Given:The length of the greenhouse $l$ $=30\ cm$.The breadth of the greenhouse $b$ $=25\ cm$.The height of the greenhouse $h$ $=25\ cm$.To do:We have to find:(i) The is the area of the glass.(ii) The tape needed for all the 12 edges.Solution:The area of the glass $=2lb+2bh+2lh$$=2{lb+bh+lh}$This implies, $=2{30\times25}+{25\times25}+{30\times25}$$=2{750+625+750}$$=(2\times2125)$$=4250\ cm^2$Therefore, The ... Read More

Given:Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions \( 25 \mathrm{~cm} \times 20 \mathrm{~cm} \times 5 \mathrm{~cm} \) and the smaller of dimensions \( 15 \mathrm{~cm} \times 12 \mathrm{~cm} \times 5 \mathrm{~cm} \). ... Read More

Given:Parveen wanted to make a temporary shelter for her car by making a box-like structure with a tarpaulin that covers all four sides and the top of the tire car (with the front face as a flap which can be rolled up). The shelter of height $2.5\ m$ with base dimensions ... Read More

Given:An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring \( 20 \mathrm{~cm}, 50 \mathrm{~cm} \) and \( 50 \mathrm{~cm} \).To do:We have to find the cloth of each colour required for the umbrella.Solution:Let the side of each triangular piece of cloth ... Read More

Given:A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being \( 9 \mathrm{~cm}, 28 \mathrm{~cm} \) and \( 35 \mathrm{~cm} \) To do: We have to find the cost of polishing the tiles at the rate \( 50 \mathrm{p} \) per ... Read More

Given:A triangle and a parallelogram have the same base and the same area.The sides of a triangle are $26\ cm, 28\ cm$ and $30\ cm$, and the parallelogram stands on the base $28\ cm$.To do:We have to find the height of the parallelogram.Solution:The sides of the triangle are $a=26\ cm, ... Read More

Given:A park, in the shape of a quadrilateral $ABCD$ has $\angle C=90^o$, $AB=9\ m$, $BC=12\ m$, $CD=5\ m$ and $AD=8\ m$. To do:We have to find the area it occupies.Solution: In $\triangle BCD$, using Pythagoras theorem, $BD^2=BC^2+CD^2$$BD^2=(12)^2+(5)^2$$BD^2=144+25=169$​$BD^2=(13)^2$$BD=13\ m$Therefore, Area of $\triangle BCD=\frac{1}{2}\times12\times5=6\times5=30\ m^2$In $\triangle ABD$, $s=\frac{1}{2}(8+9+13)=15\ m$Area of $\triangle ABD=\sqrt{s(s-a)(s-b)(s-c)}$​Area of $\triangle ... Read More

Given :The figure of the aeroplane made with coloured paper is given.To find :We have to find the total area of the colour paper used.Solution :The total area of the paper used $= part I + part II + part III + part IV + part V$Part I : It is ... Read More

Given:The quadrilateral $ABCD$ has $AB=3\ cm$, $BC=4\ cm$, $CD=4\ cm$, $DA=5\ cm$ and $AC=5\ cm$. To do:We have to find the area of the quadrilateral $ABCD$.Solution: In $\triangle ABC$, using Pythagoras theorem, $AC^2=AB^2+BC^2$$5^2=3^2+4^2$​$25=25$Therefore, Area of $\triangle ABC=\frac{1}{2}\times3\times4=6\ cm^2$In $\triangle ACD$, $s=\frac{1}{2}(5+5+4)$$=\frac{14}{2}$$=7\ cm$Area of $\triangle ACD=\sqrt{s(s-a)(s-b)(s-c)}$$=\sqrt{7(7-5)(7-5)(7-4)}$$=\sqrt{(7)(2)(2)(3)}$$=2\sqrt{21}\ cm^2$$=9.17\ cm^2$Therefore, Area of quadrilateral $ABCD=$ Area ... Read More

Given:$BC=8\ cm, \angle B=45^o$ and $AB-AC=3.5\ cm$.To do:We have to construct a $\triangle ABC$.Solution:Steps of construction:(i) Let us draw a line segment $BC$ of length $8\ cm$.(ii) Now, construct an angle $XBC$ such that $\angle XBC=45^o$(iii) Now, by taking a measure of $AB-AC=3.5\ cm$ with the compasses, let us draw ... Read More