A square ABCD is inscribed in a circle of unit radius. Semicircles are described on each side as a diameter. The area of the region bounded by the four semicircles and the circle is

A square ABCD is inscribed in a circle of unit radius. Semicircles are described on each side as a diameter. The area of the region bounded by the four semicircles and the circle is
1). 1 sq. unit
2). 2 sq. unit
3). 1.5 sq. unit
4). 2.5 sq. unit

SSC CGL Books

---

Join Telegram Group

Answer This Question
Name:
Email:
Answer :
Sum of (2+1)
Submit:

Other Questions

1. In $\triangle ABC$, D and E are points on AB and AC respectively such that $DE\parallel BC$ and DE divides the $\triangle ABC$ into two parts of equal areas. Then ratio of AD and BD is

2. Direction∶ In the following question, two equations numbered I and II are given. You have to solve both the equation and given answer. I. 3x2 - 11x - 4 = 0 II. 3y2 + 8y - 16 = 0

3. If $\frac{1}{a}(a^{2}+1)$=3, then the value of $\frac{a^{6}+1}{a^{3}}$ is

4. In a quadrilateral ABCD, the bisectors of $\angle A$ and $\angle B$ meet at O. If $\angle C$ = $70^{0}$ and $\angle D = 130^{0}$ ,then measure of $\angle AOB$ is

5. What approximate value should come in place of the question mark (?) in the following questions ? (Note : You are not expected to calculate the exact value.)

6. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer: I. 2x2 – 22x + 36 = 0 II. y2 – 8y + 15 = 0

7. In triangle PQR , points A,B and C are taken on PQ ,PR and QR resoectively such that QC = AC and CR = CB .IF $\angle QPR$ = $40^{0}$ ,then $\angle ACB$ is equal to :

8. If r = 21 and s = 18, then the value of \(\frac{{{r^3} - {s^3}}}{{{r^2}\; + \;{s^2}\; + \;rs}}\) is

9. In $\triangle ABC $,$\angle BAC$ = $90^{0}$ and $AD\bot BC$. If BD = 3 cm and CD = 4 cm, then the length of AD is

10. Direction: In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer. I. x2 – 7x + 6 = 0 II. y2 + 5y – 6 = 0