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Maths 1 Marks Question

Triangles · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 12 questions.

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12 questions · timed · auto-graded

Question 11 Mark
In the figure, altitudes $AD$ and $CE$ of $\triangle$ABC intersect each other at the point $P$. Show that: $\vartriangle PDC \sim \vartriangle BEC$
Answer
In $\triangle $$PDC$ and $\triangle $$BEC$, we have
$\angle$PDC =$\angle$$BEC ..........(1)$ [Each equal to $90^\circ$]
$\angle$DCP = $\angle$$BEC ..........(2)$ [Common angle]
In view of $(1)$ and $(2)$,
$\triangle $$PDC$ $ \sim $ $\triangle $$BEC$ $[AA$ similarity criterion$]$
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Question 21 Mark
In the figure, altitudes $AD$ and $CE$ of $\triangle$$ABC$ intersect each other at the point $P$. Show that: $\vartriangle AEP \sim \vartriangle ADB$
Answer
In $\triangle $$AEP$ and $\triangle $$ADB,$ we have
$AEP=$ $\angle$ $ADP .........(1)$ [Each equal to $90^\circ$]
$\angle$$EAP=$$\angle$$DAB ..... (2)$ [Common angle]
In view of $(1)$ and $(2),$
$\triangle $$AEP$$ \sim $$\triangle $$ADB [AA $similarity criterion]
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Question 31 Mark
In the figure, altitudes AD and CE of $\triangle$ABC intersect each other at the point P. Show that: $\vartriangle ABD \sim \vartriangle CBE$
Answer
In $\triangle $ABD and $\triangle $CBE, ....... (1) [Each equal to $90^\circ$]
$\angle$ABC = $\angle$CBE ...... (2) [Common angle]
In view of (1) and (2),
$\triangle $ABD $ \sim $ $\triangle $CBE.......[AA similarity criterion]
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Question 41 Mark
In the figure, altitudes $AD$ and $CE$ of $\triangle$$ABC$ intersect each other at the point $P$. Show that: $\vartriangle AEP \sim \vartriangle CDP$
Answer
In $\triangle $AEP and $\triangle $CDP,
$\angle$$AEP =$ $\angle$$CDF ...... (1) [$Each equal to $90^\circ$$]$
$\angle$EPA = $\angle$$DPC ...... (2)$ [vert.opp. $\angle$s$]$
In view of $(1)$ and $(2)$,
$\triangle AEP \sim \triangle CDP$ $.........[AA$ similarity criterion$]$
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Question 51 Mark
State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:
Answer
In triangle $ABC$ and $DEF$, we have
$AB = 2.5, BC = 3$
$\angle \mathrm{A}=80^{\circ}$
$EF = 6$
$DF = 5$
$\angle F=80^{\circ}$
$\frac{A B}{D F}=\frac{2.5}{5}=\frac{1}{2}$
And, $\frac{B C}{E F}=\frac{3}{6}=\frac{1}{2}$
$\angle A = \angle F$
Hence, $\vartriangle$$ABC$ $\sim \vartriangle$$DEF ($by SAS Rule$ ).$
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Question 61 Mark
State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:
Answer
In triangle $MNL$ and $QPR$, we have
$\angle M=\angle Q=70^{\circ}$
but
$\frac{M N}{P Q}=\frac{2.5}{5}=\frac{1}{2}$
$\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{MN}}{\mathrm{PQ}} = \frac{\mathrm{ML}}{\mathrm{QR}}$
Therefore, $\vartriangle$$MNL$ $\sim$ $\vartriangle$$QPR$ are similar.
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Question 71 Mark
State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:
Answer
The given triangles are not similar because the corresponding sides are not proportional.
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Question 81 Mark
State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:
Answer
From the triangle, $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}=0.5$
Hence the corresponding sides are propotional. Thus the corresponding angles will be equal.
The triangles $ABC$ and $QRP$ are similar i.e, $\vartriangle ABC \sim \vartriangle QRP$ by $SSS$ similarity.
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Question 91 Mark
State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:
Answer
From the figure:
$\angle$A = $\angle$$P = 60^\circ $
$\angle$B = $\angle$$Q = 80^\circ $
$\angle$C = $\angle$$R = 40^\circ $
Therefore, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ [By AAA similarity]
Now corresponding sides of triangles will be propotional,
$\frac{A B}{PQ}=\frac{B C}{QR}=\frac{C A}{RP}$
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Question 101 Mark
In figure DE || AC and DF || AE. Prove that $\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$
Answer
In $\triangle ABE,$ we have $DF||AE,$then
$\frac{{BD}}{{AD}} = \frac{{BF}}{{FE}}\,$$ [$By $BPT] ...... (i)$
In $\triangle ABC,$ we have $DE||AC,$ then
$\frac{{BD}}{{AD}} = \frac{{BE}}{{EC}}\,$$ [$By $BPT] ...... (2)$
From $(i)$ and $(2)$, We get
$\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$
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Question 111 Mark
State whether the following quadrilaterals are similar or not:
Answer
The given quadrilateral $PQRS$ and $ABCD$ are not similar because their corresponding sides are proportional, i.e. $1:2,$ but their corresponding angles are not equal. Or we can say that $PQRS$ is a rhombus and $ABCD$ is a square, so they cannot be similar.
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Question 121 Mark
In Figure, if $PQ || RS$, prove that $\triangle POQ \sim \triangle SOR$.
Answer
From the given figure we have,
$PQ || RS$ (Given)
So, $\angle$$P =$ $\angle$$S ($Alternate angles$)$
and $\angle$$Q =$ $\angle$$R ($Alternate angles$)$
Also, $\angle$$POQ =$ $\angle$$SOR ($Vertically opposite angles$)$
Therefore, $\triangle POQ \sim \triangle SOR$ $(AAA$ similarity criterion$)$
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