Question 12 Marks
Can you construct a $\triangle ABC$ in which $AB =5 cm, BC =4 cm$ and $AC =9 cm$ ? Give reason.
Answer
View full question & answer→No, since the sum of any two sides must always be greater than the third.
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Question 22 Marks
In the given figure, the sides BA and CA of $\triangle ABC$ have been produced to D and E such that $BA = AD$ and $CA = AE$. Prove that, $ED \| BC$.
Answer
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Question 32 Marks
AB is a line segment. AX and BY are two equal line segments drawn on opposite sides of $A B$ such that $A X \| Y B$. If $A B$ and $X Y$ intersect at $M$, prove that :
(i) $\triangle AMX \cong \triangle BMY$
(ii) AB and XY bisect each other at M .
(i) $\triangle AMX \cong \triangle BMY$
(ii) AB and XY bisect each other at M .
Answer
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Question 42 Marks
In the given figure, $PA \perp AB ; QB \perp AB$ and $PA = QB$. If PQ intersects AB at M , show that M is the mid-point of both AB and PQ .
Answer
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Question 52 Marks
In the given figure, $M$ is the mid-point of $A B$ and $C D$. Prove that $CA = BD$ and $CA \| BD$.
Answer
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Question 62 Marks
In the given figure, median AD of $\triangle ABC$ is produced. If BL and CM are perpendiculars drawn on AD and AD produced, prove that $BL = CM$.
Answer
View full question & answer→[Hint. $\triangle B L D \cong \triangle C M D$.]
Question 72 Marks
In the given figure, equal sides BA and CA of $\triangle ABC$ are produced to Q and P respectively such that $AP = AQ$. Prove that $PB = QC$.
Answer
View full question & answer→[Hint. $\triangle A P B \cong \triangle A Q C$.]
Question 82 Marks
The perpendicular bisectors of the sides of a $\triangle ABC$ meet at I. Prove that: $IA = IB = IC$.
Answer
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Question 92 Marks
ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively such that $AB = BE$ and $AD = DF$. Prove that $\triangle BEC \equiv \triangle DCF$.
Answer
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Question 102 Marks
ABCD is a parallelogram in which $\angle A$ and $\angle C$ are obtuse. Points X and Y are taken on diagonal BD such that $\angle AXD =\angle CYB =90^{\circ}$. Prove that : $XA = YC$.
Answer
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Question 112 Marks
In the given figure, ABCD is a square, $EF \| BD$ and R is the mid-point of EF.
Prove that:
(i) $BE = DF$
(ii) AR bisects $\angle BAD$
(iii) If AR is produced, it will pass through C .
Prove that:
(i) $BE = DF$
(ii) AR bisects $\angle BAD$
(iii) If AR is produced, it will pass through C .
Answer
View full question & answer→self
Question 122 Marks
In the given figure, P is a point in the interior of $\angle ABC$. If $PL \perp BA$ and $PM \perp BC$ such that $PL = PM$, prove that BP is the bisector of $\angle ABC$.
Answer
View full question & answer→[Hint. $\triangle B P L \cong \triangle B P M$.]
Question 132 Marks
In the given figure, ABCD is a parallelogram, E is the mid-point of BC . DE produced meets AB produced at L . Prove that:
(i) $AB = BL$
(ii) $AL =2 DC$.
(i) $AB = BL$
(ii) $AL =2 DC$.
Answer
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Question 142 Marks
Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD . Prove that :
(i) $\angle SAQ =\angle ABC$
(ii) $SQ = AC$.
(i) $\angle SAQ =\angle ABC$
(ii) $SQ = AC$.
Answer
View full question & answer→[Hint.
(i)$\begin{array}{l}\angle S A Q=360^{\circ}-\left(90^{\circ}+\angle B A D+90^{\circ}\right)=180^{\circ}-\angle B A D . \\
\angle A B C+\angle B A D=180^{\circ} \Rightarrow \angle A B C=180^{\circ}-\angle B A D .\end{array}$
(ii) $\triangle S A Q \cong \triangle C B A$.]
(i)$\begin{array}{l}\angle S A Q=360^{\circ}-\left(90^{\circ}+\angle B A D+90^{\circ}\right)=180^{\circ}-\angle B A D . \\
\angle A B C+\angle B A D=180^{\circ} \Rightarrow \angle A B C=180^{\circ}-\angle B A D .\end{array}$
(ii) $\triangle S A Q \cong \triangle C B A$.]
Question 152 Marks
In the given figure, in $\triangle ABC , \angle B =90^{\circ}$. If ABPQ and ACRS are squares, prove that :
(i) $\triangle ACQ \cong \triangle ABS$
(ii) $CQ = BS$.
(i) $\triangle ACQ \cong \triangle ABS$
(ii) $CQ = BS$.
Answer
View full question & answer→self
Question 162 Marks
In the given figure : $\angle BAC =\angle CDB$ and $\angle BCA =\angle CBD$. Prove that $A B=C D$.
Answer
View full question & answer→[Hint. $\triangle A B C \cong \triangle D C B$.]
Question 172 Marks
If two altitudes of a triangle are equal, prove that it is an isosceles triangle.
Answer
View full question & answer→[Hint. $\triangle A B D \cong \triangle A C E$ ]
Question 182 Marks
Which of the following pairs of triangles are congruent?
$\triangle ABC$ and $\triangle DEF$ in which : $BC = EF , AC = DF$ and $\angle C =\angle F$.
$\triangle ABC$ and $\triangle PQR$ in which : $AB = PQ , BC = QR$ and $\angle C =\angle R$.
$\triangle ABC$ and $\triangle LMN$ in which : $\angle A =\angle L =90^{\circ}, AB = LM , \angle C =40^{\circ}$ and $\angle M =50^{\circ}$
$\triangle ABC$ and $\triangle DEF$ in which : $\angle B =\angle E =90^{\circ}$ and $AC = DF$.
$\triangle ABC$ and $\triangle DEF$ in which : $BC = EF , AC = DF$ and $\angle C =\angle F$.
$\triangle ABC$ and $\triangle PQR$ in which : $AB = PQ , BC = QR$ and $\angle C =\angle R$.
$\triangle ABC$ and $\triangle LMN$ in which : $\angle A =\angle L =90^{\circ}, AB = LM , \angle C =40^{\circ}$ and $\angle M =50^{\circ}$
$\triangle ABC$ and $\triangle DEF$ in which : $\angle B =\angle E =90^{\circ}$ and $AC = DF$.
Answer
View full question & answer→A, C
Question 192 Marks
In $\triangle LMN$, if $\angle M =90^{\circ}$, name the longest side of the triangle.
Answer
View full question & answer→LN
Question 202 Marks
In each of the following figures, find the value of $x$ :
Answer
View full question & answer→(i) $x=40$
(ii) $x=34$
(ii) $x=34$
Question 212 Marks
In the given figure, AD bisects $\angle A$. If $\angle B =60^{\circ}, \angle C =$ $40^{\circ}$, then arrange $AB , BD$ and DC in ascending order of their lengths.
Answer
View full question & answer→BD=DC < AB
Question 222 Marks
In $\triangle ABC , AB =7.5 cm, BC =6.2 cm$ and $AC =5.4 cm$. Name :
(i) the least angle (ii) the greatest angle of the triangle.
(i) the least angle (ii) the greatest angle of the triangle.
Answer
View full question & answer→(i) $\angle B$
(ii) $\angle C$
(ii) $\angle C$
Question 232 Marks
The sides $A B$ and $A C$ of $\triangle A B C$ are produced to $D$ and $E$ respectively and the bisectors of $\angle CBD$ and $\angle BCE$ meet at O . If $AB > AC$, prove that $OC > OB$.
Answer
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Question 242 Marks
In the adjoining figure, $\triangle ABC$ is equilateral and D is any point onAC.
Prove that :
(i) $BD > AD$
(ii) $BD > DC$.
Prove that :
(i) $BD > AD$
(ii) $BD > DC$.
Answer
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Question 252 Marks
Can you construct a $\triangle ABC$ in which $AB =5 cm, BC =4 cm$ and $AC =9 cm$ ? Give reason.
Answer
View full question & answer→No, since the sum of any two sides must always be greater than the third.
Question 262 Marks
In the adjoining figure, $PL \perp QR ; LQ = LS$ and $LR > LQ$.
Show that $P R>P Q$.
Show that $P R>P Q$.
Answer
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Question 272 Marks
In $\triangle ABC , D$ is any point on BC .
Prove that: $AB + BC + AC >2 AD$.
Prove that: $AB + BC + AC >2 AD$.
Answer
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Question 282 Marks
In the adjoining figure, in $\triangle ABC , O$ is any point in its interior. Show that: $OB + OC < AB + AC$.
Answer
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Question 292 Marks
In a right-angled triangle, prove that the hypotenuse is the longest side.
Answer
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Question 302 Marks
In $\triangle PQR , \angle P =50^{\circ}$ and $\angle R =70^{\circ}$.
Name : (i) the shortest side (ii) the longest side of the triangle.
Name : (i) the shortest side (ii) the longest side of the triangle.
Answer
View full question & answer→(i)QR
(ii) PQ
(ii) PQ
Question 312 Marks
The perpendicular bisectors of the sides of a $\triangle A B C$ meet at I. Prove that: $I A=I B=I C$.
Answer
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Question 322 Marks
$A B C D$ is a parallelogram. The sides $A B$ and $A D$ are produced to E and F respectively such that $AB = BE$ and $AD = DF$. Prove that $\triangle BEC \cong \triangle DCF$.
Answer
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Question 332 Marks
ABCD is a parallelogram in which $\angle A$ and $\angle C$ are obtuse. Points X and Y are taken on diagonal $B D$ such that $\angle AXD =\angle CYB =90^{\circ}$. Prove that : $XA = YC$.
Answer
View full question & answer→self
Question 342 Marks
In the given figure, ABCD is a square, $EF \| BD$ and R is the mid-point of EF. Prove that :
(i) $BE = DF$
(ii) AR bisects $\angle BAD$
(iii) If AR is produced, it will pass through C .
(i) $BE = DF$
(ii) AR bisects $\angle BAD$
(iii) If AR is produced, it will pass through C .
Answer
View full question & answer→self
Question 352 Marks
In the given figure, $A B C D$ is a square and $P, Q, R$ are points on $A B, B C$ and $C D$ respectively such that $AP = BQ = CR$ and $\angle PQR =90^{\circ}$.
Prove that:
(i) $PB = QC$
(ii) $PQ = QR$
(iii) $\angle QPR =45^{\circ}$
Prove that:
(i) $PB = QC$
(ii) $PQ = QR$
(iii) $\angle QPR =45^{\circ}$
Answer
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Question 362 Marks
Equilateral triangles ABD and ACE are drawn on the sides $A B$ and $A C$ of $\triangle A B C$ as shown in the figure.
Prove that :
(i) $\angle DAC =\angle EAB$
(ii) $DC = BE$.
Prove that :
(i) $\angle DAC =\angle EAB$
(ii) $DC = BE$.
Answer
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Question 372 Marks
In the given figure, $A B C D$ is a parallelogram, $E$ is the mid-point of BC . DE produced meets AB produced at L . Prove that:
(i) $AB = BL$
(ii) $AL =2 DC$.
(i) $AB = BL$
(ii) $AL =2 DC$.
Answer
View full question & answer→self
Question 382 Marks
Squares ABPQ and ADRS are drawn on the sides AB and $A D$ of a parallelogram $A B C D$.
Prove that:
(i) $\angle SAQ =\angle ABC$
(ii) $SQ = AC$.
Prove that:
(i) $\angle SAQ =\angle ABC$
(ii) $SQ = AC$.
Answer
View full question & answer→self
Question 392 Marks
In the given figure, in $\triangle A B C, \angle B=90^{\circ}$.
If ABPQ and ACRS are squares, prove that :
(i) $\triangle ACQ \cong \triangle ABS$
(ii) $CQ = BS$.
If ABPQ and ACRS are squares, prove that :
(i) $\triangle ACQ \cong \triangle ABS$
(ii) $CQ = BS$.
Answer
View full question & answer→self
Question 402 Marks
In the given figure, ABCD is a square and $\triangle PAB$ is an equilateral triangle.
(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$
(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$
Answer
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Question 412 Marks
Answer
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Question 422 Marks
Answer
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Question 432 Marks
In the given figure : $\angle BAC =\angle CDB$ and $\angle BCA =\angle CBD$. Prove that $A B=C D$.
Answer
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Question 442 Marks
If two altitudes of a triangle are equal, prove that it is an isosceles triangle.
Answer
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Question 452 Marks
In the given figure, the line segments $A B$ and $C D$ intersect at a point $M$ in such a way that $A M=M D$ and $CM = MB$. Prove that, $AC = BD$ but AC may not be parallel to BD.
Answer
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Question 462 Marks
In the given figure, the sides $B A$ and $C A$ of $\triangle A B C$ have been produced to D and E such that $BA = AD$ and $C A=A E$. Prove that, $E D \| B C$.
Answer
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Question 472 Marks
$A B$ is a line segment. $A X$ and $B Y$ are two equal line segments drawn on opposite sides of $A B$ such that $A X \| Y B$. If $A B$ and $X Y$ intersect at $M$, prove that :
(i) $\triangle AMX \cong \triangle BMY$ (ii) AB and XY bisect each other at M .
(i) $\triangle AMX \cong \triangle BMY$ (ii) AB and XY bisect each other at M .
Answer
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Question 482 Marks
In the given figure, $PA \perp AB ; QB \perp AB$ and $PA = QB$. If $P Q$ intersects $A B$ at $M$, show that $M$ is the mid-point of both AB and PQ .
Answer
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Question 492 Marks
In the given figure, $M$ is the mid-point of $A B$ and $C D$. Prove that $CA = BD$ and $CA \| BD$.
Answer
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Question 502 Marks
In the given figure, median AD of $\triangle ABC$ is produced. If BL and CM are perpendiculars drawn on AD and AD produced, prove that $BL = CM$.
Answer
View full question & answer→self