF and AF.
4. With A and F as centres, draw arcs of radii 6 cm and 7 cm respectively. Let them cut at L.
5. Join AL and FL.
6. CALF is the required parallelogram.
Calculation of area:
Area of the parallelogram CALF = bh sq. units
= 7 x 5.9 = 41.3 sq. cm

Now, by altitude-on-hypotenuse theorem,
AB² = AD x AC gives,
10² = x × 26
\(\Rightarrow x=\frac { 100 }{ 26 } =\frac { 50 }{ 13 } units\quad and\)
BC² = CD × AC gives,
24² = y × 26
\(\Rightarrow y=\frac { 576 }{ 26 } =\frac { 288 }{ 13 } units\quad and\)
In Δ ABC, AB² + BC² = 10² + 24² = 676 = 26² = AC² Therefore, Δ ABC is a right angled triangle.

As Δ ABC is equilateral from the figure, AB = BC = AC = (x − 2) cm.
∴ From Δ BCD, by Pythagoras theorem
BD² = BC² + CD²
⇒ (x + 2)² = (x − 2)² + 8²
x² + 4x + 4 = x² − 4x + 4 + 8²
⇒ 8x = 8²
⇒ x = 8 cm
∴ The side of the equilateral Δ ABC = 6 cm and BD = 10 cm.

From Δ LMO, by Pythagoras theorem,
LM² = OL² − OM²
⇒ LM² = 13² −12² = 169 −144 = 25 = 5²
∴ LM = 5 units
From Δ NMO, by Pythagoras theorem,
MN² = ON² − OM²
= 15² −12² = 225 −144 = 81 = 9²
∴ MN = 9 units
Hence, LN = LM + MN = 5 + 9 = 14 units
Area of Δ LON = \(\frac { 1 }{ 2 } \) × base × height
= \(\frac { 1 }{ 2 } \) × LN × OM
= \(\frac { 1 }{ 2 } \) × 14 × 12
= 84 square units.