(01) A circle is inscribed in a square, An equilateral triangle of side 4 \sqrt{3} cm is inscribed in that circle. The length of the diagonal of the square (in cm) is
A) 4 \sqrt{2}\\ \\ B) 8 \\ \\ C) 8 \sqrt{2} \\ \\ D) 16 \\\ \\ Read Solution
Side of equilateral triangle=4 \sqrt{3} \ cm
Circumradius of triangle= \frac{side}{\sqrt{3}} \\ \\ = \frac{4\sqrt{3}}{\sqrt{3}} =4\ cm
We know that,
side of square= 2 * circumradius
=> 2 * 4 = 8
To find diagonal of square
Applying Pythagoras theorem,
Diagonal^{2}= 8^{2} +8^{2} \\\ \\
=> 64+64 \ = 128 \ = 8\sqrt{2} \ cm
Hence option, C) is correct
(02) The area of circle whose radius is 6 cm is trisected by two concentric circles. The radius of the smallest circle is
A) 2 \sqrt{3} \ cm \\\ \\
B) 2 \sqrt{6} \ cm \\\ \\
C) 2 \ cm \\\ \\
D) 3 \ cm \\\ \\
Read Solution
radius of circle=6 cm
Area of C circle= \pi r^{2} \\\ \\ ⟹ {\pi ( 6)^{2}} \\\ \\ ⟹ 36 \pi \ cm^{2}
Since given circle is trisected = divided into 3 equal parts
Area of each part= \frac{36\pi }{3}=12π cm²
area of each circle= \pi r^{2} \\\ \\ 12 \pi =\pi r^{2} \\\ \\ r ^{2}=12 \\\ \\ r=2 \sqrt{3} \ cm
Hence option A) is correct.
(03) A circle is inscribed in an equilateral triangle of side 8 cm. The area of the portion between the triangle and the circle is
A) 11 \ cm^{2} \\\ \\
B) 10.95 \ cm^{2} \\\ \\
C) 10 \ cm^{2} \\\ \\
D) 10.50 \ cm^{2} \\\ \\
Read Solution
given side of triangle=8 cm
inradius of circle= \frac{side}{2\sqrt{3}}\\\ \\ =\frac{8}{2\sqrt{3}} \\\ \\ =\frac{4}{\sqrt{3}}
area of circle= \pi r^{2} \\\ \\ = {\pi \left(\frac{4}{\sqrt{3}}\right)^{2}} \\\ \\ = \frac{22}{7} \times \frac{16}{3}
required area=area of equilateral triangle – area of circle
= \frac{\sqrt{3}}{4}( 8)^{2}- \frac{22}{7} \times \frac{16}{3} \\\ \\ = \frac{\sqrt{3}}{4} \times \ 64 - \frac{22}{7} \times \frac{16}{3} \\\ \\ = \ 16 \sqrt{3} -16.76 \\\ \\ =16 \times 1.73-16.76\ \\\ \\ = 27.68 - 16.76 =10.92 cm ^{2}\\\ \\ Hence 10.92\approx 10.95\Thus option B) is correct
(04) The circumference of a circle is 11 cm and the angle of a sector of the circle is 60 \degree . The area of the sector is
A) 1 \frac{29}{48} \ \ cm^{2} \\\ \\ B) 2 \frac{29}{48} \ \ cm^{2} \\\ \\ C) 1 \frac{27}{48} \ \ cm^{2} \\\ \\ D) 2 \frac{27}{48} \ \ cm^{2} \\\ \\ Read Solution
Circumference=11 cm
2 \pi r=11 \\\ \\ 2 \frac{22}{7} r=11 \\\ \\ r= \frac{7}{4} \ cm \\\ \\area of sector= \frac{\theta }{360\degree } \times \pi r^{2} \\\ \\ = \frac{60}{360} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \\\ \\ = \frac{77}{48} =1\frac{29}{48} \ cm^{2}
Hence option A) is correct.
(05) The sides of a triangle are 6 cm, 8 cm, and 10 cm. The area of the greatest square that can be inscribed in it, is
A) 18 sq. cm
B) 15 sq. cm
C) \frac{2304}{49} sq. cm
D) \frac{576}{49} sq. cm
Solution
Since the given sides of triangle denotes that it is right angled triangle.perpendicular at AC.
Hence, option D) is correct.
(06) The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circumcircle and the incircle of the triangle is
A) 50 \frac{1}{7}\ cm ^{2} \\\ \\ B) 50 \frac{2}{7}\ cm ^{2} \\\ \\ C) 75\frac{1}{7}\ cm ^{2} \\\ \\ D) 75\frac{2}{7}\ cm ^{2}\\ \\ Read Solution
Hence, option B) is correct.
(07) One acute angle of a right angled triangle is double the other.If the length of its hypotenuse is 10 cm, then its area is ?
A) \frac{25}{2}\sqrt{3} cm^{2} \\\ \\ B) 25\ cm ^{2} \\\ \\ C)25 \sqrt{3}\ cm ^{2}\\\ \\ D) \frac{75}{2} cm^{2} \\\ \\ Read Solution
hence option A) is correct.
(08) An equilateral triangle of side 6 cm has its corners cut off to form a regular hexagon. Area of this regular hexagon will be
A) 3 \sqrt{3} \\\ \\ B) 3 \sqrt{6} \\\ \\ C) 6 \sqrt{3} \\\ \\ D) \frac{5\sqrt{3}}{2}\\ \\ Read Solution
Solution
Given each sides of equilateral triangle ABC=6 cm
AD = DI = IB = 2 cm
Similarly,
AE = EF = FC = 2 cm
BH = GH = HC = 2 cm
Option (C) is the right answer
(09) The length of each side of an equilateral triangle is 14 \sqrt{3} cm. The area of the incircle (in cm) is
A) 450
B) 308
C) 154
D) 77
Hence option C) is correct.