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

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.

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.

100, 48

Let x be the wide of the screen.
From the figure,
x² + 2² = 25²
x² + 49 = 625
x² = 625 - 49
= 576
\(x=\sqrt{576}\)
= 24
The wide of the screen is 24 inches
The TV cabinet wide is 21 inches. It is not fit for the TV which has the wide screen 24 inches

Yes, the side of length 3.7 cm is the hypotenuse.

From the figure
d² = 80² + 150²
= 6400 + 22500
d² = 28900
d = 170
The distance between the helicopter and the ship is 170 m.

25, 24

Given:
DE = 7 cm, EA = 6.5 cm ㄥEDN = 100⁰ and ㄥDEA = 70⁰ and \(\overset { \_ \_ }{ DE } \) || \(\overset { \_ \_ }{ NA} \)


Steps:
1. Draw a line segment DE = 7cm.
2. Construct an angle ㄥDEX = 70⁰ at E.
3. With E as centre draw an arc of radius 6.5cm cutting EX at A.
4. Draw AY parallel to DE.
5. Construct an angle ㄥEDZ = 100⁰ at D cutting AY at N.
6. DEAN is the required trapezium.
Calculation of area:
Area of the trapezium DEAN = \(\frac { 1 }{ 2 } \) x h x (a+b) sq. units
= \(\frac { 1 }{ 2 } \) x 6.1 x (7 + 5.8) = 39.04 sq. units

Given:
CA = 9 cm, ㄥCAR = 70⁰ AR = 6 cm, and CD = 7 cm and \(\overset { \_ }{ CA } \) || \(\overset { \_ }{ DR } \)


Steps:
1. Draw a line segment CA = 9 cm.
2. Construct an angle ㄥCAX = 70⁰ at A.
3. With A as centre, draw an arc of radius 6 cm cutting AX at R.
4. Draw RY parallel to CA.
5. With C as centre, draw an arc of radius 7 cm cutting RY at D.
6. Join CD. CARD is the required trapezium.
Calculation of area:
Area of the trapezium CARD = \(\frac { 1 }{ 2 } \) x h x(a + b) sq.units
= \(\frac { 1 }{ 2 } \) x 5.6 x (9 + 11) = 56 sq. cm