5 Marks Questions - MATHS STD 7 Questions - Vidyadip

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5 Marks Questions

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15 questions · timed · auto-graded

Question 15 Marks

A $5m$ long ladder whan set against the wall of a house reaches a height of $4.8m$. How far is the foot of the ladder from the wall?

Answer

Length of ladder $A B=5 \mathrm{~m}$
and height $C A=4.8 \mathrm{~m}$
Let distance of the ladder from the wall $B C=x \mathrm{~m}$.
Now in right angled $\triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}$
$A B^2=A C^2+B C^2(B y \text { Pythagoras Theorem) }$
$\Rightarrow(5)^2=(4.8)^2+x^2$
$\Rightarrow 25=23.04+x^2$
$\Rightarrow x^2=25.00-23.04$
$\Rightarrow x^2=1.96$
$\Rightarrow x^2=(1.4)^2$
$x=1.4$
The foot of ladder are $1.4 m$ away from the wall.

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Question 25 Marks

Two poles, $18m$ and $13m$ high, stand upright in a playground. If their feet are $12m$ apart, find the distance between their tops.

Answer

Let $A B$ and $C D$ are two poles and they are $12 m$ apart.
$A B ~18 m, C D=13 \mathrm{~m}$ and $B D=12 m$
From C, draw $CE \| BD$ Then
$C E=B D=12 m$
$\text { and } A E=A B-C D=18-13=5 m$
$\text { Join } A C$
Now, in right $\triangle \mathrm{ACE}$,
$A C^2=C E^2+A E^2 \text { (By Pythagoras Theorom) }$
$A C^2=(12)^2+(5)^2$
$\Rightarrow A C^2=144+25$
$\Rightarrow A C^2=169$
$\Rightarrow A C^2=(13)^2$
$A C=13 \mathrm{~m}$
Distance between their tops $=13 \mathrm{~m}$.

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Question 35 Marks

An exterior angle of a triangle measures $110^\circ $ and its interior opposite angles are in the ratio $2 : 3$. Find the angles of the triangle.

Answer

In $\triangle\text{ABC},$ side $BC$ is produced to $D$ forming exterior $\angle\text{ACD}$

$\angle\text{ACD} = 110^\circ,\text{and}\ \angle\text{A}:\angle\text{B}=2:3$
In a triangle, exterior angles is equal to sum of its interior opposite angles. $\Rightarrow\angle\text{ACD} = \angle\text{A}+\angle\text{B}$
$\Rightarrow\angle\text{A}+\angle\text{B}=110^\circ$
But $ \angle\text{A}:\angle\text{B}=2:3$
$\therefore\angle\text{A}=\frac{2}{2+3}\times110^\circ$
$=\frac{2}{5}\times110^\circ$
$={2}\times22^\circ=44^\circ$
And $\therefore\angle\text{B}=\frac{3}{2+3}\times110^\circ$
$=\frac{3}{5}\times110^\circ$
$={3}\times22^\circ=66^\circ$ But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Sum of angles of a triangle)
$\Rightarrow44^\circ+66^\circ+\angle\text{C}=180^\circ$
$\Rightarrow110^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-110^\circ=70^\circ$ Hence, $\angle\text{A}=44^\circ,\angle\text{B}=66^\circ$ and $\angle\text{C}=70^\circ$

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Question 45 Marks

A man goes $3\ km$ due north and then 4km due east. How far is he away from his initial position?

Answer

A man goes $3 \ km$ due north and then $4 \ km$ east.
In right angled $\triangle \mathrm{OAB}$,
$\mathrm{OA}=3 \mathrm{~km}, \mathrm{AB}=4 \mathrm{~km}$
$\mathrm{OB}^2=\mathrm{OA}^2+\mathrm{AB}^2 \text { (By Pythagoras Theorom) }$
$\mathrm{OB}^2=(3)^2+(4)^2$
$\Rightarrow \mathrm{OB}^2=9+16$
$\Rightarrow \mathrm{OB}^2=25$
$\Rightarrow \mathrm{OB}^2=(5)^2$
$\mathrm{OB}=5 \mathrm{~km}$
Hence he is $5 \ km$ away from the initial position.

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Question 55 Marks

A $15\ m$ long ladder is placed against a wall to reach a window $12\ m$ high. Find the distance of the foot of the ladder from the wall.

Answer

$A B$ is a ladder and it is $15 m$ long $B$ is window and $B C=12 \mathrm{~m}$.
In right $\triangle \mathrm{ABC}$,
$\mathrm{AB}^2=A C^2+\mathrm{BC}^2 \text { (By Pythagoras Theorem) }$
$\Rightarrow(15)^2=\mathrm{x}^2+(12)^2$
$\Rightarrow 225=\mathrm{x}^2+144$
$\Rightarrow \mathrm{x}^2=81$
$\Rightarrow \mathrm{x}^2=(9)^2$
$x=9 m$
Distance of the foot of ladder from the wall $=9 \mathrm{~m}$.

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Question 65 Marks

A man goes $35m$ due west and then $12m$ due north. How far is he from the starting point?

Answer

A man starts $O$ and goes $35 m$ due west and then $12 m$ due north, then
In right $\triangle \mathrm{OAB}$,
$\mathrm{OA}=35 \mathrm{~m}, \mathrm{AB}=12 \mathrm{~m}$
$\mathrm{OB}^2=\mathrm{OA}^2+\mathrm{AB}^2$ (By Pythagoras Theorom)
$\mathrm{OB}^2=(35)^2+(12)^2$
$\Rightarrow \mathrm{OB}^2=1225+144$
$\Rightarrow \mathrm{OB}^2=1369$
$\Rightarrow \mathrm{OB}^2=(37)^2$
$\mathrm{OB}=37 \mathrm{~m}$
Hence he is $37 m$ away from the starting point.

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Question 75 Marks

AM is a median of $\triangle\text{ABC}$. Prove that $(AB + BC + CA) > 2AM$. Hint: $(AB + BM) > AM$ (In $\triangle\text{ABM}$) $(AC + MC) > AM$ (In $\triangle\text{ABM}$)Now, add the two inequalities.

Answer

Proof: $AM$ is the median of $\triangle\text{ABC}$ $M$ is mid-point of $BC$ In $\triangle\text{ABM},$
$AB + BM > AM ...(i)$
(Sum of any two sides of a triangle is greater than its third side)
Similarly in $\triangle\text{ACM},$
$AC + MC > AM ...(ii)$
Adding $(i)$ and $(ii) AB + BM + AC + MC > 2AM$
$⇒ AB + AC + BM + MC > 2AM$
$⇒ AB + AC + BC > 2AM$ Hence proved.

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Question 85 Marks

An exterior angle of a triangle is $100^\circ $ and its interior opposite angles are equal to each other. Find the measure of each angle of the triangle.

Answer

In $\triangle\text{ABC},$ side $BC$ is produced to $D$ forming exterior $\angle\text{ACD}$. $\angle\text{ACD} = 100^\circ,\text{and}\ \angle\text{A}=\angle\text{B}$
Exterior angle of a triangle is equal to the sum of its interior opposite angles.

$\angle\text{ACD} = \angle\text{A}+\angle\text{B}$ But $\angle\text{A}=\angle\text{B}$
$\Rightarrow\angle\text{A}+\angle\text{A}=\angle\text{ACD}=100^\circ$
$\Rightarrow2\angle\text{A}=100^\circ$
$\Rightarrow\angle\text{A}=50^\circ$
$\angle\text{B}=\angle\text{A}=50^\circ$ But
$ \angle\text{A}+\angle\text{B}+\angle\text{ACB} =180^\circ$ (Sum of angles of a triangle)
$\Rightarrow50^\circ+50^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow100^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-100^\circ=80^\circ$
Hence, $\angle\text{A} = 50^\circ,\text{B} = 50^\circ\text{and}\ \angle\text{C}=80^\circ$

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Question 95 Marks

If O is a point in the exterior of $\triangle\text{ABC},$ show that $2(OA + OB + OC) > (AB + BC + CA)$.Hint: Join $OA, OB, OC$.
From $\triangle\text{ABC}, OA + OB > AB$. From $\triangle\text{BOC}, OB + OC > BC$. From $\triangle\text{AOC}, OA + OC > CA$.

Answer

Given, $O$ is any point outside of the $\triangle\text{ABC}$
To prove : $2(OA + OB + OC) > (AB + BC + CA)$
Construction : join $OA, OB$ and $DC.$
Proof : In $\triangle\text{AOB},$
$OA + OB > AB ...(i)$
(Sum of any two sides of a triangle is greater than its third side)
Similarly in $\triangle\text{BOC},$
$OB + OC > BC ...(ii)$
and in $\triangle\text{COA},$
$OC + OA > CA ...(iii)$
Adding $(i), (ii)$ and $(iii),$ we get:
$OA + OB + OB + OC + OC + OA > AB + BC + CA$
$2(OA + OB + OC) > (AB + BC + CA)$
Hence proved.

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Question 105 Marks

Find the perimeter of the rectangle whose length is $40\ cm$ and a diagonal is $41\ cm.$

Answer

Given,
$A B C D$ is a rectangle and $A C$ is its diagonal.
$\mathrm{AB}=40 \mathrm{~cm} \text { and } \mathrm{AC}=41 \mathrm{~cm}$
Now in right $\triangle \mathrm{ABC}$
$A C^2=A B^2+B C^2(B y \text { Pythagoras Theorom) }$
$\Rightarrow(41)^2=(40)^2+B C^2$
$\Rightarrow 1681=1600+B C^2$
$\Rightarrow B C^2=1681-1600$
$\Rightarrow B C^2=81$
$\Rightarrow B C^2=(9)^2$
$B C=9 \mathrm{~cm}$
Now perimeter of rectangle $A B C D=2(A B+B C)=2(40+9)=2 \times 49=98 \mathrm{~cm}$.

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Question 115 Marks

A tree is broken by the wind but does not separate. If the point from where it breaks is $9m$ above the ground and its top touches the ground at a distance of $12m$ from its foot, find out the total height of the tree before it broke.

Answer

Let $A B$ be the tree which broke at $D$ and its top $A$ touches the ground at $C$.
Their $B D=5 \mathrm{~m}$ and $\mathrm{BC}=12 \mathrm{~m}$
Let $A D=x m$, then $C D=x m$
Now, in right $\triangle \mathrm{ABC}$,
$C D^2=B D^2+\mathrm{BC}^2(\text { By Pythagoras Theorom })$
$C D^2=(9)^2+(12)^2$
$\Rightarrow C D^2=81+144$
$\Rightarrow C D^2=225$
$\Rightarrow C D^2=(15)^2$
$C D=15 \mathrm{~m}$
$A D=x=15 \mathrm{~m}$
Height of the tree $A B=A D+B D=15+9=24 m$.

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Question 125 Marks

In the given figure, P is a point on the side BC of$\triangle\text{ABC}$. Prove that $(AB + BC + AC) > 2AP$ Hint: $(AB + BP) > AP$ (In $\triangle\text{ABP}) (PC + AC) > AP$ (In $\triangle\text{APC}$)Now, add the corresponding sides.

Answer

Given, In $\triangle\text{ABC}$, $P$ is a point on $BC. AP$ is joined.
To Prove: $(AB + BC + AC) > 2AP$
Proof : In $\triangle\text{ABP},$
$AB + BC + AB > AP...(i)$
is the median of $\triangle\text{ABC}$
M is mid-point of $BC$
In $\triangle\text{ABM},$
$AB + BM > AM …(i)$ (Sum of any two sides of a triangle is greater than its third side)
Similarly in $\triangle\text{ACP},$
$AC + PC > A...(ii)$
Adding $(i)$ and $(ii)$
$AB + BP + AC + PC > AP + AP$
$\Rightarrow AB + BP + PC + CA > 2AP$
$\Rightarrow AB + BC + CA > 2AP$ [$\because BC = BP + PC]$
Hence proved.

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Question 135 Marks

In the figure given alongside, find:
$i. \angle\text{ACD}$
$ii. \angle\text{ADC}$
$iii. \angle\text{DAE}$

Answer

$i.$ In $\triangle\text{ABC},$ sides $BC$ is produced to $D$ and $BA$ to $E.$
$\angle\text{CAD} = 50^\circ,\angle\text{B} = 40^\circ$ and $\angle\text{ACB} = 100^\circ$
$\angle\text{ACB} +\angle\text{ACD} = 180^\circ ($Linear pair$)$
$\Rightarrow100^\circ+\angle\text{ACD} = 180^\circ$
$\Rightarrow\angle\text{ACD} = 180^\circ-100^\circ=80^\circ$

$ii.$ In $\triangle\text{ABC},$
$\angle\text{CAD} +\angle\text{ACD}++\angle\text{ADC} = 180^\circ ($sum of angles of a triangle$)$
$\Rightarrow50^\circ+80^\circ+\angle\text{ADC} = 180^\circ$
$\Rightarrow130^\circ+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ADC}=180^\circ-130^\circ=50^\circ$
$iii.$ Now, in $\triangle\text{ABD}, BA$ is produced to $E.$
Exterior $\angle\text{DAE} =\angle\text{ACD} +\angle\text{ADC}=80^\circ+50^\circ=130^\circ$
Hence, $\angle\text{ACD}=80^\circ,\angle\text{ADC}=50^\circ$and $\angle\text{DAE}=130^\circ$

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Question 145 Marks

In the figure given alongside, $x : y = 2 : 3$ and $\angle\text{ACD}=130^\circ$. Find the values of $x, y$ and $z$.

Answer

In $\triangle\text{ABC},$ sides $BC$ is produced to $D$ forming exterior $\angle\text{ACD}$

$\angle\text{ACD} = 130^\circ,\angle\text{A} = \text{y}^\circ,\angle\text{B} = \text{x}^\circ$ and $\angle\text{ACB} = \text{z}^\circ$
$x : y = 2 : 3$
Now, in $\triangle\text{ABC},$
Exterior $\angle\text{ACD} =\angle\text{A}+\angle\text{B}$
$\Rightarrow\angle\text{A}+\angle\text{B}= 130^\circ$
But $\angle\text{A} : \angle\text{B}= 2:3$
$\therefore\angle\text{B}=\frac{130^\circ\times2}{2+3}$
$=\frac{130^\circ\times2}{5}$
$=26^\circ\times2$
$=52^\circ$ and $\angle\text{A}=\frac{130^\circ\times3}{2+3}$
$=\frac{130^\circ\times3}{5}$
$=26^\circ\times3$
$=78^\circ$ But, $\angle\text{A} +\angle\text{B}++\angle\text{ACB} = 180^\circ$ (sum of angles of a triangle)
$\Rightarrow78^\circ+52^\circ+\angle\text{ACB} = 180^\circ$
$\Rightarrow130^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-130^\circ=50^\circ$
Hence, $\angle\text{x}=52^\circ,\angle\text{y}=78^\circ$and $\angle\text{z}=50^\circ$

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Question 155 Marks

In the given figure. $DE \| BC$. If $\angle\text{C}=65^\circ$ and $\angle\text{B}=55^\circ$, find
$i. \angle\text{ADE}$
$ii. \angle\text{AED}$
$iii. \angle\text{C}$

Answer

$i. DE \| BC$
$\therefore\angle\text{ABC}=\angle\text{ADE}=55^\circ ($corresponding angles$)$
$ii.$ Sum of the angles of any triangle is $180^\circ$.
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-(65^\circ-55^\circ)=60^\circ$

$DE \| BC$
$\therefore \angle\text{AED}=\angle\text{ACB}=60^\circ ($corresponding angles$)$
$iii.$ We have found in point $(ii)$ that $\angle\text{C}=60^\circ$

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