5 Marks Questions

Question 15 Marks

A pole $6 m$ high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point $P$ on the ground is $60^{\circ}$ and the angle of depression of the point $P$ from the top of the tower is $45^{\circ}$. Find the height of the tower and the distance of point $P$ from the foot of the tower. $($Use $\sqrt{3}=1.73)$

Answer

Let $B C$ be the pole and $A B$ be the tower of height ' $h$ ' $m$.
$\tan 45^{\circ}=1=\frac{h}{x}$
$\Rightarrow h=x-\cdots \text { (i) }$
$\tan 60^{\circ}=\sqrt{3}=\frac{h+6}{x}$
$\Rightarrow h+6=x \sqrt{3}....(ii)$
$\text { Solving (i) (ii) to get }$
$h=3(\sqrt{3}+1)=8.19$
$\text { and } x=8.19$
Therefore, the height of tower is $8.19 m$ and the distance of point $P$ from the foot of the tower is $8.19 m$

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Question 25 Marks

In the given figure $PA , QB$ and $RC$ are each perpendicular to $AC$ . If $AP =x, BQ =y$ and $CR =z$, then prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$

Answer

$\Delta PAC \sim \Delta QBC$
$\therefore \frac{x}{y}=\frac{A C}{B C} \text { or } \frac{y}{x}=\frac{B C}{A C} \text {..... (i) }$
$\Delta RCA \sim \triangle QBA$
$\therefore \frac{z}{y}=\frac{A C}{A B} \text { or } \frac{y}{z}=\frac{A B}{A C}......(ii)$
Adding $(i)$ and $(ii)$
$\frac{y}{x}+\frac{y}{z}=\frac{B C+A B}{A C}$
$\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{1}{y}$

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Question 35 Marks

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer

Correct figure, given, to prove and construction Correct proof

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Question 45 Marks

In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.

Answer

Here $n =40$,

$S_9=\frac{9}{2}[2 a+8 d]=153 \Rightarrow a+4 d=17......(i)$
and $S_{40}-S_{34}=687$ or $a_{35}+a_{36}+a_{37}+a_{38}+a_{39}+a_{40}=687$
$\Rightarrow 6 a +219 d=687$ or $2 a +73 d=229.....$ (ii)
solving (i) and (ii) to get $a =5, d=3$
Also, $S _{40}=\frac{40}{2}(10+39 \times 3)=2540$

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Question 55 Marks

The sum of first and eighth terms of an $A.P.$ is $32$ and their product is $60$ . Find the first term and common difference of the $A.P.$ Hence, also find the sum of its first $20$ terms.

Answer

$a+a_8=32$
$\Rightarrow 2 a+7 d=32 .....(i)$
$a \times a_8=60$
$\Rightarrow a(a+7 d)=60......(ii)$
Solving $(i) \ (ii)$, we get
$a=2 \text { or } a=30$
and $d=4$ or $d=-4$
First term and common difference of $A.P.$ are $2$ and $4$ or $30$ and $-4$ respectively.
$\text { Now, for } a=2 \ d=4$
$S_{20}=10(4+76)=800$
$\text { and for } a=30 \ d=-4$
$S_{20}=10(60-76)=-160$

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Question 65 Marks

An arc of a circle of radius 21 cm subtends an angle of $60^{\circ}$ at the centre. Find :
(i) the length of the arc.
(ii) the area of the minor segment of the circle made by the corresponding chord.

Answer

(i) Length of the arc $AB =2 \times \frac{22}{7} \times 21 \times \frac{60}{360}$
$=22 cm$
(ii) Area of sector OALB $=\frac{22}{7} \times 21 \times 21 \times \frac{60}{360}=231 cm^2$
Area of $\triangle OAB =\frac{\sqrt{3}}{4} \times 21 \times 21=\frac{441 \sqrt{3}}{4} cm^2$
Area of minor segment $=\left(231-\frac{441 \sqrt{3}}{4}\right) cm ^2$
or $(231-190.95)=40.05 cm^2$

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