Practice Set 7.4 Mensuration Class 10th Mathematics Part 2 MHB Solution

Practice Set 7.4
  1. In figure 7.43, A is the centre of the circle. ∠ ABC = 45° and AC = 7√2 cm. Find the…
  2. In the figure 7.44, O is the centre of the circle. M (arc PQR) = 60° OP = 10 cm. Find…
  3. In the figure 7.45, if A is the centre of the circle. ∠ PAR = 30°, AP = 7.5, find the…
  4. In the figure 7.46, if O is the centre of the circle, PQ is a chord. ∠ POQ = 90°, area…
  5. A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of…
Practice Set 7.4
Question 1.

In figure 7.43, A is the centre of the circle. ∠ ABC = 45° and AC = 7√2 cm. Find the area of segment BXC.



Answer:

From the property, we know that, If two sides of a triangle are equal then their corresponding angles are also equal.


So, as AB = AC,


⇒ ∠ ABC = ∠ ACB = 45°


As the sum of angles of a triangle is equal to 180°

⇒∠ ABC + ∠ ACB + ∠ BAC = 180°

⇒45° + 45° + ∠ BAC = 180°

⇒90° + ∠ BAC = 180°

⇒ ∠ BAC = 90°




⇒ AT = 49 sq.cm


Area of the sector = one-fourth of a circle




⇒ AS = 77 sq.cm


Area of the shaded region, AR = AS - AT


⇒ AR = 77 – 49


⇒ AR = 28 sq.cm


∴ The area of the shaded region is 28 sq. cm.


Question 2.

In the figure 7.44, O is the centre of the circle. M (arc PQR) = 60° OP = 10 cm.

Find the area of the shaded region. (π = 3.14, √3 = 1.73)



Answer:

Since the angle subtended at centre is 60°

And by the property, if two sides of a triangle are equal then their corresponding angles are also equal.


⇒ ∠ ORP = ∠ OPR


As the sum of all internal angles of a triangle is equal to 180°


⇒ ∠ ORP = ∠ OPR = 60°


⇒ Δ OPR is an equilateral triangle.




⇒ AT = 43.25 sq. cm


Area Of Sector (O-PQR), AS is given as:




⇒ AS = 52.33 sq.cm


Area of shaded region, AR = As – AT


⇒ AR = 52.33 – 43.25


⇒ AR = 9.08 sq.cm


∴ Area of shaded region is 9.08 sq.cm



Question 3.

In the figure 7.45, if A is the centre of the circle. ∠ PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)



Answer:

Radius of circle, r = 7.5 cm


∠ PAR = θ = 30°


Area(A – PQR), AS:




⇒ AS = 14.71 sq.cm


Also,





⇒ AT = 14.06 sq. cm


Area of segment PQR, AR = AS - AT


⇒ AR = 14.71 – 14.06 = 0.6562 sq.cm


∴ Area of shaded region is 0.6562 sq.cm



Question 4.

In the figure 7.46, if O is the centre of the circle, PQ is a chord. ∠ POQ = 90°, area of shaded region is 114 cm2, find the radius of the circle. (π = 3.14)



Answer:

Area Of shaded region, AR = 114 sq.cm


Area of sector (O-PRQ),AS = one-fourth of area of circle




Area of Shaded Region, AR = AS – AT





⇒ r = 20 cm


∴ Radius of the circle is 20 cm



Question 5.

A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14)


Answer:


Radius of circle, r = 15cm


Central angle, θ = 60°


Since the angle subtended at centre is 60°


And by the property, if two sides of a triangle are equal then their corresponding angles are also equal.


⇒ ∠ OQP = ∠ OPQ


As the sum of all internal angles of a triangle is equal to 180°


⇒ ∠ OQP = ∠ OPQ = 60°


⇒ Δ OPQ is an equilateral triangle.




AT = 97.32 sq.cm




⇒ AR = 117.75 – 97.32


⇒ AR = 20.43 sq.cm


Now,




⇒ AS = 706.5 – 20.43


⇒ AS = 686.07 sq.cm


∴ The area of minor segment and major segment is 20.43 sq.cm and 686.07 sq.cm respectively

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