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

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.

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.

From the figure AC is to be found.
By using Pythagoras theorem,
AC² = AB² + BC² = 282 + 212 = 784 + 441 = 1225 = 352
∴ AC = 35 km

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.

Let \(\triangle\) ABC is a isosceles triangle.
Draw AD perpendicular to BC, we get two right angled triangles.
In  \(\triangle\)ABD
13² = x² + h²        ..........(1)
In \(\triangle\) ADC,
13² = h² +(24 - x)²  .........(2)
From (1) and (2)
\( x^{2}+\mathrm{h}^{2} =(24-x)^{2}+\mathrm{h}^{2} \)
\(x^{2} =(24-x)^{2} \\ x^{2} =576-48 x+x^{2} \)
\(0 = 576-48 x \\ \therefore \ 48 x =576 \)
\(x =\frac{576}{48} \\ x =12 \)
Put x = 12 in (1), we get 13² = x² +h²
\( 13^{2} =12^{2}+h^{2} \)
\(169 =144+h^{2} \)
\(\therefore h^{2} =169-144 \)
\(h^{2} =25 \)
\(h =\sqrt{25}\)
= 5
Height of the isosceles triangle is 5 cm.

Given:
DE = 8 cm, ES = 5.5 cm, KS = 5 cm, KD = 6 cm and \(\overset { \_ \_ }{ DE } \) || \(\overset { \_ \_ }{ KS } \)


Steps:
1. Draw a line segment DE = 8cm.
2. Mark the point A on DE such that DA = 5 cm.
3. With A and E as centres, draw arcs of radii 6 cm and 5.5 cm respectively. Let them cut at S. Join AS and ES.
4. With D and S as centres, draw arcs of radii 6 cm and 5 cm respectively. Let them cut at K. Join DK and KS.
5. DESK is the required trapezium.
Calculation of area:
Area of the trapezium DESK = \(\frac { 1 }{ 2 } \) x h x (a + b) sq. units
= \(\frac { 1 }{ 2 } \) x 5.5 x (8 + 5) = 35.75 sq. cm

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

Given:
BO = 7cm, OA = 6cm, BA = 10cm,
TA = 6 cm and \(\overset { - }{ BO } \) || \(\overset { - }{ TA } \)


​​​​​​Steps:
1. Draw a line segment BO = 7 cm.
2. With B and O as centres, draw arcs of radii 10cm and 6cm respectively and let them cut at A.
3. Join BA and OA.
4. Draw AX parallel to BO
5. With A as centre, draw an arc of radius 6cm cutting AX at T.
6. Join BT. BOAT is the required trapezium.
Calculation of area:
Area of the trapezium BOAT = \(\frac { 1 }{ 2 } \) x h x (a+b) sq units
= \(\frac { 1 }{ 2 } \) x 5.9 x (7+6) = 38.35 sq. cm