Here, Height (AD) = 3 cm and Area of LlABC = 30 cm2
But Area of a triangle =\(\frac{1}{2}\) \(\times\)b\(\times\)h
\(\therefore\) \(\frac{1}{2}\) \(\times\) b \(\times\)h=30 or \(\frac{1}{2}\)\(\times\)BC\(\times\)3=30
or BC =\(\frac{30\times2}{3}\)=20 cm
Therefore, BC = 20 cm.

Since, a and b are two different numbers.
Let a = 15 and b = 10.
\(\therefore\) \(\frac{a-b}{a+b}\)=\(\frac{15-10}{15+10}\)=\(\frac{5}{25}=\frac{1}{5}\)
and \(\frac{a+b}{a-b}\)=\(\frac{15+10}{15-10}=\frac{25}{5}\)=5
So, (a + b) always greater than (a - b) when denominator is less number is greater,
then \(\left(\frac{a+b}{a-b}\right)>\left(\frac{a-b}{a+b}\right)\)

Yes, RS II PQ II TU. Since, two lines parallel to a given line are also parallel to each other.

Yes,we can draw ΔABC.
Here, the side AC, ㄥA and ㄥB of ΔABC are given. But to draw the triangle, we required ㄥC.
In ΔABC, by angle sum property, we have
ㄥA + ㄥB + ㄥC = 180° ⇒ 60° + 50° + ㄥC = 180°
⇒ 110° + ㄥC = 180°  ⇒ ㄥC = 180° -110° = 70°
Now, we have side AC, m ⇒ A and mㄥC.
So, we can draw MBC, by ASA criterion.

With respect to the given line(s) of symmetry, the other hole(s) are marked in the figures given below:

60 m²

114

Area of circle = \(\pi\)r²
\(\Rightarrow\) \(\pi\)r² = 38.5 cm²
\(\Rightarrow\)\(\frac { 22 }{ 7 } \) r² = 38.5
\(\Rightarrow\) r² = \(\frac { 38.5\times 7 }{ 22 } \)
\(\Rightarrow\) r² = 12.25
\(\Rightarrow\) r = 3.5cm
Circumference = 2\(\pi \)r
= 2 x \(\frac { 22 }{ 7 } \)x 3.5
= 22 cm

All lines of symmetry for the given figure will be

Yes,we can slightly modify the construction of a line parallel to a given line through a point not on the line to use the idea of equal corresponding angles instead of equal alternate angles. For modifying this construction, we change only Step IV, VI and VII and new steps are given below:
Step IV With A as centre and radius same as in Step III draw an arc EF to cut BA (extended) at P.
Step V Place the pointed tip of the compasses at C and adjust the. opening so that the pencil tip is at D.
Step VI With the same opening as in Step V and with Pas centre, draw an arc which cut the arc EF at Q.
Step VII Join AQ to draw a line m, which is parallel to the given line I.
Here, \(\angle ABC,\angle PAQ\) are corresponding angles,

Therefore, I II m.

There are two examples as given below:
Example 1. Let the length of a rectangle = 5 cm and breadth of a rectangle = 3 cm
\(\therefore\) Area of a rectangle = Length X Breadth
                                = 5 X 3=15 cm²
and perimeter of a rectangle = 2 (Length + Breadth)
                                         = 2 ( 5 + 3) = 2 X 8 = 16 cm
Again, let the length of a rectangle = 6 cm and breadth of a rectangle = 5 cm
\(\therefore\) Area of a rectangle = Length X Breadth = 6 X 5 = 30 cm² 
and perimeter of a rectangle = 2 (Length + Breadth)
                                         = 2 (6 + 5) = 2 X 11 = 22 cm
Thus, for both rectangles, area increases as the perimeter Increases.
Example 2. Let the side of a square = 4 ern
\(\therefore\) Perimeter of square = 4 X Side = 4 x 4 = 16 cm
and area of square = (Side X Side)
                           = 4 X 4 = 16 sq ern
Again, let the side of a square = 5 cm
\(\therefore\) Perimeter of square = 4 X Side = 4 x 5 = 20 cm
and area of square = Side X Side
                           = 5 x 5 = 25 sq cm
Thus, for both squares, area increases as the perimeter Increases.

ABCD is the rhombus such that its diagonals AC and BD intersect at O. Here BD = 12 cm.

Suppose AC = x cm.
Since, the area of a rhombus =\(\frac{1}{2}\)\(\times\)[Product of 2 its diagonals]
\(\therefore\)Area of the rhombus ABCD =\(\frac{1}{2}\)\(\times\) 12 \(\times\) x cm²
2 But the area of the rhombus ABCD = 96 cm²
\(\Rightarrow\)\(\frac{1}{2}\)\(\times\)12 \(\times\) x = 96
\(\Rightarrow\)x=\(\frac{96\times2}{12}\)cm \(\Rightarrow\)x=8\(\times\)2cm
\(\Rightarrow\) x=16 cm
Since, the diagonals of a rhombus, bisect each other at right angles.
\(\therefore\)\(\angle \)COD = 90°, OC =\(\frac{1}{2}\)\(\times\) 16 cm=8 cm
and Od=\(\frac{1}{2}\)\(\times\) 12 cm =6 cm
Now, in right \(\Delta\)COD, we have
\(\Rightarrow\)OD² + OC² = CD² \(\Rightarrow\) 6² + 8² = CD²
\(\Rightarrow\)36 + 64 = CD² \(\Rightarrow\)100 = CD²
\(\Rightarrow\) 10² = CD² \(\Rightarrow\) CD = 10 cm.
Since, all the sides of rhombus are equal and perimeter of a rhombus = 4 x Side.
\(\therefore\) Perimeter ofthe rhombus = 4 x 10 em = 40 cm

Let ABCD be the rectangular park of sides 40 m and 25 m, and the shaded region represents the path 2.5 m wide.
Now, PQ = (40 + 2.5 + 2.5) m = 45 m
PS = (25 + 2.5 + 2.5) m = 30 m
\(\therefore\)Area of rectangle ABCD = I x b = 40 m x 25 m
= 1000 m2
Area of rectangle PQRS = 45 m x 30 m = 1350 m²
So, Area of the path = [Area of rectangle PQRS] - [Area of rectangle ABCD]
= 1350 m² - 1000 m² = 350 m²

Sequence is \(\frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -5 }{ 6 } \\ \)
L.C.M.of 5, 10 and 6=30
\(\Rightarrow \frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -5 }{ 6 } \)
\(\Rightarrow -\frac { 3\times 6 }{ 5\times 6 } ,\frac { 7\times 3 }{ -10\times 3 } ,-\frac { 5\times 5 }{ 6\times 5 } \)
\(\Rightarrow -\frac { 18 }{ 30 } ,\frac { 21 }{ 30 } ,\frac { 25 }{ 30 } \)
Since \(-\frac { 25 }{ 30 } <\frac { 7 }{ -10 } <\frac { -3 }{ 5 } \)
Hence sequence in ascending order is
\(\frac { -5 }{ 6 } <\frac { 7 }{ -10 } <\frac { -3 }{ 5 } \).