[2 Mark Question Answer]

Question 12 Marks

$AC$ and $BD$ are two perpendicular diameters of a circle with centre $O$. If $AC = 16 \ cm$, calculate the area and perimeter of the shaded part. (Take $\pi = 3.14).$

Answer

Given radius of circle is 8 cm.
Area of shaded part = $\frac{1}{2} \pi r^2=\frac{1}{2} \times 3.14 \times 8^2$
Area of shaded part $= 100.48 cm^2$
The perimeter of shaded part $= \pi r x 4r$
The perimeter of shaded part $= 3.14 x 8 + 4 x 8$
The perimeter of shaded part $= 25.12 + 32$
The perimeter of shaded part $= 57.12 cm.$

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Question 22 Marks

The diagram represents the area swept by wiper of a car. With the dimension given in figure, calculate the shaded swept by the wiper.

Answer

Area of shaded portion $=\frac{\theta}{360} \times \pi\left(21^2-7^2\right) cm ^2$
Area of shaded portion $={\frac{30^{\circ}}{360}}^{\circ} \times \frac{22}{7} \times\left(21^2-7^2\right) cm ^2$
Area of shaded portion $=\frac{22}{12 \times 7} \times 392 cm ^2$
Area of shaded portion $= 102.67 cm^2$

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Question 32 Marks

$AC$ and $BD$ are two perpendicular diameter of a circle $ABCD.$ Given that the area of shaded portion is $308\ cm^2$ calculate:
(i) The length of $AC$ and
(ii) The circumference of circle

Answer

(i) Let r be the radius of the circle.
Then $\frac{1}{2} \pi r^2=308$
$r^2=308 \times 2 \times \frac{7}{22}=14^2$
$ r=14\ cm $
Now, $AC = 2r = 28\ cm$
(ii) Circumference $= 2\pi r$
Circumference $= \frac{22}{7} \times 28$
Circumference $= 88\ cm$

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Question 42 Marks

In the figure alongside, OAB is a quadrant of a circle. The radius $OA = 3.5 cm$ and $OD = 2 cm$. Calculate the area of the shaded $22$ portions.

Answer

Area of the shaded portion = Area of Quadrant - Area of Δ OAD
Area of the shaded portion $=\frac{1}{4} \pi(3.5)^2-\frac{1}{2} \times 3.5 \times 2$
Area of the shaded portion = $\frac{22}{7} \times \frac{1}{4} \times 3.5 \times 3.5-3.5$
Area of the shaded portion = $6.125 cm^2$

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Question 52 Marks

Four equal circles, each of radius $5 cm$, touch each other as shown in the figure. Find the area included between them. (Take $π= 3.14$)

Answer

Required area = Area of square ABCD - Area of 4 quadrants
Required area = ( 10 cm x 10 cm ) - $\frac{1}{4} \pi r^2$
Required area $=100-=\frac{1}{4} \times \frac{22}{7} \times 5 \times 5$
Required area $= 100 - 78.57 cm^2$
Required area $= 21.43 cm^2$

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Question 62 Marks

Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.

Answer

Steps of construction:
(i) Draw AB = 6 cm and cut arc of 4 cm from A and B these arcs intersect at C join AC and BC.
(ii) Draw the bisector (internal) of ∠ C and mark the point P, taking CP = 5 cm
(iii) Draw a line EF parallel to AB at a distance of 5 cm.
(iv) Take P as centre cut two points on line EF as PQ and PR are each equal to 5 cm.

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Question 72 Marks

Take a point O on the plane at the paper. With O as center draw a circle of radius 3 cm. Take a point P on this circle and draw a tangent at P.

Answer

Steps of construction:
(i) Take a point O on the plane at the paper and draw a circle at radius 3 cm.
(ii) Take a point P on the circle and join OP.
(iii) Construction ∠ OPT = 90°(iv) Produce TP to T' obtain the required tangent TPT'.

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Mathematics [2 Mark Question Answer]

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