Given:
AI = 6 cm,
IM = 5 cm,
AM = 9 cm,
MS = 6.5 cm.
Rough Diagram

Steps:
1. Draw aline segment AI = 6 cm.
2. With A and I as centres draw arcs of radius
9 cm and 5 cm respectively and let them cut at M.
3. Join AM and IM.
4. Draw MX parallel to AI.
5. With M as centre, draw an arc of radius6.5 cm cutting MX at S.
6. |oin AS. AIMS is the required trapezium
Calculation of Area:
Area of the trapezium AIMS
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 4.8 \times(6+6.5) \)
\(=1 / 2 \times 4.8 \times 12.5=30 \mathrm{sq} . \mathrm{cm} . \)
2. Given:
CU = 7cm; o
CE = 6cm; TE = 5cm.
Rough Diagram 

Steps: 
1. Draw a line segment CU = 7cm
2. Constract an angle o at c
3. With C as centre, draw an arc of radius 6 cm cutting CX at E.
4. Draw EY parallel to CU.
5. With E as centre, draw an arc of radius 5 cm cutting EY at T.
6. Join TLI CUTE is the required trapezium.
Calculation of Area:
Area of the trapezium CUTE
\(=1 / 2 \times \mathrm{h} \times(\mathrm{a}+\mathrm{b}) \)
\(=1 / 2 \times 6 \times(5+7) \)
\(=1 / 2 \times 6 \times 12=36 \text { sq. } \mathrm{cm} . \)
3. Given:
AR = 7 cm; RM = 6.5 cm
ZRAY = 100^o; ZARM = 60^o
Rough Diagram
​​​​​
Steps: -
1. Draw a line segment AR 7 cm.
2. Construct an angle o at R. S.
3. With R as centre draw an arc of radius 6.5 cm cutting RX at M.
4. Draw MZ parallel to AR.
5. Construct an angle ZRAY = 100^o A cutting MZ at Y.
6. ARMY is the required trapezium
Calculation of Area:
Area of the trapezium ARMY
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 5.6 \times(7+7) \)
\(=1 / 2 \times 5.6 \times 14=39.2 \text { sq. cm } \)
4. Given:
CI = 7 cm ; IT = 5.5 cm
TY = 4 cm ; YC = 6cm.
Rough Diagram

Steps:
1..Draw a line segment CI = 7 cm.
2. Mark the point A on CI such that CA = 4 cm.
3. With A and I as centres, draw arcs of radii 6 cm and S.S cm. Let them cut at T. join AT and IT.
4. With C and T as centres, draw arcs of radii 6 cm and 4 cm respectively. Let them cut at Y. Join TY and CY.
5. CITY is the required trapezium.
Calculation of Area:
Area of the trapezium CITY
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 6 \times(4+4) \)
\(=1 / 2 \times 6 \times 8=24 \text { sq. } \mathrm{cm} . \)

Let ABCD be a rhombus. The diagonals of the rhombus meet at O.
Since \(\triangle\) AOB is a right angled triangle
AB² = AO² + OB² 
= 82 + 62
= 64 + 36
= 100
\(\mathrm{AB}=\sqrt{100}=10\)
​​​​​​The perimeter of the rhombus = 4 x AB
= 4 x 10
= 40cm

By using Pythagoras theorem,
AP² = AK² + PK² = AK² + AK² (since it is an isosceles A) = 2AK²
Hence it is proved 

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

100, 48