SChand CLASS 9 Chapter 16 Area of plane figures Exercise 16(D)

 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