616 cm²

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.

(i) Here, Base (b) = 4 cm and Height (h) = 3 cm
\(\therefore\) Area of triangle ABC=\(\frac{1}{2}\) \(\times\)b\(\times\)h=\(\frac{1}{2}\)\(\times\)4\(\times\)3 cm² = 6 cm²

Area of \(\Delta\) =  \(\frac { 1 }{ 2 } \) x base x height
= \(\frac { 1 }{ 2 } \)x 25 x 14
= 25 x 7 = 175 cm²

Area of parallelogram = base x height
In the question,
AB = base = 8 cm.
and height = 5.2 cm.
\(\therefore\)       Area = 8 x 5.2
= 41.6cm²

64 m²

Rs 720.

Given, radius = 49 cm
\(\therefore\) Area of a circle = \(\pi\)r² = \(\pi\) x (49)²
                          = \(\frac{22}{7}\) x 49 x 49 = 22 x 7 x 49
                         = 154 x 49 = 7546 cm².

Given, radius = 21 cm
\(\therefore\) Area of a circle = \(\pi\)r² = \(\pi\)(21)² = \(\frac{22}{7}\) x 21 x 21
                          = 22 x 3 x 21 = 1386 cm².

770 cm²

In the given parallelogram ABCD, DC = Base = 8 cm and AE = Height = 6cm
\(\therefore\) Area of the parallelogram = DC x AE = 8 X 6 cm² = 48 cm².

99 m²

2500 m.

Perimeter of the rectangle
= 2\(\times\)(l+b)
⇒ 130 = 2\(\times\)(l + 30)
⇒ l+30 = \(\frac { 130 }{ 2 } \)
⇒ l+30= 65
⇒ l = 65 - 30 = 35 cm
Hence, the length of the rectangle is 35 cm.
Area of the rectangle
= l\(\times\)b
= 35\(\times\)30 = 1050 cm²
Hence, the area of the rectangle is 1050 cm²

We need to find the area of rectangular swimming pool to cover it with plastic sheer.