Question

The diameter of a circle whose area is equal to the sum of areas of two circles of radii 24 cm and 7 cm respectively is ​

Answer 1

Answer:

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Answer 2

Answer:

r1 = 24cm

area1 = π $r^{2}$

       $=\frac{22}{7} \times {24}^{2}\\\\=\frac{22}{7} \times 576 \\\\=\frac{22\times576}{7} \\\\=\frac{12672}{7}cm^{2}$

r2 = 7cm

area2 = π $r^{2}$

       $=\frac{22}{7} \times {7}^{2}\\\\=\frac{22}{7} \times 49 \\\\=\frac{22\times7(7)}{7} \\\\=22 \times 7 \\\\=154 cm^{2}$

area  of required circle = area of first circle with radius 24cm +  area of second circle with radius 7cm

$= \frac{12672}{7} +154\\\\=\frac{12672}{7} +\frac{154(7)}{1(7)}\\\\=\frac{12672}{7} +\frac{1078}{7}\\\\=\frac{12672 + 1078}{7}\\\\=\frac{13750}{7}cm^{2}$

$area = \frac{13750}{7} cm^{2} \\\\ pi \times r^{2} = \frac{13750}{7}\\\\r^{2} =\frac{13750}{7} \times \frac{7}{22} \\\\r^{2} = \frac{1250(11)}{2(11)} \\\\r^{2} =  \frac{625(2)}{2} \\\\r^{2} =625\\\\r^{2} = (25)^{2}\\\\r = 25cm$                                        

radius = 25 cm

∴ diameter = 25 × 2

∴ diameter = 50cm

Step-by-step explanation: