DAV Class 8 Maths Chapter 14 Brain Teasers
Mensuration Brain Teasers
1. A. Tick (✓) the correct option.
(a)
The volume of a cube of side 0.01 m in cm3 is—
\(
\begin{aligned}
(i)\ &\ 0.001 \\
(ii)\ &\ 1 \\
(iii)\ &\ 0.0001 \\
(iv)\ &\ 0.000001
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{Side} &= 0.01\ \text{m} \\ &\implies 1\ \text{cm} \\[6pt] \text{Volume} &= (\text{side})^3 \\ &= 1 \ \text{cm}^3 \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{1} \)
(b)
If the height of a cylinder is halved, its volume will be—
\(
\begin{aligned}
(i)\ &\ \frac{1}{2}\ \text{times} \\[6pt]
(ii)\ &\ \frac{1}{3}\ \text{times} \\[6pt]
(iii)\ &\ 2\ \text{times} \\[6pt]
(iv)\ &\ 3\ \text{times}
\end{aligned}
\)
Solution
\[ \begin{aligned} \text{volume of cylinder} & =\pi r^2 h \\ \\ \text{if } h &= \frac{1}{2}h \\[6pt] V &= \frac1{2}\pi r^2 h \end{aligned} \]Answer \( {\color{orange}(i)}\ \color{red}{\dfrac{1}{2}\ \text{times}} \)
(c)
The area of a trapezium having two parallel sides as 10 cm and 12 cm and height as 4 cm is—
\(
\begin{aligned}
(i)\ &\ 42\ \text{cm}^2 \\
(ii)\ &\ 44\ \text{cm}^2 \\
(iii)\ &\ 46\ \text{cm}^2 \\
(iv)\ &\ 48\ \text{cm}^2
\end{aligned}
\)
Solution
\[ \begin{aligned} \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\ \\ &= \frac{1}{\cancel2} \times (10 + 12) \times \cancel4^2 \\[6pt] &= 22 \times 2 \\[6pt] \text{Area} & = 44\ \text{cm}^2 \end{aligned} \]Answer \( {\color{orange}(ii)}\ \color{red}{44\ \text{cm}^2} \)
(d)
The number of faces a tetrahedron has—
\(
\begin{aligned}
(i)\ &\ 14 \\
(ii)\ &\ 12 \\
(iii)\ &\ 6 \\
(iv)\ &\ 4
\end{aligned}
\)
Answer \( {\color{orange}(iv)}\ \color{red}{4} \)
(e)
If the length of side of a cube is doubled, then the ratio of volumes of new cube and original cube is—
\(
\begin{aligned}
(i)\ &\ 1:2 \\
(ii)\ &\ 2:1 \\
(iii)\ &\ 4:1 \\
(iv)\ &\ 8:1
\end{aligned}
\)
Solution
\[ \begin{aligned} \color{green} \text{Volume of cube} &= \color{green}l^3 \\ \\ \text{Original cube } &= l \\ \text{New cube } &= 2l \\ \text{Volume of New cube} &= \text{Volume of Original cube} \\[6pt] & = \frac{\text{Volume of New cube}}{\text{Volume of Original cube}} \\[6pt] & = \frac{(2l)^3}{(l)^3} \\[6pt] & = \frac{8l^3}{l^3} \\[6pt] & = \frac{8\cancel{l^3}}{\cancel{l^3}} \\[6pt] & = \frac{8}{1} \\[6pt] & = 8:1 \\[6pt]\end{aligned} \]Answer \( {\color{orange}(iv)}\ \color{red}{8:1} \)
B. Answer the following questions.
(a) Find the number of edges of a polyhedron having 20 faces and 12 vertices.
Solution
\[ \begin{aligned} \textbf{Euler’s formula: } \ & F + V - E = 2 \\[6pt] V &= 12\\ F &= 20\\[4pt] 20 + 12 - E &= 2 \\ 32 - E &= 2 \\ E &= 32 - 2 \\ E&= \color{green}{30} \end{aligned} \]
Answer Number of edges \(=\ \color{red}{30}\)
(b) Find the surface area of a cube whose volume is \(729 \text{ m}^3\).
Solution
\[ \begin{aligned} \color{green} \text{Volume of cube} &= 729 \\ \color{green} \color{green}l^3 &= 729 \\ l &= \sqrt[3]{729} \\ l &= 9m \\ \\\color{green} \text{Surface area} &= \color{green} 6l^2 \\ &= 6\times 9^2 \\ &= 6\times 81 \\ &= \color{green}{486}\ \text{m}^2 \end{aligned} \]
Answer Surface area \(=\ \color{red}{486\ \text{m}^2}\)
(c) Volume of a cylinder is \(2376\ \text{cm}^3\). If the diameter of its base is \(12\) cm, find its height.
Solution
\[ \begin{aligned} \text{Diameter} & = 12 \ cm\\ \text{Radius} & = 6 \ cm\\ \color{green} \text{Volume of a cylinder} &= 2376 \ cm^3 \\ \color{green} \pi r^2 h & = 2376 \\[6pt] \frac{22}{7}\times 6^2 \times h & = 2376 \\[6pt] h & = \frac{\cancel{2376}^{ \ \cancel{66}^3} \times 7}{\cancel{22}_1 \times \cancel{36}_1}\\[6pt] h &= 3 \times 7 \\ h &= \color{green}{21}\ \text{cm} \end{aligned} \]
Answer Height \(=\ \color{red}{21\ \text{cm}}\)
(d) Draw the solid represented by the given Top/Front/Side views.
Solution
(e) Find the total surface area of a cuboid of length \(10\) cm, breadth \(8\) cm and height \(6\) cm.
Solution
\[ \begin{aligned} \color{green} \text{TSA of cuboid} &= 2(lb+bh+hl) \\ &= 2(10\times 8 + 8\times 6 + 6\times 10)\\ &= 2(80 + 48 + 60)\\ &= 2(188) \\ &= \color{green}{376}\ \text{cm}^2 \end{aligned} \]
Answer Total surface area \(=\ \color{red}{376\ \text{cm}^2}\)
2. Find the area of given quadrilateral.
Solution
\[ \begin{align*} \text{Diagonal }(d) &= 6 \, \text{cm} \\ h_1 &= 3.8 \, \text{cm} \\ h_2 &= 7.2 \, \text{cm} \\ \\ \color{green} \text{Area of the quadrilateral } ABCD &= \color{green} \frac{1}{2} \times d \times (h_1 + h_2) \\[6pt] &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {\cancel{6}}^{\color{green}3} \times (3.8 + 7.2) \\[6pt] &= 3 \times 11 \\[6pt] \text{Area} &= 33 \, \text{cm}^2 \end{align*} \]
Answer Area of the quadrilateral ABCD \( = \color{red} 33 \, \text{cm}^2 \)
3. A field is in the shape of a trapezium. One of its parallel sides is twice the other parallel side and distance between them is 100 m. If area of the field is \(10500 \ m^2 \), find the length of the parallel sides.
Solution
\[ \begin{align*} a &= x \\ b &= 2x \\ h &= 100 \, \text{m} \\ \\ \color{green} \text{Area of the trapezium} &= 10500 \ m^2 \\ \\ \color{green} \frac{1}{2} \times (a + b) \times h &= 10500 \\ \\ \frac{1}{\cancel2_1} \times (x + 2x) \times \cancel{100}^{50} &= 10500 \\ \\ 3x \times 50 &= 10500 \\ \\ x &= \frac{\cancel{10500}^{\cancel{210}^{70}}}{\cancel3_1 \times \cancel{50}_1} \\ \\ x &= 70 \ m \\ \\ \color{magenta} \textbf{Parallel} & \color{magenta} \textbf{ Sides} \\ a &= 70 \ m \\ b &= 140 \ m \end{align*} \]
Answer Parallel sides are \( \color{red}{70\ \text{m}}\) and \(\color{red}{140\ \text{m}}\)
4. Two parallel sides \(DC\) and \(AB\) of a trapezium are 12 cm and 36 cm respectively. If the non-parallel sides are each 15 cm, find the area of the trapezium.
Solution
\[ \begin{align*} \text{Draw } & DF \parallel CB \\ \implies DFBC & \text{ is a parallelogram} \\ \text{In } &\triangle ADF \\ DA = DF & = 15 \, cm \\ AB &= 36 \, \text{cm} \\ AF &= 36 - 12 \implies 24 \, \text{cm} \\ DE \perp AF \implies & \text{E is the midpoint of AF} \\[6pt] EF = \frac{AF}{2} \\[6pt] = \frac{24}{2} \\[6pt] \implies EF = 12 \, cm \\ \\ \text{In } \triangle DEF & \text{ (right angled triangle)} \\ (DE)^2 + (EF)^2 &= (DF)^2 \\ (DE)^2 + 12^2 &= 15^2 \\ (DE)^2 + 144 &= 225 \\ (DE)^2 &= 225 - 144 \\ (DE)^2 &= 81 \\ DE &= \sqrt{81} \\ DE &= 9 \, cm \\ \\\text{In trapezium } & DABC \\ a &= 12 \, \text{cm} \\ b &= 36 \, \text{cm} \\ h &= 9 \, \text{cm} \\\\[6pt] \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\[6pt] &= \frac{1}{2} \times (12 + 36) \times 9 \\[6pt] &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {{\cancel{48}}^{\color{green}24}} \times 9 \\[6pt] &= 24 \times 9 \\[6pt] \text{Area} &= 216 \, \text{cm}^2 \end{align*} \]
Answer Area of the trapezium \(=\ \color{red}{216 \ \text{cm}^2}\)
5. The length, breadth and height of a cuboidal box are \(2 \ \text{m}\ 10 \ \text{cm}\), \(1 \ \text{m}\) and \(80 \ \text{cm}\) respectively. Find the area of canvas required to cover this box.
Solution
\[ \begin{aligned} l&=2\ \text{m}\ 10\ \text{cm} \implies 2.1\ \text{m}\\ b&=1\ \text{m} \\ h&=80\ \text{cm} \implies 0.8\ \text{m} \\[6pt] \color{green}\text{T.S.A of cuboid} &= \color{green}2 \times (lb+bh+hl) \\ &= 2 \times \Big[(2.1\times1) + (1\times0.8) + (0.8\times2.1)\Big] \\ &= 2 \times (2.1+0.8+1.68) \\ &= 2 \times 4.58\\ &=9.16 \ \text{m}^2 \end{aligned} \]Answer Canvas required \(=\ \color{red}{9.16 \ \text{m}^2}\)
6. A metallic pipe is \(0.7 \ \text{cm}\) thick. Inner radius of the pipe is \(3.5 \ \text{cm}\) and length is \(5 \ \text{dm}\). Find its Total Surface Area. [Hint: TSA \(=\) inner CSA \(+\) outer CSA \(+\) area of two rims]
Solution
\[ \begin{aligned} \text{Inner radius } (r)&=3.5 \ \text{cm}\\ \text{thickness} &=0.7 \ \text{cm} \\ \text{Outer Radius } (R)&=3.5+0.7 \implies 4.2 \ \text{cm} \\ h&=5 \ \text{dm} \implies 50 \ \text{cm} \\[6pt] \color{green}\text{TSA} &= \color{green} \underbrace{2\pi rh}_{\text{inner CSA}} +\underbrace{2\pi Rh}_{\text{outer CSA}} +\underbrace{2\pi(R^2-r^2)}_{\text{two rims}} \\[6pt] & = \color{green} 2\pi h(r+R) + 2\pi(R^2-r^2) \\[6pt] & = \color{green} 2\pi [h(r+R) + (R^2-r^2)] \\[6pt] &= 2 \times \frac{22}{7} \times \left[ 50 \times (3.5 + 4.2)+ (4.2^2-3.5^2)\right] \\[6pt] &= \frac{44}{7} \times \left[ (50 \times 7.7) + (17.64-12.25)\right] \\[6pt] &= \frac{44}{7} \times \left[ 385 + 5.39 \right] \\[6pt] &= \frac{44}{\cancel7_1} \times \cancel{390.39}^{55.77} \\[6pt] &= 44 \times 55.77 \\[6pt] &= 2453.88 \ \text{cm}^2 \end{aligned} \]Answer T.S.A of the pipe \(=\ \color{red}{2453.88 \ \text{cm}^2}\)
7. A cylinder of curved surface area \(1250 \ \text{m}^2\) is formed from a rectangular metallic sheet. If the length of the sheet is double its breadth, find the dimensions of the sheet.
Solution
\[ \begin{aligned} \text{Let breadth}&=x \\ \text{length} &=2x \\[6pt] \color{green}\text{Area of the rectangular sheet} &= \color{green} \color{green}\text{CSA of cylinder} \\[6pt] l \times b &= 1250 \\[6pt] 2x \times x &= 1250 \\[6pt] 2x^2 &= 1250 \\[6pt] x^2 &= \frac{1250}{2} \\[6pt] x^2 &= 625 \\[6pt] x & = \sqrt{625}\\[6pt] x & = 25 \\[6pt] \therefore\ \text{Breadth} &= 25 \ \text{m}\\ \text{Length} &= 50 \ \text{m} \end{aligned} \]Answer Dimensions of the sheet \(Length = {\color{red}{50 \ m}} , Breadth = {\color{red}{25 \ m}}\)
8. Diameter and length of a roller are 84 cm and 120 cm respectively. In how many revolutions can the roller level a playground of area \(1584 \ \text{m}^2\)?
Solution
\[ \begin{aligned} d &= 84 \ cm \\ r &= 42 \ cm \\ h &= 120 \ cm \\[6pt] \color{green}\text{CSA} &= \color{green} 2\pi r h \\[6pt] &= 2\times \frac{22}{\cancel7_1} \times \cancel{42}^6 \times 120 \\[6pt] &= 44 \times 6 \times 120 \\[6pt] &= 44 \times 720 \\[6pt] \color{green}\text{CSA}&= 31680 \ cm^2 \\[6pt] \text{Area of the playground} &= 1584 \ m^2 \\ 1 \ m^2 & = 10000 \ cm^2 \\ & = 1584 \times 10000 \\ \color{green}\text{Area of the playground}& = 15840000 \ cm^2 \\[6pt] \color{green} \text{Revolutions} &= \color{green} \frac{\text{Area of the playground}}{\text{CSA}} \\[6pt] &= \frac{15840000}{31680} \\[6pt] \color{green} \text{Revolutions} &= 500 \end{aligned} \]Answer Number of revolutions \(= \color{red}{500}\)
9. Water is pouring into a reservoir at the rate of 60 litres per minute. If the volume of the reservoir is \(108 \ m^3\), find the number of hours it will take to fill the reservoir.
Solution
\[ \begin{aligned} \text{Rate of water pouring into a reservoir} &= 60 \ \text{L/min} \\[4pt] \text{Volume of reservoir} &= 108 \ \text{m}^3 \\[4pt] 1 \ \text{m}^3 &= 1000 \ \text{L} \\[4pt] \text{Volume in litres} &= 108 \times 1000 \\ &= 108000 \ \text{L} \\[8pt] \color{green}\text{Time taken to fill the reservoir} &= \color{green}\frac{\text{Volume}}{\text{Rate}} \\[6pt] &= \frac{108000}{60} \\[6pt] &= 1800 \ \text{min} \\[6pt] \text{Time in hours} &= \frac{1800}{60} \\[6pt] \color{green}\text{Time taken to fill the reservoir}&= 30 \ \text{hours} \end{aligned} \]
Answer Time to fill the reservoir \(= \color{red}{30 \ \text{hours}}\)
10. An iron pipe is \(21 \ \text{m}\) long and its exterior diameter is \(8 \ \text{cm}\). If the thickness of the pipe is \(1 \ \text{cm}\) and iron weight is \(8 \ \text{g/cm}^3\), find the weight of the pipe.
Solution
\[ \begin{aligned} \text{Length }(h) &= 21 \ \text{m} \implies 2100 \ \text{cm} \\ \text{Outer diameter }(D) &= 8 \ \text{cm} \\ \text{Outer radius }(R) &= 4 \ \text{cm} \\ \text{Thickness} &= 1 \ \text{cm} \\ \text{Inner radius }(r) &= 3 \ \text{cm} \\[6pt] \color{green}\text{Volume of metal} &= \color{green}\pi h(R^2-r^2) \\[6pt] &= \pi \times 2100 \times (4^2-3^2) \\[6pt] &= \frac{22}{\cancel7_1} \times \cancel{2100}^{300} \times (16-9) \\[6pt] &= 22 \times 300 \times 7 \\[6pt] &= 6600 \times 7 \\ &= 46200 \ \text{cm}^3 \\[8pt] \text{Weight} &= \text{Volume} \times \text{density} \\ &= 46200 \times 8 \ \text{g} \\ &= 369600 \ \text{g} \\[6pt] 1000 \ g &= 1 \ kg \\[6pt] &= \frac{369600}{1000} \ \text{kg} \\[6pt] \color{green} \text{Weight of the pipe} & = \color{green}{369.6 \ \text{kg}} \end{aligned} \]
Answer Weight of the pipe \(= \color{red}{ 369.6 \ \text{kg}}\)
11. A well is dug \(20 \ \text{m}\) deep and it has a diameter \(7 \ \text{m}\). The earth which is so dug out is spread evenly on a rectangular plot \(22 \ \text{m}\) long and \(14 \ \text{m}\) broad. What is the height of the platform formed?
Solution
\[ \begin{aligned} \color{magenta}\textbf{Well} \\ \text{Depth }(h) &= 20 \ \text{m} \\ \text{Diameter} &= 7 \ \text{m} \\ r &= \frac{7}{2} \ \text{m} \\[6pt] \color{magenta}\textbf{Cuboid} \\ l &= 22 \ \text{m} \\ b &= 14 \ \text{m} \\ h &= ? \\ \color{green}\text{Volume of Cuboid} &= \color{green}\text{Volume of well} \\ \color{green} l \times b \times h &= \color{green}\pi r^2 h \\ 22 \times 14 \times h &= \frac{22}{7}\times \left(\frac{7}{2}\right)^2 \times 20 \\[6pt] 22 \times 14 \times h &= \frac{22}{\cancel7_1}\times \frac{\cancel{49}^7}{\cancel4_1} \times \cancel{20}^5 \\[6pt] h &= \frac{\cancel{22} \times \cancel7^1 \times 5}{\cancel{22} \times \cancel{14}_2} \\[6pt] h &= \frac{5}{2} \\[6pt] \text{Height} &= 2.5 \ \text{m} \end{aligned} \]
Answer Height of the platform \(= \color{red}{2.5 \ \text{m}}\)
12. The volume of a right circular cylinder is \(448\pi \ \text{cm}^3\) and height \(7 \ \text{cm}\). Find the lateral surface area and total surface area.
Solution
\[ \begin{aligned} h&=7 \ \text{cm} \\ \color{green}\text{Volume of cylinder} &= 448\pi \ \text{cm}^3 \\[6pt] \color{green}\pi r^2 h &= 448\pi \\[6pt] \color{green}\pi \times r^2 \times 7 &= 448 \times \pi \\[6pt] r^2 &= \frac{\cancel{448}^{64} \times \cancel{\pi}}{\cancel{\pi} \times \cancel{7}_1} \\[6pt] r^2 &= 64 \\[6pt] r &= \sqrt{64} \\[6pt] r &= 8 \ \text{cm} \\[8pt] \color{green}\text{CSA} &= \color{green}2\pi r h \\[6pt] &= 2 \times \frac{22}{\cancel7} \times 8 \times \cancel7 \\[6pt] &= 44 \times 8 \\[6pt] \color{green}\text{CSA} &= 352 \ \text{cm}^2 \\[6pt] \color{green}\text{TSA} &= \color{green} 2\pi r(r+h) \\[6pt] &= 2 \times \frac{22}{7} \times 8 \times (7+8) \\[6pt] &= \frac{44}{7} \times 8 \times 15 \\[6pt] &= \frac{44 \times 120}{7} \\[6pt] &= \frac{5280}{7} \\[6pt] \color{green}\text{TSA} &= 754.28 \ \text{cm}^2 \end{aligned} \]
Answer Lateral surface area \(=\ \color{red}{352 \ \text{cm}^2}\), Total surface area \(=\ \color{red}{754.28 \ \text{cm}^2}\)
13. Verify Euler’s formula for (i) Square pyramid, (ii) Triangular prism, (iii) Rectangular prism.
Solution
Euler’s Formula: \(\color{green}{F + V - E = 2}\)
(i) Square Pyramid
\[ \begin{aligned} F &= 5 \\ V &= 5 \\ E &= 8 \\[4pt] \implies& \color{green}{F + V - E} = 2 \\ &= 5 + 5 - 8 \\ &= \color{green}{2} \end{aligned} \](ii) Triangular Prism
\[ \begin{aligned} F &= 5 \\ V &= 6 \\ E &= 9 \\[4pt] \implies& \color{green}{F + V - E} = 2 \\ &=5 + 6 - 9 \\ &= \color{green}{2} \end{aligned} \](iii) Rectangular Prism (Cuboid)
\[ \begin{aligned} F &= 6 \\ V &= 8 \\ E &= 12 \\[4pt] \implies & \color{green}{F + V - E} = 2 \\ &= 6 + 8 - 12 \\ & = \color{green}{2} \\ \end{aligned} \]Answer Euler’s formula is verified.