4 Marks Questions

Question 14 Marks

Prove The Theorem : The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Answer

Given : In $\triangle \mathrm{ABC}$, bisector of $\angle \mathrm{C}$ interesects seg $A B$ in the point $E$.To prove : $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{CA}}{\mathrm{CB}}$

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Question 24 Marks

Prove The Theorem : When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

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Question 34 Marks

In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

Answer

PROOF:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\Rightarrow \frac{A D}{D C}=\frac{A B}{C B} \ldots . .(1) $
$\text { and } \frac{A E}{E B}=\frac{A C}{C B} \ldots . .(2)(\because, B D \text { and } C E \text { are angle bisectors of } \angle B \text { and } \angle C \text { respectively.) } $
$\text { Now, } \because \operatorname{seg} A B \cong \operatorname{seg} A C$
$\Rightarrow A B=A C$
$\Rightarrow \frac{A B}{C B}=\frac{A C}{C B} $
$\Rightarrow R \cdot H . S \text { of }(1) \&(2) \text { are equal. } $
$\Rightarrow L \cdot H \cdot S \text { of }(1) \&(2) \text { will be equal. } $
$\therefore \text { Equating } L . H \cdot S \text { of }(1) \&(2), \text { we get- }$
$\Rightarrow \frac{A D}{D C}=\frac{A E}{E B} $
$\Rightarrow E D \| B C(B y \text { converse basic proportionality theorem) }$

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Question 44 Marks

In fig 1.78, bisectors of ∠B and ∠C of ΔABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find $\frac{AX}{XY}.$

Answer

By Bisector Theorem $\Rightarrow \frac{ AX }{ XY }=\frac{ AB }{ BY }$
$\Rightarrow$ And $\frac{ AX }{ XY }=\frac{ AC }{ CY }$
Equating (1) & (2), we get-
$\Rightarrow \frac{A B}{B Y}=\frac{A C}{C Y} $
$\Rightarrow \frac{A B}{A C}=\frac{B Y}{C Y} $
$\Rightarrow \frac{A B+A C}{A C}=\frac{B Y+C Y}{C Y} $
$= \frac{B C}{C Y} $
$\Rightarrow \frac{5+4}{4}=\frac{6}{C Y} $
$\Rightarrow C Y=\frac{24}{9} $
$\Rightarrow C Y=\frac{8}{3} $
Now $\frac{A X}{X Y}=\frac{A C}{C Y} $
$\Rightarrow \frac{A X}{X Y}=\frac{4}{8} $
$\Rightarrow \frac{A X}{X Y}=\frac{3}{2} $

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Question 54 Marks

In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.
AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.

Answer

coming soon

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Question 64 Marks

Given : In $\triangle \mathrm{ABC}$, bisector of $\angle \mathrm{C}$ interesects seg $A B$ in the point $E$.To prove : $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{CA}}{\mathrm{CB}}$

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Question 74 Marks

When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

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Question 84 Marks

Diagonals of a quadrilateral $\mathrm{ABCD}$ intersect in point $\mathrm{Q}$. If $2 \mathrm{QA}=\mathrm{QC}$, $2 \mathrm{QB}=\mathrm{QD}$, then prove that $\mathrm{DC}=2 \mathrm{AB}$.
Given :
$ 2 Q A=Q C$
$2 Q B=Q D $
To prove : $\mathrm{CD}=2 \mathrm{AB}$

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Question 94 Marks

In $\Delta$ABC, seg DE || side BC. If 2A ($\Delta$ADE) = A($\square$DBCE). Show that BC = $\sqrt{3} \times$ DE

Answer

self

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Question 104 Marks

In $\triangle ABC , \angle ABC =90^{\circ} . \triangle PAB , \triangle QAC$ and $\triangle RBC$ are the equilateral triangles constructed on sides $A B, A C$ and $B C$ respectively. Prove that : $A (\triangle P A B)+ A (\triangle R B C)= A (\triangle Q A C)$.

Answer

self

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Question 114 Marks

In the adjoining figure, seg AB || side DC, $OD =x OB =x-3, OC$ $=x-5, OA =3 x-19$ Find the value of $x$.

Answer

$x=8, x=9$

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Question 124 Marks

In the adjoining figure, if $\Delta$ABN = $\Delta$ACM, show that $\Delta$AMN $\sim$ $\Delta$ABC.

Answer

self

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Question 134 Marks

In the adjoining figure, seg CE $\perp$ side AB, seg AD $\perp$ side BC. Prove that (i) AEP ~ CDP (ii) AEP ~ ADB

Answer

self

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Question 144 Marks

Prove The Theorem : The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Answer

Given : In $\triangle \mathrm{ABC}$, bisector of $\angle \mathrm{C}$ interesects seg $A B$ in the point $E$.To prove : $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{CA}}{\mathrm{CB}}$

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Question 154 Marks

Prove The Theorem : When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

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Question 164 Marks

In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

Answer

PROOF:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\begin{array}{l}
\Rightarrow \frac{A D}{D C}=\frac{A B}{C B} \ldots .(1) \\
\text { and } \frac{A E}{E B}=\frac{A C}{C B} \ldots .(2)(\because, B D \text { and } C E \text { are angle bisectors of } \angle B \text { and } \angle C \text { respectively.) } \\
\text { Now, } \because \operatorname{seg} A B \cong \operatorname{seg} A C \\
\Rightarrow A B=A C \\
\Rightarrow \frac{A B}{C B}=\frac{A C}{C B} \\
\Rightarrow \text { R.H.S of (1) \& (2) are equal. } \\
\Rightarrow \text { L.H.S of (1) \& (2) will be equal.| }
\end{array}$
(2) ( $\because, B D$ and $C E$ are angle bisectors of $\angle B$ and $\angle C$ respectively.)
$\Rightarrow$ R.H.S of $(1) \&(2)$ are equal.
$\Rightarrow$ L.H.S of (1) \& (2) will be equal.|
$\therefore$ Equating L.H.S of $(1) \&(2)$, we get-
$\Rightarrow \frac{ AD }{ DC }=\frac{ AE }{ EB }$
$\Rightarrow E D \| B C$ (By converse basic proportionality theorem)

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Question 174 Marks

In fig 1.78, bisectors of ∠B and ∠C of ΔABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find $\frac{A X}{X Y}$

Answer

By Bisector Theorem-
$\begin{array}{l}
\Rightarrow \frac{A X}{X Y}=\frac{A B}{B Y} \ldots \ldots \\
\Rightarrow \text { And, } \frac{A X}{X Y}=\frac{A C}{C Y} .
\end{array}$
Equating (1) \& (2), we get-
$\begin{array}{l}
\Rightarrow \frac{ AB }{ BY }=\frac{ AC }{ CY } \\
\Rightarrow \frac{ AB }{ AC }=\frac{ BY }{ CY } \\
\Rightarrow \frac{ AB + AC }{ AC }=\frac{ BY + CY }{ CY } \\
=\frac{ BC }{ CY } \\
\Rightarrow \frac{5+4}{4}=\frac{6}{ CY } \\
\Rightarrow C Y=\frac{24}{9} \\
\Rightarrow C Y=\frac{8}{3}
\end{array}$
Now, $\frac{A X}{X Y}=\frac{A C}{C Y}$
$\begin{aligned}
\Rightarrow & \frac{A X}{X Y}=\frac{4}{\frac{8}{3}} \\
\Rightarrow & \frac{A X}{X Y}=\frac{3}{2}
\end{aligned}$

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Question 184 Marks

In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.
AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.

Answer

Given seg. PA, seg 2B, seg. RC, Seg SD are perpendicular to line AD.
BB = 60 , BC = 0.7 , CD = 80 Ps = 280
P2,2R, RS=?
Perpendicular to the same line are parallel to each other
$\therefore$ seg. $P A \|$ seg. $2 B \|\ $ seg. $R C \|$ seg. $s D$
$\therefore \frac{P Q}{A B}=\frac{Q R}{B C}=\frac{R S}{C D}=\frac{P S}{A D} \dots \text { (1) }$ .... property of 3 or more parallel lines & their transversal.
Also AD= AB+BC+CD.....(A-B-C-D)
=60+70+80
=210.... (2)
$\frac{PQ}{60}=\frac{Q R}{70}=\frac{R S}{80}=\frac{4}{3}$...... From 1&2
$\begin{aligned} \therefore & \frac{P2}{60}=\frac{4}{3} \\ P2 & =\frac{60 \times 4}{3} \\ P2 & =80 \text { units }\end{aligned}$
$\begin{aligned} \therefore \frac{2 R}{70} & =\frac{4}{3} \\ 2 R & =\frac{70 \times 4}{3} \\ & =\frac{280}{3} \text { units }\end{aligned}$
$\begin{aligned} \frac{R S}{80} & =\frac{4}{3} \\ R S & =\frac{80 \times 4}{3} \\ & =\frac{320}{3} \text { units }\end{aligned}$

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Question 194 Marks

Given : In $\triangle \mathrm{ABC}$, bisector of $\angle \mathrm{C}$ interesects seg $A B$ in the point $E$.To prove : $\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{CA}}{\mathrm{CB}}$

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Question 204 Marks

When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

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Question 214 Marks

Diagonals of a quadrilateral $\mathrm{ABCD}$ intersect in point $\mathrm{Q}$. If $2 \mathrm{QA}=\mathrm{QC}$, $2 \mathrm{QB}=\mathrm{QD}$, then prove that $\mathrm{DC}=2 \mathrm{AB}$.
Given :
$
\begin{array}{l}
2 Q A=Q C \\
2 Q B=Q D
\end{array}
$
To prove : $\mathrm{CD}=2 \mathrm{AB}$

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Question 224 Marks

Answer

Given: In △ABC, ∠ABC = 90°. Equilateral triangles △PAB, △QAC and △RBC are constructed on sides AB, AC and BC respectively.
To prove: A(△PAB) + A(△RBC) = A(△QAC)
Since △PAB is equilateral,
A(△PAB) = $ \frac{\sqrt{3}}{4} AB^{2} $
Since △RBC is equilateral,
A(△RBC) = $ \frac{\sqrt{3}}{4} BC^{2} $
Since △QAC is equilateral,
A(△QAC) = $ \frac{\sqrt{3}}{4} AC^{2} $
In right angled △ABC at B,
$ AB^{2} + BC^{2} = AC^{2} $ (by Pythagoras theorem)
Multiplying both sides by $ \frac{\sqrt{3}}{4} $,
$ \frac{\sqrt{3}}{4} AB^{2} + \frac{\sqrt{3}}{4} BC^{2} = \frac{\sqrt{3}}{4} AC^{2} $
So,
A(△PAB) + A(△RBC) = A(△QAC).
Hence proved.

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Question 234 Marks

In the adjoining figure, seg CE $\perp$ side AB, seg AD $\perp$ side BC. Prove that (i) AEP ~ CDP (ii) AEP ~ ADB

Answer

Given: In triangle ABC, CE ⟂ AB and AD ⟂ BC.
(i) To prove: △AEP ~ △CDP
∠AEP = 90° (since CE ⟂ AB)
∠CDP = 90° (since AD ⟂ BC)
∠APE = ∠CPD (vertically opposite angles)
Therefore, by AA similarity criterion,
△AEP ~ △CDP.
(ii) To prove: △AEP ~ △ADB
∠AEP = 90° (since CE ⟂ AB)
∠ADB = 90° (since AD ⟂ BC)
∠APE = ∠ABD (common angle at A)
Therefore, by AA similarity criterion,
△AEP ~ △ADB.

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Question 244 Marks

In the adjoining figure, seg AB || side DC, $OD =x OB =x-3, OC$ $=x-5, OA =3 x-19$ Find the value of $x$.

Answer

$x=8, x=9$

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Question 254 Marks

In the adjoining figure, if $\Delta$ABN = $\Delta$ACM, show that $\Delta$AMN $\sim$ $\Delta$ABC.

Answer

Triangles $\triangle A B N$ and $\triangle A C M$ lie on the same base line AC.
Given that their areas are equal, triangles on the same base and between the same parallels have equal altitudes.
Hence, the perpendicular distances of points B and M from line AC are equal.
Therefore, segment $B M \| A C$.
Now, in triangle $\triangle A B C$, a line through point M is drawn parallel to side AC, meeting side AC at point N.
By the Basic Proportionality Theorem, the line parallel to one side of a triangle divides the other two sides proportionally.
Thus,
$\frac{A M}{A B}=\frac{A N}{A C}$
Therefore, the corresponding sides of triangles $\triangle A M N$
$\triangle A B Care proportional.$
Hence, by the AA similarity criterion,

Thus, $\triangle A M N \sim \triangle A B C$
Hence, the given statement is proved.

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Question 264 Marks

In $\Delta$ABC, seg DE || side BC. If 2A ($\Delta$ADE) = A($\square$DBCE). Show that BC = $\sqrt{3} \times$ DE

Answer

Since segment $D E \| B C$ , triangle $\triangle A D E$ and triangle$\triangle A B C$. are similar.
Let
$A(\triangle A D E)=x$
Given, $2 A(\triangle A D E)=A(\square D B C E)$
$\Rightarrow A(\square D B C E)=2 x$
Now, $A(\triangle A B C)=A(\triangle A D E)+A(\square D B C E)$
$A(\triangle A B C)=x+2 x=3 x$
For similar triangles, the ratio of areas is equal to the square of the ratio of corresponding sides.
Hence,
$\frac{A(\triangle A D E)}{A(\triangle A B C)}=\left(\frac{D E}{B C}\right)^2$
Substituting the values,
$\frac{x}{3 x}=\left(\frac{D E}{B C}\right)^2$
$\frac{1}{3}=\left(\frac{D E}{B C}\right)^2$
Taking square root on both sides,
$\frac{D E}{B C}=\frac{1}{\sqrt{3}}$
Therefore,$B C=\sqrt{3} D E$
The given statement is proved.

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