Question

A) In the given figure (1), 'O' is the centre and PT is the diameter of the semicircle PQRST.
If PQ=QR and PTR=50° then find:
1) ZRQT
II) III) _POQ
IV) Prove that OQ II TR

B) I) SPTO is a parallelogram SP is produced to point 'E' so that PE=SP. Prove that EO bisects
PT. (figure for this question is not given)

II) In the given figure - (ii) , XYZW is a trapezium in which XY II WZ.P and Q are the midpoints
of XW and YZ respectively. WQ and XY when produced meet at point E prove that;
1] WQ=QE
2] PR II XY

Answer 1

Answer:

Solutions4u

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Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Areas Related to Circles Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.

If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, calculate the diameter of the larger circle (in cm). (2012D)

Solution:

πR2 = πr21 + πr22

πR2 = π(r21 + πr22) [ r1 = \frac{10}{2} = 5cm, r2 = \frac{24}{2} = 12 cm]

R2 = 52 + 122 = 25 + 144

R2 = 169 = 13 cm

∴ Diameter = 2(13) = 26 cm

Question 2.

The circumference of a circle is 22 cm. Calculate the area of its quadrant (in cm2). (2012OD)

Solution:

Circumference of a circle = 22 cm 2πr = 22 cm

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles 1

Question 3.

If the difference between the circumference and the radius of a circle is 37 cm, then using π = \frac{22}{7}, calculate the circumference (in cm) of the circle. (2013D)

Solution:

2πr – r = 37 ⇒ r(2π – 1) = 37

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles 2

Question 4.

If he is taken as \frac{22}{7}, calculate the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution. (2013OD)

Solution:

Radius (r) = \frac{35}{2}

Required distance = Perimeter = 2πr

= 2 × \frac{22}{7} × \frac{35}{7} cm = 110 cm or 1.1 m

Question 5.

In Figure , find the area of the shaded region. (2011OD)

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles 3

Solution:

Area of shaded region

Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles 4