5 Marks Question - Maths STD 10 Questions - Vidyadip

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29 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

The area of an equilateral triangle is $49\sqrt{3}\text{cm}^2.$ taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. $\big[\text{Take }\sqrt{3}=1.73\big]$

Answer

Area of equilateral triangle $\text{ABC}=49\sqrt{3}\text{cm}^2$

Let a be its side $\therefore\frac{\sqrt{3}}{4}\text{a}^2=49\sqrt{3}$ Or $\text{a}^2=49\times4$ $\therefore\text{a}=7\times2$ $\Rightarrow\text{a}=14\text{cm}$ Area of sector $\text{BDF}=\pi\text{r}^2\times\frac{\theta}{360^\circ}$ $=\frac{22}{7}\times7\times7\times\frac{60}{360}\text{cm}$ $=\frac{11\times7}{3}\text{cm}^2=\frac{77}{3}\text{cm}^2$ Area pf sector BDF = Area of sector CDE = Area of sector AEF Sum of area of all the sectors $=\frac{77}{3}\times3\text{cm}^2=77\text{cm}^2$ $\therefore$ Shaded area = Area of $\triangle\text{ABC}$ - Sum of area of all sectors $=49\sqrt{3}-77\text{cm}^2=(84.77-77.00)\text{cm}^2$ $=7.77\text{cm}^2$

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Question 25 Marks

A racetrack is in the form of a ring whose inner circumference is 352m and outer circumference is 396m. Find the Width and the area of the track.

Answer

Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:
$2\pi\text{r}=352$
$\Rightarrow\text{r}=\frac{352}{2\pi}$
Also,
$2\pi\text{R}=396$
$\Rightarrow\text{R}=\frac{396}{2\pi}$
Width of the track $=(\text{R}-\text{r})$
$=\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\text{m}$
$=\frac{1}{2\pi}(396-352)\text{m}$
$=\Big(\frac{1}{2}\times\frac{7}{22}\times44\Big)\text{m}$
$=7\text{m}$
Area of the track $=\pi\big(\text{R}^2-\text{r}^2\big)$
$=\pi\big(\text{R}+\text{r}\big)\big(\text{R}-\text{r}\big)$
$=\Big[\pi\Big(\frac{396}{2\pi}+\frac{352}{2\pi}\Big)\times\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\Big]\text{m}^2$
$=\Big(\pi\times\frac{748}{2\pi}\times7\Big)\text{m}^2$
$=\frac{748}{2}\times7\text{m}^2$
$=2618\text{m}^2$

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Question 35 Marks

With the vertices $A, B$ and $C$ of a tringle $A B C$ as centres, arcs are drawn with radii $5 cm$ each as shown in the given figure. If $A B=14 cm, B C=48 cm$ and $C A=50 cm$ then find the area of the shaded region. [Use $\pi=3.14$ ]

Answer

Area of sector $= \frac{\angle\text{A}}{360^\circ}\times\pi\text{r}^2+\frac{\angle\text{B}}{360^\circ}\times\pi\text{r}^2+\frac{\angle\text{C}}{360^\circ}\times\pi\text{r}^2$
$=\frac{3.14\times25}{360^\circ}(\angle\text{A}+\angle\text{B}+\angle\text{C})$
$=\frac{3.14\times25}{360^\circ}\times180^\circ$
$=39.25\text{cm}^2$
Since, 14cm, 48cm and 50cm is a Pythagorean triplet, it is a right $\triangle$
Area $=\frac{1}{2}\times48\times14=336\text{cm}^2$
$\therefore$ Area of shaded portion $= (336 - 39.25)cm^2$
$= 296.75cm^2$

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Question 45 Marks

In the given figure, $ABCD$ is a rectangle with $\text{AB}=80\text{cm}$ and $\text{BC}=70\text{cm},$ $\angle\text{AED}=90^\circ$ and $\text{DE}=42\text{cm}$ A semicircle is dn wn, taking BC as diameter. Find the area o the shaded region.

Answer

Length of rectangle $ABCD = AB = 80cm$
Breadth of rectangle $ABCD = BC = 70cm$
$\therefore$ Area of rectangle $ABCD = AB \times BC = 80 \times 70 = 5600cm^2$
In right-angled $\triangle\text{AED},$
$AE^2 = (AD^2 - DE^2) = (70^2 - 42^2) = (70 + 42) = 112 \times 28 = 4 \times 28 \times 28$
$\Rightarrow AE = 2 \times 28 = 56cm$
$\therefore$ Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times42\times56=1176\text{cm}^2$
Area of semi-circle $=\frac12\pi\times\Big(\frac{70}{2}\Big)^2$
$=\Big\{\frac{1}{2}\times\frac{22}{7}\times35\times35\Big\}\text{cm}^2=1925\text{cm}^2$
Thus, Area of the shaded region
= Area of rectangle ABCD $- ($Area of $\triangle\text{AED}$ + Area of semi-circle$)$
$= 5600 - (1176 + 1925)$
$= 5600 - 3101$
$= 2499cm^2$

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Question 55 Marks

If three circles of radius a each, drawn such that each touches the other two, prove that the area included between included between them is equal to $\frac{4}{25}\text{a}^2.$ $\big[\text{Take } \sqrt{3}=1.73\text{ and}\ \pi=3.14\big]$

Answer

Let A, B, C, be the centres of these circles. joint AB, BC, CA
Required area = $\big($area of $\triangle\text{ABC}$ with each side 2$\big)$
-3(area of sector with r = a cm, $\theta$ = 60°)
$=\Big[\frac{\sqrt{3}}{4}\times(2\text{a})^2-\frac{3\pi\text{a}^2\times60}{360}\Big]$
$=(1.73\text{a}^2-1.57\text{a}^2)$
$=0.16\text{a}^2$
$=\frac{16}{100}\text{a}^2$
$=\Big(\frac{4}{25}\text{a}^2\Big)\text{sq. units}$

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Question 65 Marks

The inside perimeter of a running track shown in the shown in the figure is 400m. the length of the straigth portions is 90m, and the ends are semicicles. if the track. Also find the length of the outer boundary of the track. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Length of the inner curved portion
= (400 - 290)m
= 220m
Let the radius of each inner curved part be r
Then, $\frac{22}{7}\times\text{r}=110\text{m}$
$\text{r}=\Big(110\times\frac{7}{22}\Big)\text{m}=35\text{m}$
Inner radius = 35m, outer radius = (35 + 14) = 49m
$\therefore$ Area of the track = (area of 2 rectangles each 90m 14m) + (area pf circular ring with R = 49m, r = 35m)
$\Big[2\times90\times14+\frac{22}{7}\big\{(49)^2-(35)^2\big\}\Big]\text{m}^2$
$\Big[2520+\frac{22}{7}(49+35)(49-35)\Big]\text{m}^2$
$\big[2520+3696\big]\text{m}^2=6216\text{m}^2$
Length of outer boundary of the track
$\Big[2\times90+2\times\frac{22}{7}\times49\Big]\text{m}=488\text{m}$

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Question 75 Marks

In the given figure, from a rectangular region $ABCD$ with $AB = 20cm$, a right triangle $AED$ with $AE = 9cm$ and $DE = 12cm$, is cut off. On the other end, taking $BC$ as diameter, a semicircle is added on outside the region. Find the area of the shaded region. $\big[\text{Use }\pi=3.14\big]$

Answer

In right-angled $\triangle\text{AED},$
$AD^2 = DE^2 + AE^2 = 12^2 + 9^2 = 144 + 81 = 225$
$\Rightarrow\text{AD}=\sqrt{225}=15\text{cm}$
Now, Area of $\triangle\text{AED}=\frac12\times\text{DE}\times\text{AE}=\frac12\times12\times9=54\text{cm}^2$
Length of rectangle ABCD = AB = 20cm
Breadth of rectangle ABCD = AD = 15cm
$\therefore$ Area of rectangle $ABCD = AB \times BC = 20 \times 15 = 300cm^2$
Area of semi-circle $=\frac{1}{2}\pi\times\Big(\frac{15}{2}\Big)^2=\Big\{\frac12\times3.14\times7.5\times7.5\Big\}\text{cm}^2=88.3125\text{cm}^2$
Thus, Area of rectangle ABCD + Area of semi-circle - Area of $\triangle\text{AED}$
$= 300 + 88.31 - 54$
$= 334.31cm^2$

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Question 85 Marks

Three semicircles each of diameter 3cm, a circle of diameter 4.5,cm and a semicircle of radius 4.5cm are drawn in the given figure. Find the area of the shaded region. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Area of semi-circle PQR $=\frac{1}{2}\pi\Big(\frac92\Big)^2$
$=\frac{81\pi}{8}\text{cm}^2$
Area of region circle, $\text{A}=\pi\Big(\frac94\Big)^2$
$=\frac{81\pi}{16}\text{cm}^2$
Area of region (B + C) $=\pi\Big(\frac32\Big)^2$
$=\frac{9\pi}{4}\text{cm}^2$
Area of region $\text{D}=\frac{1}{2}\pi\Big(\frac{3}{2}\Big)^2$
$=\frac{9\pi}{8}\text{cm}^2$
Area of shaded region = Area of semicircle - Area of circle - Area of region (B + C) + Area of region D
$=\frac{18\pi}{8}-\frac{81\pi}{16}-\frac{9\pi}{4}+\frac{9\pi}{8}$
$=\frac{63\pi}{16}\text{cm}^2$
$=\frac{99}{8}\text{cm}^2$

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Question 95 Marks

A square tank has an area of $1600 cm^2$. There are four semicircular plots around it. Find the cost of turfing the plots at Rs. $12.50$ per $m ^2$. [Take $\pi=3.14$ ]

Answer

Let am be the side of the square.
Area of the square
$=\text{a}^2$
Thus. we have
$\text{a}^2=1600$
$\Rightarrow\text{a}=40$
Area of the plots = 4(Area of the semicircle of radius 20m)
$=\Big|4\Big(\frac{1}{4}\pi\text{r}^2\Big)\Big|\text{m}^2$
$=\Big|\Big(\frac{1}{2}\times3.14\times20\times20\Big)\Big|\text{m}^2$
$=2512\text{m}^2$
$\therefore$ Costofturfmgtheplotsat$12.50 per m^2 = Rs. (2512 \times 12.50)$
$= Rs. 31400$

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Question 105 Marks

A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35cm then find the total area of the design. $\big[\text{Use }\sqrt{3}=1.732\text{ and }\pi=3.14\big]$

Answer

ABCDEF is a hexagon
$\therefore\angle\text{AOB}=60,$ Radius = 35cm
Area of sector AOB
$=\pi\text{r}^2\times\frac{60^\circ}{360^\circ}=\frac{\pi\times35\times35}{6}\text{cm}^2$
$=\frac{3.14\times35\times35}{6}\text{cm}^2$
$=641.083\text{cm}^2$
Area of $\triangle\text{AOB}=\frac{\sqrt{3}}{4}\times\text{r}^2=\frac{\sqrt{3}}{4}\times35\times35\text{cm}^2$
$=530.425\text{cm}^2$
Area of segment $APB = (641.083 = 530.425)cm^2 = 110.658cm^2$^
Area of design (shaded area) $= 6110.658m^2 = 663.948m^2$
$= 663.948m^2$

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Question 115 Marks

Three equal circles, each of radius 6cm, touch one another as shown in the figune. find the area enclosed between them. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.732.\big]$

Answer

Let A, B, C, be the centres of these circles. joint $AB, BC, CA$
Required area = $\big($area of $\triangle\text{ABC}$ with each side a = 12cm$\big)$
-3(area of sector with r = 6, $\theta$ = 60°)
$=\Big[\frac{\sqrt{3}}{4}\times(12)^2-3\times\Big(3.14\times(6)^2\times\frac{60}{360}\Big)\Big]$
$=\Big[\frac{\sqrt{3}}{4}\times12\times12-3\times3.14\times6\Big]\text{cm}$
$=(36\times1.73-56.52)\text{cm}^2$
$=(62.28-56.52)\text{cm}^2$
$=5.76\text{cm}^2$
The area enclosed $= 5.76cm^2$

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Question 125 Marks

In the given figure, $O$ is the centre of the circle with $\text{AC}=24\text{cm},\ \text{AB}=7\text{cm}$ and $\angle\text{BOD}=90^\circ$ Find the area of shaded region. $\big[\text{Use }\pi=3.14\big]$

Answer

In right-angled $\triangle\text{BAC},$
$CB^2 = AC^2 + AB^2 = 24^2 + 7^2 = 576 + 49 = 625$
$\Rightarrow\text{CB}=\sqrt{325}=25\text{cm}$
$\Rightarrow\text{OC}=\frac12\text{CB}=\frac{25}{2}\text{cm}=12.5\text{cm}$ = radius of the circle
Now, area of $\triangle\text{BAC}=\frac12\times\text{AC}\times\text{AB}=\frac{1}{2}\times24\times7=84\text{cm}^2$
Area of the circle$ = 3.14 \times 12.5 \times 12.5 = 490.625cm^2$
Now, area of the shaded region
= Area of the circle - Area of $\triangle\text{BAC}$ - Area of quadrant $COD$
$= (490.625 - 84 - 122.66)cm^2$
$= 283.96cm^2$

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Question 135 Marks

In the given figure, the side of suare is 28cm and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Side of square = 28cm and radius of each circle $=\frac{28}{2}\text{cm}$
Area of the shaded region = Area of the square + Area of the two circle - Area of the quadrants
$=(28)^2+2\times\pi\times\Big(\frac{28}{2}\Big)^2-2\times\frac14\times\pi\times\Big(\frac{28}{2}\Big)^2$
$=(28)^2+\frac32\times\pi\times\Big(\frac{28}{2}\Big)^2$
$=(28)^2\Big(1+\frac32\times\frac{22}{7}\times\frac12\times\frac{1}{2}\Big)$
$=(28)^2\Big(1+\frac{33}{28}\Big)$
$=(28)^2\times\frac{61}{28}$
$=28\times61$
$=1708\text{cm}^2$

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Question 145 Marks

PQRS is a diameter of a circle of radius 6cm. The lengths PQ, QR an RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS= 12cm, find the perimeter and area of the shaded region. $\big[\text{Take }\pi=3.14\big]$

Answer

PS = 12cm
PQ = QR = RS = 4cm, QS = 8cm
Perimeter = arc PTS + arc PBQ + arc QES
$=(\pi\times6+\pi\times2+\pi\times4)\text{cm}$
$=12\pi \ \text{cm}$
$=12\pi=12\times3.14\text{cm}$
$=37.68\text{cm}$
Area of shaded region = (area of the semi cirlce PBQ) + (area of semicircle PTS) - (area of semicircle QES)
$=\Big[\frac{1}{2}\pi\times(2)^2\times\pi\times(6)^2-\frac{1}{2}\times\pi\times(4)^2\Big]\text{cm}^2$
$=\big[2\pi+18\pi-8\pi\big]=12\pi \ \text{cm}^2=(12\times3.14)\text{cm}^2$
$=37.68 \ \text{cm}^2$

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Question 155 Marks

In the given figure, two concentric circles with centre $O$ have radii $21\ cm$ and $42\ cm$. If $\angle\text{AOB}=60^\circ,$ find the area of the shaded region. $\Big[\text{Use }\pi=\frac{22}{7}.\Big]$

Answer

Radii of the concentric circles = 21cm and 42cm
Area between the circles $=\pi\text{R}^2-\text{r}^2=\Big(\frac{22}{7}\Big)\big(42^2-21^2\big)=4158\text{cm}^2$
Angle subtended by the arc in the inner circle $= 60^\circ $
Area of the sector in the inner circle
$=\Big(\frac{60^\circ}{360^\circ}\Big)\times\pi\text{R}^2=\Big(\frac{60^\circ}{360^\circ}\Big)\times\Big(\frac{22}{7}\Big)\times(42)^2=924\text{cm}^2$
Area of the portion of the sector in between the circles $= 924 - 231 = 693cm^2$
Area of the shaded portion = (Area between the circles) - (Area of the portion of the sector in between the circles)
$= 4158 - 693$
$= 3465$
Therefore, the area of the shaded region is $3465cm^2$

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Question 165 Marks

In the given figure, $ABCD$ is a rectangle of Dimensions $21\ cm \times 14\ cm$ A semicircle is drawn with $BC$ as diameter. Find the area and the perimeter of the shaded region in the figure.

Answer

Area of shaded region = Area of rectangle - Area of semi-circle
$= (21 \times14) -\frac{1}{2}\times\pi\times7\times7$
$= 294 - 77$
$= 217cm^2$
Perimeter of shaded region $= 21 + 14 + 21 + \pi\times7$
$= 56 + 22$
$= 78cm.$

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Question 175 Marks

A child draws the figuie of an aeroplane as shown. Here, the wings $ABCD$ and FGHI are parallelograms, the tail $DEF$ is an sosceles triangle, the cockpit $CKI$ is a semicircle and $CDFI$ is a square. In the given figure, $\text{BP}\perp\text{CD},\ \text{HQ}\perp\text{FI}$ and $\text{EL}\perp\text{DF}.$ if $\text{CD}=8\text{cm},\ \text{BP}=\text{HQ}=4\text{cm}$ and $\text{DE}=\text{EF}=5\text{cm},$ find the area of the whole figure. $\big[\text{Take }\pi=3.14\big]$

Answer

Given $\text{BP}\perp\text{CD}, \ \text{HQ}\perp\text{FI}$ and $\text{EL}\perp\text{DF},$
$DE = 8cm, BP = HQ = 4cm$ and $DE = EF = 5cm$
Area of parallelogram $ABCD = BP \times DC$
$= 4 \times 8 = 32cm^2$
Area of parallelogram $FGHI = FI \times HQ$
$= 8 \times 4 = 32cm^2$
Area of semicircle $\text{CKI}=\frac{1}{2}\pi\text{r}^2$
$-\frac{1}{2}\times3.14\times(4)^2=25.12\text{cm}^2$
Area of isosceles $\triangle\text{DEF}=\frac{1}{4}\text{b}\sqrt{4\text{a}^2-\text{b}^2}$
$=\frac{1}{4}(8)\sqrt{4(5)^2-(8)^2}=2\sqrt{100-64}$
$=3\sqrt{36}=12\text{cm}^2$
Area of square $CDFI =$ (side)$^2 = (8)^2 = 64cm^2$
Area of whole figure = area of $||^{gm} ABCD +$ area of $||^{gm} FGHI +$ area of semi - circle $CKI +$ area of $\triangle\text{DEF}$ + area of square $CDFI$
$= (32 + 32 + 25.12 + 12 + 12 + 64)cm^2$
$= 165.12cm^2$

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Question 185 Marks

A chord of a circle of radius 10cm subtends a right angle at the centre. Find the area of the minor segment. $\big[\text{Use }\pi=3.14\big]$

Answer

Let O be the center of the circle and AB be the chord.
Now, Area of the minor segment
= Area of sector OACBO - Area of $\triangle\text{OAB}$
$=3.14\times10\times10\times\frac{90}{360}-\frac12\times10\times10$
$=78.5-50$
$=28.5\text{cm}^2$

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Question 195 Marks

A chord of a circle of radius 30cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segments. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.732\big]$

Answer

Let AB be the chord of a circle with centre O and radius 30cm such that $\angle\text{AOB}=60^\circ$
Area of the sector OACBO $=\frac{\pi\text{r}^2\theta}{360}$
$=\Big(3.14\times30\times30\times\frac{60}{360}\Big)\text{cm}^2$
$=471\text{cm}^2$
Area of $\triangle\text{OAB}=\frac{1}{2}\text{r}^2\sin\theta$
$=\Big(\frac{1}{2}\times30\times30\times\sin60^\circ\Big)\text{cm}^2$
$=(225\times1.732)\text{cm}^2$
$=389.7\text{cm}^2$
Area of the minor segment = (Area of the sector OACBO) - (Area of $\triangle$OAB)
$=(471-389.7)\text{cm}^2$
$=81.3\text{cm}^2$
Area of the major segment = (Area of the circle) - (Area of the minor segment)
$=\big|(3.14\times30\times30)-81.3\big|\text{cm}^2$
$=2744.7\text{cm}^2$

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Question 205 Marks

The radius of a circular garden is $100 m $. There is a road $10 m$ wide, running all around it . Find the area of the road and the cost of levelling it at $₹ 20$ per $m ^2$. [Use $\pi=3.14$ ]

Answer

Area of the road
= Ar(circular region of radius 110m) - Ar(circular region of radius 100m)
$=\pi(110)^2-\pi(100)^2$
$=\pi(110^2-100)^2$
$=3.14\times(110+100)(110-100)$
$=3.14\times210\times10$
$=6954\text{m}^2$
Cost of levelling per $m^2 = ₹ 20$
$\therefore$ Cost of levelling $6954m^2 = ₹ 20 \times 6594 = ₹ 131880$

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Question 215 Marks

The perimeter of the quadrant of a circle is 25cm. find its area. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$

Answer

Let r the radius of the quadrant.
Perimeter of the quadrant $=\Big(2\text{r}+\frac{2\pi\text{r}\times90}{360}\Big)\text{cm}$
$=\Big(2\text{r}+\frac{\pi\text{r}}{2}\Big)\text{cm}$
$=\Big(2\text{r}+\frac{22\text{r}}{7\times2}\Big)\text{cm}$
$=\Big(2\text{r}+\frac{11\text{r}}{7}\Big)\text{cm}$
$=\Big(\frac{14\text{r}+11\text{r}}{7}\Big)\text{cm}$
$=\frac{25\text{r}}{7}\text{cm}$
Given, perimeter of a quadrant = 25cm
$\Rightarrow\frac{25\text{r}}{7}=25$
$\Rightarrow\text{r}=7\text{cm}$
$\therefore$ Area of the quadrant $=\frac14\pi\text{r}^2=\Big(\frac14\times\frac{22}{7}\times7\times7\Big)\text{cm}^2$
$=\frac{77}{2}\text{cm}^2=38.5\text{cm}^2$

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Question 225 Marks

In the given figure, $\triangle\text{ABC}$ is right-angled at A. Find the area of the shaded region if AB = 6cm, BC = 10cm and O is the centre of the incircle of $\triangle\text{ABC}.\ \big[\text{Take }\pi=3.14\big]$

Answer

In $\triangle\text{ABC}, \ \angle\text{A}=90^\circ, \ \text{AB}=6\text{cm}, \ \text{BC}=10\text{cm}$
$\text{BC}^2=\text{AC}^2+\text{AB}^2$
$\therefore\text{AC}^2=\text{BC}^2-\text{AB}^2=10^2-6^2=100-36=64$
$\therefore\text{AC}=8\text{cm}$
Area of $\triangle\text{ABC}=\frac{1}{2}\times\text{AC}\times\text{AB}=\frac{1}{2}\times8\times6\text{cm}^3=24\text{cm}^3$
Let r be the radius of circle of centre O
Area of $\triangle\text{OCB}=\frac{1}{2}\times10\times\text{r} \ \text{cm}^2=5\text{r} \ \text{cm}^2$
Area of $\triangle\text{OAB}=\frac{1}{2}\times6\times\text{r} \ \text{cm}^2=3\text{r} \ \text{cm}^2$
Area of $\triangle\text{OCA}=\frac{1}{2}\times8\times\text{r} \ \text{cm}^2=4\text{r} \ \text{cm}^2$
Area of $\big(\triangle\text{OCB}+\triangle\text{OAB}+\triangle\text{OCA}\big)$ = Area of $\triangle\text{ABC}$
$\therefore5\text{r}+3\text{r}+4\text{r}=24$
Or $12\text{r}=24 \ \therefore\text{r}=2\text{cm}$
$\therefore$ Area of in circle $=\pi\text{r}^2=3.14\times2\times2\text{cm}^2$
$=12.56\text{cm}^2$
⇒ Shaded area = Area of $\triangle\text{ABC}$ - Area of in circle
$=(24-12.56)\text{cm}^2=11.44\text{cm}^2$

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Question 235 Marks

ABCD is a field in the shape of a trapezium, $\text{AD}||\text{BC},\ \angle\text{ABC}=90^\circ$ and $LADC 60^\circ .$ $\angle\text{ADC}=60^\circ.$ Four sectors are formed with centres $A, B, C$ and $D$, as shown in the figure. The radius of each sector is $14\ m$. Find the following.

Total area of the four sector,
Area of the remaining portion, given that $AD =55m, BC = 45m$ and $AB = 30m,$

Answer

Total area of 4 sectors $=\Big\{\frac{22}{7}\times(14)^2\times\Big(\frac{90}{360}+\frac{90}{360}+\frac{120}{360}+\frac{60}{360}\Big)\Big\}\text{m}^2$

$=\frac{1}{2}\times(55+45)\times30$
$=100\times15$
$=1500\text{m}^2$
Now, area of the remaining portion
= Ar(trapezium ABCD) - Total area of 4 sectors
$= (1500 - 616)m^2$
= 884m2$=\Big\{22\times2\times14\times\Big(\frac{1}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{6}\Big)\Big\}\text{m}^2$
$=\Big\{616\times\frac{3+3+4+2}{12}\Big\}\text{m}^2$
$=\Big\{616\times\frac{12}{12}\Big\}\text{m}^2$
$=616\text{m}^2$ Area of traprzium ABCD $=\frac{1}{2}\times(\text{AD}+\text{BC})\times\text{AB}$

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Question 245 Marks

In the given figure, $PQ =24 cm, PR =7 cm$ and $O$ is the centre of the circle. Find th area of the shaded region.. $\big[\text{Take }\pi=3.14\big]$

Answer

In $\triangle\text{PQR}, \ \angle\text{P}=90, \ \text{PQ}=24\text{cm}, \ \text{PR}=7\text{cm}$
$\therefore\text{QR}^2=\text{RP}^2+\text{PQ}^2=7^2+24^2$
$=49+576=625$
$\therefore\text{QR}=25\text{cm}$
Area of semicircle
$=\frac{1}{2}\times\pi\times\Big(\frac{25}{2}\Big)^2$
$=\frac{1}{2}\times3.14\times\frac{25\times25}{4}\text{cm}^2$
$=\frac{625\times3.14}{8}=245.31\text{cm}^2$
Area of $\triangle\text{PQR}=\frac{1}{2}\times7\times24\text{cm}^2=84\text{cm}^2$
$\therefore$ shaded area $= 245.31 - 84 = 161.31cm^2$

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Question 255 Marks

In the given fingure, $\triangle\text{ABC}$ is right-angled at A. semicircles are drawn on AB, AC and BC as diameter, it is given that AB = 3cm and AC = 4cm. find the area of the shaded region.

Answer

Area pf shaded region = Area of $\triangle\text{ABC}$ + Area of semi-circle APB + Area of semi circle AQC - Area of semi circle BAC
Now, Area of a $\triangle\text{ABC}=\frac{1}{2}\times3\times4=6\text{cm}^2 \ ...(1)$
Area of semi-circle $\text{APB}=\frac{1}{2}\pi\text{r}^2=\frac{1}{2}\pi\times\Big(\frac{3}{2}\Big)^2=\frac{9}{8}\pi\ ...(2)$
Area of semi-circle $\text{AQC}=\frac{1}{2}\pi\text{r}^2_2$
$=\frac{1}{2}\pi\Big(\frac{4}{2}\Big)^2=2\pi\text{cm}^2 \ ...(3)$
Further in $\triangle\text{ABC}, \ \angle\text{A}=90$
$\therefore\text{BC}^2=\text{AB}^2+\text{AC}^2=9+16=25$
$\therefore\text{BC}=5$
Area of semi-circle $\text{BAC}=\frac{1}{2}\pi\Big(\frac{5}{2}\Big)^2=\frac{25}{8}\pi \ ...(4)$
Adding (1), (2), (3) and subtracting (4)
$\therefore$ Area of shaded region $=6+\frac{9}{8}\pi+2\pi-\frac{25}{8}\pi$
$=6+\frac{25}{8}\pi-\frac{25}{8}\pi=6\text{cm}^2$

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Question 265 Marks

If the diameters of the concentric circles shown in the figure below are in the ratio 1 : 2: 3 then find the ratio of the areas of three regions. $\big[\text{Use }\pi=3.14\big]$

Answer

Let the diameters of concentric circles be k, 2k and 3k.
$\therefore$ Radius of concentric circles are $\frac{\text{k}}{2},\text{k}$ and $\frac{3\text{k}}{2}.$
$\therefore$ Area of inner circle, $\text{A}_1=\pi\Big(\frac{\text{k}}{2}\Big)^2=\frac{\text{k}^2\pi}{4}$
$\therefore$ Area of middle region $\text{A}_2=\pi(\text{k})^2-\frac{\text{k}^2\pi}{4}=\frac{3\text{k}^2\pi}{4}$
$\big[\because$ area of middle region, $=\pi(\text{R}^2-\text{r}^2)\big]$
Where R is radius of outer and r is radius r is radius of innerring
and area of outer region, $\text{A}_3=\pi\Big(\frac{3\text{k}}{2}\Big)^2-\pi\text{k}^2$
$=\frac{9\pi\text{k}^2}{4}-\pi\text{k}^2=\frac{5\pi\text{k}^2}{4}$
$\therefore$ Required ratio $\text{A}_1:\text{A}_2:\text{A}_3$
$=\frac{\text{k}^2\pi}{4}:\frac{3\text{k}^2\pi}{4}:\frac{5\pi\text{k}^2}{4}=1:3:5$

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Question 275 Marks

Four cows are tethered at the four corners of a square field of side 50m such that the each can graze the maximum unshared area. What area will be left unqrazed? $[\text{Take }\pi=3.14]$

Answer

Let r be the radius of the circle.
Thus. we have:
$\text{r}=\frac{50}{2}\text{m}$
$=25\text{m}$
Area left ungrazed = (Area of the square) - 4(Area of the sector where r = 25m and $\theta$ = 90°)
$=\Big|(50\times50)-4\Big(3.14\times25\times25\times\frac{90}{360}\Big)\Big|\text{m}^2$
$=\Big|2500-\Big(4\times\frac{1962}{4}\Big)\Big|\text{m}^2$
$=(2500-1962.5)\text{m}^2$
$=537.5\text{m}^2$

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Question 285 Marks

In the given figure, a circle is inscribed in an equilateral triangle $ABC$ of side $12\ cm$. Find the radius of inscribed circle and the area of the shaded region. $\big[\text{Use}\sqrt{3}=1.73\text{ and }\pi=3.14\big]$

Answer

Since $\text{AD}\perp\text{BC},$ is the mid-point of BC.
$\therefore\text{DC}=\frac12\times12=6\text{cm}$
In right-angled $\triangle\text{ADC},$
$AD^2 = AC^2 - DC^2 = 12^2 - 6^2 = 144 - 36 = 108$
$\Rightarrow\text{AD}=\sqrt{108}=6\sqrt{3}\text{cm}$
$\Rightarrow\text{OD}=\frac{6\sqrt{3}}{3}\Rightarrow2\sqrt{3}\text{cm}$
So, the radius of inscribed circle is $2\sqrt{3}\text{cm}. $
$=\frac{\sqrt{3}}{4}\times12\times12-3.14\times\big(2\sqrt{3}\big)^2$
$=36\sqrt{3}-37.68$
$= 62.28 - 37.68$
$=24.6\text{cm}^2$

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Question 295 Marks

On a circular table cover of radius 42cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design. $\Big[\text{Use }\sqrt{3}=1.73\text{ and }\pi=\frac{22}{7}\Big]$

Answer

Let O be the center of the circumcircle.
Join OB and draw $\text{AD}\perp\text{BC}.$
Then. OB = 42cm and $\angle\text{OBD}=30^\circ$
In $\triangle\text{OBD},$
$\sin30^\circ=\frac{\text{OD}}{\text{OB}}$
$\Rightarrow\frac12=\frac{\text{OD}}{42}$
$\Rightarrow OD = 21cm$
Now, $BD^2 = OB^2 - OD^2 = 42^2 - 21^2 = (42 + 21)(42 - 21) = 63 \times 21$
$\Rightarrow\text{BD}=\sqrt{63\times21}=\sqrt{3\times21\times21}=21\sqrt{3}\text{cm}$
$\Rightarrow\text{BC}=2\times21\sqrt{3}=42\sqrt{3}\text{cm}$
Now, area of the shaded region
= Area of the circle - Area of an equilateral $\triangle\text{ABC}$
$=\frac{22}{7}\times42\times42-\frac{\sqrt{3}}{4}\times42\sqrt{3}\times42\sqrt{3}$
$=(5544 - 2288.79)\text{cm}^2$
$= 3255.21\text{cm}^2$

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