Question : A right-angled isosceles triangle is inscribed in a semi-circle of radius 7 cm. The area enclosed by the semi-circle but exterior to the triangle is:
Option 1: 14 cm2
Option 2: 28 cm2
Option 3: 44 cm2
Option 4: 68 cm2
Correct Answer: 28 cm2
Solution : Area of the semi-circle = $\frac{1}{2} \pi r^2$ = $\frac{1}{2}\times \frac{22}{7}\times 7^2$ = 77 cm2 The altitude of the isosceles triangle OC is $r$ = 7 cm. The base of the isosceles triangle AB is the diameter = $2 r$ = 14 cm. Area of the Isosceles triangle ABC = $\frac{1}{2}$ × Base × Height = $\frac{1}{2}\times 14\times 7$ = 49 cm2 So, the area enclosed by the semi-circle but exterior to the triangle = 77 - 49 = 28 cm2 Hence, the correct answer is 28 cm2.
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Question : The area of a circle is 1386 cm2. What is the radius of the circle? [Use $\pi= \frac{22}{7}$]
Option 1: 7 cm
Option 2: 14 cm
Option 3: 18 cm
Option 4: 21 cm
Question : ABC is a right-angled triangle with AB = 6 cm and BC = 8 cm. A circle with centre O has been inscribed inside $\triangle ABC$. The radius of the circle is:
Option 1: 1 cm
Option 2: 2 cm
Option 3: 3 cm
Option 4: 4 cm
Question : What is the total surface area of a solid right circular cylinder of radius 7 cm and height 8 cm?$(\pi=\frac{22}{7})$
Option 1: 560 cm2
Option 2: 660 cm2
Option 3: 850 cm2
Option 4: 760 cm2
Question : A string of length 24 cm is bent first into a square and then into a right-angled by keeping one side of the square fixed as its base. Then the area of a triangle equals to:
Option 1: 24 cm2
Option 2: 60 cm2
Option 3: 40 cm2
Option 4: 28 cm2
Question : The circumference of a circle exceeds its diameter by 60 cm. The area of the circle is: (Take $\pi=\frac{22}{7}$ )
Option 1: 1078 cm2
Option 2: 616 cm2
Option 3: 536 cm2
Option 4: 346.5 cm2
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