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Geometry and Mensuration: Test 2
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Question 1 |
The area of a rectangular field is 460 square metres. If the length is 15 per cent more than the breadth, what is breadth of the rectangular field?
15 metres | |
26 metres | |
34.5 metres | |
none |
Question 1 Explanation:
Let the breadth = b cm. Therefore, length = 1.15b cm.
Thus, $ \begin{array}{l}1.15{{b}^{2}}=460\\=>b=20\,cm\end{array}$
Thus, $ \begin{array}{l}1.15{{b}^{2}}=460\\=>b=20\,cm\end{array}$
Question 2 |
A sector of a circle of radius 15 cm has the angle 120°. It is rolled up so that two bounding radii are joined together to form a cone. The volume of the cone is
$ \displaystyle \left( 250\sqrt{2}\pi \right)\,c{{m}^{3}}$ | |
$ \displaystyle \left( 500\sqrt{2} \right)\frac{\pi }{3}\,c{{m}^{3}}$ | |
$ \displaystyle \left( 250\sqrt{2} \right)\frac{\pi }{3}\,c{{m}^{3}}$ | |
$ \displaystyle \left( 1000\sqrt{2} \right)\frac{\pi }{3}\,c{{m}^{3}}$ |
Question 2 Explanation:
The circumference of the sector
$ \begin{array}{l}\frac{1}{3}\times 2\pi 15=10\pi \\2\pi r=10,r=5\\Radius\,\,of\,cone=5\pi \,cm\\Lateral\,\,Height\,=15\,cm\\Height=\sqrt{{{15}^{2}}-{{25}^{2}}}\\=10\sqrt{2\,}\,cm\\Volume\,of\,cone=\frac{1}{3}\pi {{r}^{2}}h\\=\frac{1}{3}\pi \times 25\times 10\sqrt{2}\\=\left( 250\sqrt{2} \right)\frac{\pi }{3}c{{m}^{3}}\end{array}$
$ \begin{array}{l}\frac{1}{3}\times 2\pi 15=10\pi \\2\pi r=10,r=5\\Radius\,\,of\,cone=5\pi \,cm\\Lateral\,\,Height\,=15\,cm\\Height=\sqrt{{{15}^{2}}-{{25}^{2}}}\\=10\sqrt{2\,}\,cm\\Volume\,of\,cone=\frac{1}{3}\pi {{r}^{2}}h\\=\frac{1}{3}\pi \times 25\times 10\sqrt{2}\\=\left( 250\sqrt{2} \right)\frac{\pi }{3}c{{m}^{3}}\end{array}$
Question 3 |
A toy is in the shape of a hemisphere surmounted by a cone. If radius of base of the cone is 3 cm and its height is 4 cm, the total surface area of the toy is___Πcm2
33 | |
42 | |
66 | |
56 |
Question 3 Explanation:
$ \displaystyle \begin{array}{l}Slant\,Height=l=\sqrt{{{4}^{2}}+{{3}^{2}}}=5\\Surface\,area\,of\,hemi\,sphere\\=2\pi {{\left( 3 \right)}^{2}}=18\pi \,\,c{{m}^{2}}\\Surface\,\,\,\,area\,\,\,of\,\,\,cone\,\,=\pi \,rl\\=\pi \,\times 3\times 5=15\,\pi \,\,c{{m}^{2}}\\Therefore\,required\,area=18\pi +15\pi =33\pi cm{}^{2}\end{array}$
Question 4 |
What will be the cost of gardening 1 metre broad boundary around a rectangular plot having perimeter of 340 metres at the rate of Rs.10 per square metre?
Rs.3, 400/- | |
Rs.1, 700/- | |
Rs.3, 440/- | |
Cannot be determined |
Question 4 Explanation:
Given Perimeter is = 340 m
Let the length and breadth of the plot are l and b respectively.
Now total length = l+2
Total width = b+2
So are of the boundary = [(l+2)(b+2) – lb]
2(l+b)+4
344
Total cost = Rs. 3440
Let the length and breadth of the plot are l and b respectively.
Now total length = l+2
Total width = b+2
So are of the boundary = [(l+2)(b+2) – lb]
2(l+b)+4
344
Total cost = Rs. 3440
Question 5 |
What will be the area (in square metres) of 1.5 metre wide garden developed around all the four sides of a rectangular field having area equal to 300 square metres and breadth equal to three-fourth of the length?
96 | |
105 | |
114 | |
Cannot be determined |
Question 5 Explanation:
Let Length = p cm and breadth = 3p/4 cm.
Total area = 300 = 3p2/4 => p= 20 m.
Width = 15 m.
Area of garden = {(20 + 3) × (15 + 3) - 20 × 15} = (23 × 18 - 20 × 15) = 414 - 300 = 114 m.
Correct option is (c)
Total area = 300 = 3p2/4 => p= 20 m.
Width = 15 m.
Area of garden = {(20 + 3) × (15 + 3) - 20 × 15} = (23 × 18 - 20 × 15) = 414 - 300 = 114 m.
Correct option is (c)
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