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Maths 3 Marks Question

Triangles · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 68 questions.

Questions · Page 2 of 2

3 Marks Question

Question 513 Marks
The diagonals of a quadrilateral ABCD intersect each other at the point O such that  $\frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}$. Show that ABCD is a trapezium.
Answer
Given: The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}$
To prove: ABCD is trapezium.
Construction: Through O draw a line OE||BA intersecting AD at E.
Proof: In $\triangle DBA$$\because OE||BA$
 
$\therefore \frac{{DO}}{{BO}} = \frac{{DE}}{{AE}} \Rightarrow \frac{{CO}}{{AO}} = \frac{{DE}}{{AE}}$
$\because \frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}\,[Given]$
$\Rightarrow \frac{{DO}}{{BO}} = \frac{{CO}}{{AO}} \Rightarrow \frac{{AO}}{{CO}} = \frac{{AE}}{{DE}}$.........[Taking reciprocals]
$\therefore $In $\triangle ADC$
OE $\parallel$ CD ...........[By converse basic proportionality theorem]
But OE $\parallel$ BA
BA $\parallel$ CD........[By construction]
The quadrilateral ABCD is a Trapezium.
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Question 523 Marks

Give two different examples of each of the following:

  1. similar figures.
  2. non-similar figures.
Answer
  1. Two examples of similar figures are:
    1. Two equilateral triangles with sides 1 cm and 2 cm respectively
    2. Two squares with sides 1 cm and 2 cm respectively
  2. Now two examples of non-similar figures are:
    1. Trapezium and square
    2. Triangle and parallelogram
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Question 533 Marks
Give two different examples of each of the following:
  1. similar figures.
  2. non-similar figures.
Answer
  1. Two examples of similar figures are:
    1. Two equilateral triangles with sides 1 cm and 2 cm respectively
    2. Two squares with sides 1 cm and 2 cm respectively
  2. Now two examples of non-similar figures are:
    1. Trapezium and square
    2. Triangle and parallelogram
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Question 543 Marks
In the given figure, the line segment XY is parallel to side AC of $\triangle$ABC and it divides the triangle into two parts of equal areas. Find the ratio $\frac{\mathrm{A} \mathrm{X}}{\mathrm{AB}}$.
Answer
Since XY $\parallel$ AC
​$\therefore $​ ​$\angle$​BXY = ​$\angle$​BAC
​$\angle$​BYX = ​$\angle$BCA​ [Corresponding angles]
​$\therefore $​ ​$\triangle $​BXY ​$\cong$​ $\triangle $BAC [AA similarity]
​$\therefore $​ ​​  $\frac{{ar\left( {\triangle BXY} \right)}}{{ar\left( {\triangle BAC} \right)}} = \frac{{B{X^2}}}{{B{A^2}}}$
But ar(​$\triangle $​BXY)=ar(XYCA)
​$\therefore $​ 2(​$\triangle $​BXY)=ar($\triangle $BXY)+ar(XYCA)
= ar($\triangle $BAC)
$\therefore \frac{{ar\left( {\triangle BXY} \right)}}{{ar\left( {\triangle BAC} \right)}} = \frac{1}{2}$
$\therefore \frac{{B{X^2}}}{{B{A^2}}} = \frac{1}{2}$
​$\Rightarrow $​ $\therefore \frac{{BX}}{{BA}} = \frac{1}{{\sqrt 2 }}$
​$\therefore $​$\frac{{BA - BX}}{{BA}} = \frac{{\sqrt 2 - 1}}{{\sqrt 2 }}$
​$\Rightarrow $​$\frac{{AX}}{{AB}} = \frac{{\sqrt 2 - 1}}{{\sqrt 2 }} = \frac{{2 - \sqrt 2 }}{2}$
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Question 553 Marks
In Fig. CM and RN are respectively the medians of $\triangle$ABC and $\triangle$PQR. If $\triangle$ABC $\sim\triangle$PQR, prove that:
  1. $\triangle$AMC $\sim\triangle$PNR
  2. $\frac{C M}{R N}=\frac{A B}{P Q}$
  3. $\triangle$CMB $\sim\triangle$RNQ
Answer
  1. $\triangle$ABC $\sim\triangle$PQR (Given)
    So, $\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$ ...(1) (corresponding sides of similar triangles are proportional)
    and $\angle$A = $\angle$P, $\angle$B = $\angle$Q and $\angle$C = $\angle$R ...(2)
    But AB = 2 AM and PQ = 2 PN (As CM and RN are medians)
    So, from (1),
    i.e., $\frac{A M}{P N}=\frac{C A}{R P}$ ...(3)
    Also, ∠ MAC = ∠ NPR [From (2)] ...(4)
    So, from (3) and (4),
    $\triangle$AMC $\sim\triangle$PNR (SAS similarity criterion) ...(5)
  2. From (5), $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{CA}}{\mathrm{RP}}$  ...(6) (corresponding sides of similar triangles are proportional)
    But $\frac{C A}{R P}=\frac{A B}{P Q}$ [From (1)] ...(7)
    Therefore,$\frac{C M}{R N}=\frac{A B}{P Q}$ [From (6) and (7)] ...(8)
  3. Again,$\frac{A B}{P Q}=\frac{B C}{Q R}$  [From (1)]
    Therefore $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ [From (8)] ...(9)
    Also, $\frac{C M}{R N}=\frac{A B}{P Q}=\frac{2 B M}{2 Q N}$
    i.e., $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BM}}{\mathrm{QN}}$ ...(10)
    i.e.,$\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BM}}{\mathrm{QN}}$ [From (9) and (10)]
    Therefore, $\triangle$CMB $\sim\triangle$RNQ (SSS similarity criterion)
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Question 563 Marks

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Answer
We have,

Height of girl = 90 cm = 0.9 m
Height of lamp-post = 3.6 m
Speed of girl = 1.2 m/sec
Time taken = 4 sec.
$\therefore$ Distance moved by girl (CQ) = Speed $\times$ Time
= 1.2 $\times$ 4
= 4.8 m
Let length of shadow (AC) = x cm
In $\Delta$ABC and $\Delta$APQ
$\angle ACB = \angle AQP$ [Each 90°]
$\angle BAC = \angle PAQ$ [Common]
Then, $\Delta$ABC ~ $\Delta$APQ [By AA similarity]
$\therefore \frac{{AC}}{{AQ}} = \frac{{BC}}{{PQ}}$ [Corresponding parts of similar $\Delta$ are proportional]
$ \Rightarrow \frac{x}{{x + 4.8}} = \frac{{0.9}}{{3.6}}$
$\Rightarrow \frac{x}{{x + 4.8}} = \frac{1}{4}$
$\Rightarrow$ 4x = x + 4.8
$\Rightarrow$ 4x - x = 4.8
$\Rightarrow$ 3x = 4.8
$\Rightarrow x = \frac{{4.8}}{3} = 1.6m$
$\therefore$ Length of shadow = 1.6 m
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Question 573 Marks
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Answer
We have,

Height of girl = 90 cm = 0.9 m
Height of lamp-post = 3.6 m
Speed of girl = 1.2 m/sec
Time taken = 4 sec.
$\therefore$ Distance moved by girl (CQ) = Speed $\times$ Time
= 1.2 $\times$ 4
= 4.8 m
Let length of shadow (AC) = x cm
In $\Delta$ABC and $\Delta$APQ
$\angle ACB = \angle AQP$ [Each 90°]
$\angle BAC = \angle PAQ$ [Common]
Then, $\Delta$ABC ~ $\Delta$APQ [By AA similarity]
$\therefore \frac{{AC}}{{AQ}} = \frac{{BC}}{{PQ}}$ [Corresponding parts of similar $\Delta$ are proportional]
$ \Rightarrow \frac{x}{{x + 4.8}} = \frac{{0.9}}{{3.6}}$
$\Rightarrow \frac{x}{{x + 4.8}} = \frac{1}{4}$
$\Rightarrow$ 4x = x + 4.8
$\Rightarrow$ 4x - x = 4.8
$\Rightarrow$ 3x = 4.8
$\Rightarrow x = \frac{{4.8}}{3} = 1.6m$
$\therefore$ Length of shadow = 1.6 m
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Question 583 Marks
ABCD is a trapezium with AB || DC. E and F are two points on non-parallel sides AD and BC respectively, such that EF is parallel to AB. Show that $\frac{AE}{ED}=\frac{BF}{FC}$
Answer
Given, In trapezium ABCD,
$AB || DC$ and $EF || DC$
To prove $\frac{AE}{ED}=\frac{BF}{FC}$
Construction: Join AC to intersect EF at G.

Proof Since, $AB || DC$ and $EF || DC$
$EF || AB$ [since, lines parallel to the same line are also parallel to each other ]...... (i)
In $\triangle ADC$, $EG || DC$ [$\because $ $EF || DC$]
By using basic proportionality theorem,
$\frac{AE}{ED}=\frac{AG}{GC}$ ....(ii)
In $\triangle ABC$, $ GF || AB$ [$\because $ $EF || AB$ from (i)]
By using basic proportionality theorem ,
$\frac{CG}{AG}=\frac{CF}{BF}$ or $\frac{AG}{GC}=\frac{BF}{CF}$ [ On taking reciprocal of the terms]............. (iii)
From Equations (ii) and (iii), we get
$\frac{AE}{ED}=\frac{BF}{FC}$
Hence Proved.
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Question 593 Marks
ABCD is a trapezium with AB || DC. E and F are two points on non-parallel sides AD and BC respectively, such that EF is parallel to AB. Show that $\frac{AE}{ED}=\frac{BF}{FC}$ 
Answer
Given, In trapezium ABCD,
$AB || DC$ and $EF || DC$ 
To prove $\frac{AE}{ED}=\frac{BF}{FC}$ 
Construction: Join AC to intersect EF at G.

Proof Since, $AB || DC$ and $EF || DC$ 
$EF || AB$ [since, lines parallel to the same line are also parallel to each other ]...... (i)
In $\triangle ADC$, $EG || DC$ [$\because $ $EF || DC$]
By using basic proportionality theorem,
$\frac{AE}{ED}=\frac{AG}{GC}$ ....(ii)
In $\triangle ABC$, $ GF || AB$ [$\because $ $EF || AB$  from (i)]
By using basic proportionality theorem ,
$\frac{CG}{AG}=\frac{CF}{BF}$ or $\frac{AG}{GC}=\frac{BF}{CF}$ [ On taking reciprocal of the terms]............. (iii)
From Equations (ii) and (iii), we get
$\frac{AE}{ED}=\frac{BF}{FC}$ 
Hence Proved.
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Question 603 Marks
$O$ is any point inside a rectangle $ABCD$ see Fig. Prove that $OB^2 + OD^2 = OA^2 + OC^2 .$
Answer
Through $O,$ draw $PQ \| BC$ so that $P$ lies on $AB$ and $Q$ lies on $DC.$
Now, $PQ \| BC$
Therefore, $PQ \perp AB$ and $PQ \perp DC  ( \angle B =  90^{\circ}$ and $\angle C = 90^\circ )$
So, $\angle BPQ = 90^\circ$ and $\angle CQP = 90^\circ$
Therefore, $BPQC$ and $APQD$ are both rectangles.
Now, from $\triangle OPB$
$OB^2 = BP^2 + OP^2 ...(1)$
Similarly, from $\triangle OQD,$
$OD^2 = OQ^2 + DQ^2 ...(2)$
From $\triangle OQC,$ we have
$OC^2 = OQ^2 + CQ^2...(3)$
and from $\triangle OAP,$ we have
$OA^2 = AP^2 + OP^2 ...(4)$
Adding $(1)$ and $(2),$
$OB^2 + OD^2 = BP^2 + OP^2+ OQ^2 + DQ^2$
$= CQ^2 + OP^2 + OQ^2 + AP^2 ($As $BP = CQ$ and $DQ = AP)$
$= CQ^2 + OQ^2 + OP^2 + AP^2$
$OB^2 + OD^2= OC^2 + OA^2 [$From $(3)$ and $(4)]$
Hence proved.
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Question 613 Marks
$O$ is any point inside a rectangle $ABCD$ see Fig. Prove that $OB^2 + OD^2 = OA^2 + OC^2$
.
Answer
Through $O,$ draw $PQ \| BC$ so that $P$ lies on $AB$ and $Q$ lies on $DC.$
Now, $PQ \| BC$
Therefore, $PQ \perp AB$ and $PQ \perp  DC ( \angle B = 90^\circ$ and $\angle C = 90^\circ )$
So, $\angle BPQ = 90^\circ$ and $\angle CQP = 90^\circ$
Therefore, $BPQC$ and $APQD$ are both rectangles.
Now, from $\triangle OPB$
$OB^2 = BP^2 + OP^2 ...(1)$
Similarly, from $\triangle OQD,$
$OD^2 = OQ^2 + DQ^2 ...(2)$
From $\triangle OQC,$ we have
$OC^2 = OQ^2 + CQ^{2 }...(3)$
and from $\triangle OAP,$ we have
$OA^2 = AP^2 + OP^2 ...(4)$
Adding $(1)$ and $(2),$
$OB^2 + OD^2 = BP^2 + OP^2+ OQ^2 + DQ^2$
$= CQ^2 + OP^2 + OQ^2 + AP^2 ($As $BP = CQ$ and $DQ = AP)$
$= CQ^2 + OQ^2 + OP^2 + AP^2$
$OB^2 + OD^{2 }= OC^2 + OA^2 [$From $(3)$ and $(4)]$
Hence proved.
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Question 623 Marks
$BL$ and $CM$ are medians of ​$\triangle $​ABC$ right angled at $A.$ Prove that $4(BL^2+ CM^2) = 5BC^2$
Answer

$BL$ and $CM$ are medians of a $\triangle ABC$ in which $\angle  A=90^\circ$
From $\triangle  ABC, BC^2 = AB^2 + AC^2 ....(i)$
From right angled $ \vartriangle ABL,$
$BL^2 = AL^2+ AB^2$
i.e $B{L^2} = {\left( {\frac{{AC}}{2}} \right)^2} + A{B^2}$
$\Rightarrow  4BL^2= AC^2+ 4AB^2 .....(ii)$
From right angled  $\triangle CMA,$
$CM^2 = AC^2+ AM^2$
i.e $C{M^2} = A{C^2} + {\left( {\frac{{AB}}{2}} \right)^2}[$mid$-$point$]$
$\Rightarrow  C{M^2} = A{C^2} + \frac{{A{B^2}}}{4}$
$\Rightarrow  4C{M^2} = 4A{C^2} + A{B^2} .....(iii)$
Adding $(ii)$ and $(iii),$ we get
i.e. $4(BL^2 + CM^{2)} = 5(AC^2+ AB^2) = 5BC^2 [$From $(i)]$
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Question 633 Marks
$BL$ and $CM $ are medians of ​$\triangle ​ABC$ right angled at $A.$ Prove that $4(BL^2+ CM^2) = 5BC^2$​​​​​​​
Answer

$BL$ and $CM$ are medians of a $\triangle ABC$ in which $\angle A=90^\circ$
From $\triangle ABC, BC^2 = AB^2 + AC^2 ....(i)$
From right angled $ \triangle ABL,$
$BL^2 = AL^2+ AB^2$
i.e $B{L^2} = {\left( {\frac{{AC}}{2}} \right)^2} + A{B^2}$
$\Rightarrow 4BL^2= AC^2+ 4AB^2 .....(ii)$
From right$-$angled $\triangle CMA,$
$CM^2 = AC^2+ AM^2$
i.e $C{M^2} = A{C^2} + {\left( {\frac{{AB}}{2}} \right)^2}[$mid$-$point$]$
$\Rightarrow  C{M^2} = A{C^2} + \frac{{A{B^2}}}{4}$
$\Rightarrow  4C{M^2} = 4A{C^2} + A{B^2} .....(iii)$
Adding $(ii)$ and $(iii),$ we get
i.e. $4(BL^2 + CM^{2)} = 5(AC^2+ AB^2) = 5BC^2 [$From $(i)]$
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Question 643 Marks
In the given figure, if $AD \bot BC$, prove that $AB^2 + CD^2 = BD^2 + AC^2.$
Answer

In right angled $\triangle BDA$,
By pythagoras theorem
$AB^2 = AD^2 + BD^2 ...(i)$
And in right angled $\triangle CDA$ ,
By pythagoras theorem
$AC^2 = CD^2 + AD^2 ...(ii)$
On subtracting Eq$(ii)$ from Eq$(i) ,$ we get
$AB^2 - AC^2 = [AD^2 + BD^2] - [CD^2 + AD^2]$
$AB^2 - AC^2 = AD^2 + BD^2 - CD^2 - AD^2$
$AB^2 - AC^2 = BD^2 - CD^2$
$\therefore AB^2 + CD^2 = BD^2 + AC^2$​​​​​​​
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Question 653 Marks
In the given figure, $\angle$ ACB = 90° and CD $\perp$ AB. prove that $\frac { B C ^ { 2 } } { A C ^ { 2 } } = \frac { B D } { A D }$.
Answer

 ​​​​​​In $\triangle $ACD and $\triangle $ ABC $\angle A = \angle A $ (Common)  $\angle ADC = \angle ACB$ ( each 90°) Thus, By AA similarity criteria  $\triangle ADC \sim \triangle ACB$ Thus, $\frac {AD} {AC} = \frac {AC} {AB} $ $\Rightarrow AC^2 = AD ×AB$... (i)  Similarly, $$$\triangle CDB \sim \triangle ACB $ And, $\frac {DC} {BC} = \frac {BC} {AB} $ $\Rightarrow BC^2 = DB ×AB$... (ii)  Dividing (ii) by (i)  $\frac {BC^2 } {AC^2 } = \frac {DB} {AD} $ Hence Proved  $$
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Question 663 Marks
If a line intersects the sides AB and AC of a $\triangle$ABC at D and E respectively and is parallel to BC, prove that $\frac{A D}{A B}=\frac{A E}{A C}$ (see figure).
Answer
DE || BC (Given)
therefore $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$    (by BPT)
or $\frac{D B}{A D}=\frac{E C}{A E}$     (by taking reciprocal on both sides)
or $\frac{D B}{A D}+1=\frac{E C}{A E}+1$   (add 1 on both sides)
we get  $\frac{A B}{A D}=\frac{A C}{A E}$
So $\frac{A D}{A B}=\frac{A E}{A C}$
Hence proved
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Question 673 Marks
If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
Answer
Image
This theorem can be proved by taking two triangles $ABC$ and $DEF$ such that $\frac{ AB }{ DE }=\frac{ BC }{ EF }=\frac{ CA }{ FD }(<1)($ see Fig. 6.26):
Cut $DP = AB$ and $DQ = AC$ and join $PQ$.
It can be seen that $\frac{ DP }{ PE }=\frac{ DQ }{ QF }$ and $PQ \| EF$ (How?)
So, $\angle P =\angle E$ and $\angle Q =\angle F$.
Therefore,
$\frac{ DP }{ DE }=\frac{ DQ }{ DF }=\frac{ PQ }{ EF }$
So,
$\frac{ DP }{ DE }=\frac{ DQ }{ DF }=\frac{ BC }{ EF } \quad \text { (Why?) }$
So,
$BC = PQ \quad \text { (Why?) }$
Thus,
$\Delta ABC \cong \Delta DPQ$
(Why ?)
So,
$\angle A =\angle D , \angle B =\angle E \text { and } \angle C =\angle F \text { (How ?) }$
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Question 683 Marks
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar
Answer
Image
This criterion is referred to as the SAS (Side-Angle-Side) similarity criterion for two triangles.
As before, this theorem can be proved by taking two triangles $ABC$ and DEF such that $\frac{ AB }{ DE }=\frac{ AC }{ DF }(<1)$ and $\angle A =\angle D$ (see Fig. 6.28). Cut DP = AB, DQ $= AC$ and join $PQ$.
$\text { Now, } PQ \| EF \text { and } \triangle ABC \cong \triangle DPQ $
$\text { So, } \angle A =\angle D , \angle B =\angle P \text { and } \angle C =\angle Q $
$\text { Therefore, } \Delta ABC \sim \Delta DEF $
$\text { We now take some examples to illustrate the use of these criteria. }$
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3 Marks Question - Page 2 - Maths STD 10 Questions - Vidyadip