1. AR = 6 cm; RT = 5 cm and 
Rough diagram


Steps:
1. Diaw a liqe segment AR = 6 cm.
2. Make an angle o at R on AR
3. With R as centre, draw an arc of radius 5 cm cutting RX at T.
4. With A and T as centres draw arcs of radii 5 cm and 5 cm respectively. Let them cut at S.
5. ]oin AS and TS
6. ARTS is the required parallelogram
Calculation of area:
Area of the Parallelogram ARTS
= bh sq. units
= 6 x 4.2 = 28.2 sq. cm
2. Given: CA = 6 cm; AP = 8 cm and CP = 5.5 cm.
Rough diagram

Steps:
1. Draw a line segment CA = 6 cm.
2. With C and A as centres, draw an arc of radius  5.5 cm and 8 cm respectively. Let them cut at p.
3. ]oin CP and AP.
4. With A and P as centres draw an arcs of radius 5.5 cm and 6 cm respectively. Let them cut at M.
5. Join AM and PM.
6. CAMP is the required parallelogram.
Calculation of area:
Area of the Parallelogram CAMP '
= bh sq. units
= 6 x 5.4 = 32.4 sq. cm
3. Given: ER = 10cm ; AN = 7 cm and o
Rough diagram

Steps:
1. Draw a line segment ER = 10 cm.
2. Mark O the mid point of an angle 4. With O as centre and 3.5 cm arcs on XY on either side of .gn. tet the arcs cut OX at N and OY at A.
5. EARN is the required parallelogram.
Calculation of area:
Area of the Parallelogram EARN
= bh sq. units
= 7 x 4.8 = 33.6 sq. cm
4. Given: GA,=7.5 cm; GI = 9 cm and o
Rough diagram

Steps:
1. Draw a line segment GA = 7.5 cm.
2. Make an angle IGAI = 100^o at A on GA.
3. With G as centre, draw an arc of radius 9 cm cutting AX at I and join GI.
4. With G as centre drawer an arc of radius equal to the length of AI.
5. With I as centre, draw an arc of radius 7.5 cm. Let both arcs cut at N.
6. Join IN and GN.
7. GAIN is the required parallelogram
Calculation of area:
Area of the Parallelogram GAIN
= bh sq. units .
= 7.5 x 3.7 = 27.75 sq. cm

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
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 6 \times(4+4) \)
\(=1 / 2 \times 6 \times 8=24 \text { sq. } \mathrm{cm} . \)

24 cm

\(\triangle\) AFI is a right angled triangle
(AI)² = (AF)² + (FI)²
25² = (AF)² + 15²
625 = (AF)² + 225
(AF)² = 625 - 225
= 400
\(\mathrm{AF}=\sqrt{400}\)
AF = 20ft
\(\triangle\) FRI is a right angled triangle.
(FI)² + (FR)² = (RI)²
(FR)² + 15² = 17²
(FR)² + 225 = 289
(FR)² = 289 - 225
= 64
\(A R=\sqrt{64}\)
= 8ft
AR = AF+FR
= 20 + 8
= 28 ft
AR = 28 ft

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

4. 25 ft

From the figure
x² = 20² + 15²
= 440 +225
= 625
\(\therefore x=\sqrt{625}\)
x = 25
The length of the support cable is 25 ft.

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.

25cm, 16cm, 9cm

(i) 12 cm

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

30 inches

25, 24

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.

(i) From the figure,
x² = 9² + 40²
= 81 + 1600
= 1681
\(x=\sqrt{1681}\)
x = 41
(ii) From the figure
34² = 30² + y²
1156 = 900 + y²
y = 1156 - 900
y² = 256
\(y=\sqrt{256}\)
y = 16
(iii) From the figure
39² = 36² + z²
1521 = 1296 + Z²
z² = 1521- 1296
z² = 225
\(z=\sqrt{225}\)
z = 15

(i) 8,15,17
8² + 152 = 64+ 225
= 289
= 17²
i.e., 8² + 15² = 17² 
8, 15, 17 are the sides of a right-angled triangle
(ii) 12,13, t5
12² + 13² = 144+ 164
= 313 
15² = 225
\(\text { i.e., } 12^{2}+13^{2} \neq 15^{2}\)
12, 13, 15 are not the sides of a right  angled triangle
(iii) 30,40,50 
30² + 40² = 900 + 1600
= 2500 ;
= 50²
i.e,, 30² + 40² = 502
30 ,40,50 are the sides of a right angled triangle.
(iv) 9, 40,41
9² + 40² = 81 + 1600
= 1681 
= 412 
i.e., 9² + 40² = 41²
9, 40, 41 are the sides of a right angled triangle
(v) 24,45,51
24² + 45² = 576 + 2025
= 2601 = 51²
24,45,51 are the sides of a right angled triangle