Exercise 16 D
Question-1
Find the area of the trapezium if the:
(i) Parallel sides are $3 \mathrm{~cm}$ and $6 \mathrm{~cm}$ and perp. distance between them is $10 \mathrm{~cm}$.
(ii) Parallel sides are $25 \mathrm{~m}$ and $33 \mathrm{~m}$ and perp. distance between them is $20 \mathrm{~m}$.
Answer-1
(i) Area of the trapezium $=1 / 2 \times(3+6) \times 10 \mathrm{~cm}^{2}$
$\begin{aligned}&=9 \times 10 / 2 \mathrm{~cm}^{2} \\&=45 \mathrm{~cm}^{2}\end{aligned}$
(ii) Area of the trapezium $=1 / 2 \times(25+33) \times 20 \mathrm{~m}^{2}$
$=580 \mathrm{~m}^{2}$
Question-2
The area of a trapezium is $240 \mathrm{~m}^{2}$ and the sum of the parallel sides is $48 \mathrm{~m}$. Find the height.
Answer-2
Let, ' $h$ ' be the height of the trapezium
$1 / 2 \times 48 \times h=240$
Or, $h=10 m$
Question-3
The parallel sides of a trapezium are $4.36 \mathrm{~cm}$ and $3.18 \mathrm{~cm}$ and area is $18.85 \mathrm{~cm}^{2}$. Find the distance between the parallel sides.
Answer-3
Let ' $h$ ' be the distance between the parallel side,
$(1 / 2) \times(4.36+3.18) \times h=18.85$
Or, $7.54 \times \mathrm{h}=37.70$
Or, $h=37.7 / 7.54$
$=5 \mathrm{~cm}$
Question-4
The area of a trapezium is $475 \mathrm{~cm}^{2}$ and the height is $19 \mathrm{~cm}$. Find its two parallel sides if one side is $4 \mathrm{~cm}$ greater than the other.
Answer-4
Let, one side is $=x \mathrm{~cm}$
another side is $=(x+4) \mathrm{cm}$
$\therefore 1 / 2 \times 19 \times(2 x+4)=475$
$2 x+4=50$
$2 x=46$
$x=23$
So, two parallel sides are $23 \mathrm{~cm}, 27 \mathrm{~cm} \mathrm{~cm}$.
Question-5
The parallel sides of a trapezium are in the ratio $2: 5$ and the distance between the parallel sides is $10 \mathrm{~cm}$. if the area of the trapezium is $350 \mathrm{~cm}^{2}$, find the lengths of its parallel sides.
Answer-5
Let, the common ratio be $x$
So, sides of the trapezium are $2 x$ and $5 x$
$\therefore 1 / 2 \times 10 \times 7 x=350$
Or, $x=10$
So, lengths of parallel sides are $=20 \mathrm{~cm}$ and $50 \mathrm{~cm}$.
Question-6
In the given figure, $A D=B C=5 \mathrm{~cm}, A B=7 \mathrm{~cm}$. The parallel sides $A B, D C$ are $4 \mathrm{~cm}$ apart $D C$ $=\mathrm{cm}$. Find $x$ and the area of the trapezium $A B C D$.
Answer-6
$A B C D$ is a trapezium in which $A B=7 \mathrm{~cm}$,
$A D=B C=5 \mathrm{~cm}$,
$\mathrm{DC}=\mathrm{x} \mathrm{cm}$
$A B / / D C$,
Distance between $A B$ and $D C$ is $4 \mathrm{~cm}$
To find:
Value of $x=?$
In $\Delta \mathrm{BFC}, \angle \mathrm{BFC}=90^{\circ}$
$\mathrm{BC}^{2}=\mathrm{BF}^{2}+\mathrm{FC}^{2}($ using Phythagorean theorem $)$
$=>5^{2}=4^{2}+\mathrm{FC}^{2}$
$=>\mathrm{FC}^{2}=5^{2}-4^{2}$
$=>\mathrm{FC}^{2}=25-16=9$
$=>\mathrm{FC}=\sqrt{9}=\sqrt{3}^{2}=3 \mathrm{~cm}$
$=>\mathrm{DC}=\mathrm{DE}+\mathrm{EF}+\mathrm{FC}$
$=>\mathrm{DC}=3+7+3=13 \mathrm{~cm}$
$\mathrm{Now}$
$\mathrm{Area}=1 / 2(\mathrm{sum}$ of $/ / \mathrm{side}) \times \mathrm{h}$
$=1 / 2(7+13) \times 4$
$=1 / 2\langle 20 \times 4\rangle$
$=40 \mathrm{~cm}^{2}$
Question-7
The parallel sides of a trapezium are $7.5 \mathrm{~cm}, 3.9 \mathrm{~cm}$, and the other sides are each $2.6 \mathrm{~cm}$.
Find its area.
Answer-7
the parallel sides of a trapezium are $7.5 \mathrm{~cm}$ and $3.9 \mathrm{~cm}$
other sides are each $2.6 \mathrm{~cm}$
$=>$ isosceles trapezium
if we draw perpendicular from vertex of $3.9 \mathrm{~cm}$ side to $7.5 \mathrm{~cm}$ side
distance between Perpendicular Drawn \& nearest vertex of $7.5 \mathrm{~cm}$ would be
$=(7.5-3.9) / 2$
$=1.8 \mathrm{~cm}$
Perpendicular height Would be $=\sqrt{2} .6^{2}-1.8^{2}$
$=\sqrt{3} .52$
$=1.876 \mathrm{~cm}$
Area of Trapezium $=(1 / 2)(7.5+3.9)(1.876)$ $=10.69 \mathrm{~cm}^{2}$
Question-8
given figure show cross section of a concrete structure with measurement given below calculate area of cross section
Answer-8
(IMAGE TO BE ADDED)
Question-9
In the figure find:
(i) $\mathrm{AB}$
(ii) area of the trapezium $A B C D$.
Answer-9
(i) $D C^{2}=D E^{2}+E C^{2}$
$(10)^{2}=6^{2}+E C^{2}$
$E C^{2}=100-36=64$
$\mathrm{EC}=8$
$\therefore \mathrm{AB}=8 \mathrm{~cm}$
(ii) Area $=1 / 2 \times(8+2) \times 8$
$=4 \times 10$ $=40 \mathrm{~cm}^{2}$
Question-10
The cross-section of a tunnel perpendicular to its length is a trapezium $A B C D$ as shown in the figure. $A M=B N ; A B=4.4 \mathrm{~m} ; C D=3 \mathrm{~m}$. The height of the tunnel is $2.4 \mathrm{~m}$. The tunnel is 50 $\mathrm{m}$ long.
Calculate:
(i) the cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per $m^{2}$
(ii) the cost of paving the floor at the rate of Rs. 18 per $\mathrm{m}^{2}$
Answer-10
The cross-section of a tunnel is of the trapezium-shaped $A B C D$ in which $A B=7 \mathrm{~m}, C D=5$ $\mathrm{m}$ and $\mathrm{AM}=\mathrm{BN}$. The height is $2.4 \mathrm{~m}$ and its length is $40 \mathrm{~m}$.
(i) $A M=B N=(7-5) / 2=2 / 2=1 \mathrm{~m}$
$\therefore \ln \Delta \mathrm{ADM}$,
$\mathrm{AD}^{2}=\mathrm{AM}^{2}+\mathrm{DM}^{2} \quad \ldots[$ Using Pythagoras theorem $]=1^{2}+(2.4)^{2}$
$=1+5.76$
$=6.67$
$=(2.6)^{2}$
$\mathrm{AD}=2.6 \mathrm{~m}$
Perimeter of the cross- section of the tunnel $=(7+2.6+2.6+5) \mathrm{m}=17.2 \mathrm{~m}$
Length $=40 \mathrm{~m}$
$\therefore$ The internal surface area of the tunnel ( except the floor )
$=(17.2 \times 40-40 \times 7) \mathrm{m}^{2}$
$=(688-280) \mathrm{m}^{2}$
$=408 \mathrm{~m}^{2}$
Rate of painting $=$ Rs. 5 per $\mathrm{m}^{2}$
Hence, total cost of painting = Rs. $5 \times 408=$ Rs. 1632
(ii) Area of floor of tunnel $=1 \times b=40 \times 7=280 \mathrm{~m}^{2}$
Rate of cost of paving $=$ Rs. 18 per $\mathrm{m}^{2}$
Total cost $=280 \times 18=$ Rs. 5040
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