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:
DC = 8 cm, UK = 6 cm and ㄥDOU = 110⁰


Steps:
1. Draw a line segment DC = 8 cm.
2. Mark O the midpoint of \(\overset { \_ \_ }{ DC } \).
3. Draw a line \(\overset { \_ \_ }{ XY } \) through O which makes ㄥDOY = 110⁰.
4. With O as centre and 3 cm as radius draw two arcs on \(\overset { \_ \_ }{XY } \)on either sides of \(\overset { \_ \_ }{DC } \). Let the arcs cut \(\overset { \_ \_ }{ OX } \)at K and \(\overset { \_ \_ }{OY } \) at U
5. Join \(\overset { \_ \_ }{DU } \) , \(\overset { \_ \_ }{UC } \), \(\overset { \_ \_ }{CK } \) and \(\overset { \_ \_ }{KD } \).
6. DUCK is the required parallelogram.
Calculation of Area:
Area of the parallelogram DUCK = bh sq.units
= 5.8 x 3.9 = 22.62sq.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

Given:
BI = 6.5 cm, IR = 5 cm and ㄥBIR = 70⁰


Steps:
1. Draw a line segment BI = 6.5 cm.
2. Make an angle ㄥBIX = 70⁰ at I on \(\overset { \_ \_ }{ BI } \).
3. With I as centre, draw an arc of radius 5 cm cutting IX at R.
4. With B and R as centres, draw arcs of radii 5 cm and 6.5 cm respectively. Let them cut at D.
5. Join BD and RD.
6. BIRD is the required parallelogram.
Calculation of area:
Area of the parallelogram BIRD = bh sq. units
= 6.5 × 4.7 = 30.55 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.

Here, the hypotenuse is 29 m. One side of the right triangle is 20 m. let the other side be ‘l’ m Therefore, by Pythagoras theorem,
l² = 29² − 20² = 841 − 400 = 441 = 21²
∴ l = 21m
Therefore, the area rectangular plot of land = l × b square units. = 20 × 21 = 210 m².

Now Δ ABC is right angled.
Therefore, by Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = 8² +15² = 64 + 225 = 289
AC² = 17²
⇒ AC = 17m.
Therefore, the length of the diagonal of the two intersecting roads is 17 m.

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.

The ladder, wall and the ground form a right triangle with the ladder as the hypotenuse. From the figure, by Pythagoras theorem,
20² = 16² + x²
⇒ 400 = 256 + x²
⇒ x² = 400 − 256 = 144 = 12²
⇒ x = 12 feet
Therefore, the base (foot) of the ladder is 12 feet away from the wall.

Take a = 5, b = 12 and c = 13
Now, a² + b² = 5² +12² = 25 +144 = 169 = 13² = c²
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.

a) Δ ABC is right angled as AB ⊥ AC at A.
b) AB and AC are legs of Δ ABC.
c) CB is called as the hypotenuse.
d) ㄥB + ㄥC = 90⁰ and equal angles are opposite to equal sides. Hence, ㄥB = ㄥC = \(\frac { { 90 }^{ 0 } }{ 2 } \) = 45⁰