[4 marks sum]

Question 14 Marks

In the given figure, sides AB and AC of $\triangle ABC$ have been produced to D and E respectively. If $\angle CBD =x^{\circ}$ and $\angle BCE$ $=y^{\circ}$ such that $x>y$, show that $AB > AC$.

Answer

[Hint. $x>y \Rightarrow(180-y)>(180-x)$.]
In the given figure, $AB = AC$. Show that $AD > AB$.

View full question & answer→

Question 24 Marks

In a right-angled triangle, prove that the hypotenuse is the longest side.

Answer

[Hint. In a right triangle, the angle opposite to the hypotenuse is a right angle and other two angles are acute.]

View full question & answer→

Question 34 Marks

In the given figure, side $A B$ of $\triangle A B C$ is produced to $D$ such that $BD = BC$.
If $\angle A =60^{\circ}$ and $\angle B =50^{\circ}$, prove that :
(i) $AD > CD$
(ii) $AD > AC$.

Answer

[Hint. $\angle C=70^{\circ}, \angle C B D=130^{\circ}$ and $\angle B D C=\angle B C D=25^{\circ}$.]

View full question & answer→

Question 44 Marks

In the given figure, $\angle ABC =66^{\circ}, \angle DAC =38^{\circ}$. CE is perpendicular to AB and AD is perpendicular to BC .
Prove that: $CP > AP$.

Answer

self

View full question & answer→

Question 54 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.

Answer

(i) $\angle B$
(ii) $\angle C$

View full question & answer→

Question 64 Marks

In $\triangle LMN$, if $\angle M =90^{\circ}$, name the longest side of the triangle.

Answer

View full question & answer→

Question 74 Marks

In the given figure, $BD \| CE ; AC = BC , \angle ABD =20^{\circ}$ and $\angle ECF =70^{\circ}$. Find $\angle GAC$.

Answer

130
[Hint. $\angle D B C=\angle E C F=70^{\circ} \quad($ Corres. $\left.\angle s)\right]$

View full question & answer→

Question 84 Marks

In the given figure, AC is the bisector of $\angle A$. If $AB = AC$, $AD = CD$ and $\angle ABC =75^{\circ}$, find the values of $x$ and $y$.

Answer

x=30, y=120

View full question & answer→

Question 94 Marks

In the adjoining figure, $AB = AC$. If $DB \perp BC$ and EC $\perp BC$, prove that :
(i) $BD = CE$
(ii) $AD = AE$.

Answer

[Hint. $\triangle A B D \cong \triangle A C E$.]

View full question & answer→

Question 104 Marks

In the given figure, $AD = AE$ and $\angle BAD =\angle CAE$. Prove that: $AB = AC$.

Answer

Hint :
[Prove that $\triangle A B D \cong \triangle A C E$.$\angle A D E=\angle A E D \Rightarrow 180^{\circ}-\angle A D E=180^{\circ}-\angle A E D$
$\Rightarrow \angle A D B=\angle A E C .]$

View full question & answer→

Question 114 Marks

If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.

Answer

self

View full question & answer→

Question 124 Marks

In an isosceles triangle, prove that the altitude from the vertex bisects the base.

Answer

self

View full question & answer→

Question 134 Marks

In a $\triangle A B C, A B=A C$. If the bisectors of $\angle B$ and $\angle C$ meet AC and AB at points D and E respectively, show that :
(i) $\triangle DBC \equiv \triangle ECB$
(ii) $BD = CE$.

Answer

self

View full question & answer→

Question 144 Marks

Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

Answer

self

View full question & answer→

Question 154 Marks

In the given figure, side BA of $\triangle ABC$ has been produced to D such that $CD = CA$ and side CB has been produced to E . If $\angle BAC =106^{\circ}$ and $\angle ABE =128^{\circ}$, find $\angle BCD$.

Answer

$\angle BCD =54$

View full question & answer→

Question 164 Marks

In the given figure, $AB = AD ; CB = CD ; \angle A =42^{\circ}$ and $\angle C=108^{\circ}$, find $\angle ABC$.

Answer

105
[Hint. Join BD.]

View full question & answer→

Question 174 Marks

In the given figure, $\triangle ABC$ is an equilateral triangle whose base BC is produced to D such that $BC = CD$. Calculate :
(i) $\angle ACD$
(ii) $\angle ADC$.

Answer

(i) $\angle ACD =120$
(ii) $\angle ADC =30$

View full question & answer→

Question 184 Marks

In the given figure, sides $A B$ and $A C$ of $\triangle A B C$ have been produced to D and E respectively. If $\angle CBD =x^{\circ}$ and $\angle BCE$ $=y^{\circ}$ such that $x>y$, show that $AB > AC$.

Answer

self

View full question & answer→

Question 194 Marks

In the given figure, $AB > AC$. If BO and CO are the bisectors of $\angle B$ and $\angle C$ respectively, prove that $BO > CO$.

Answer

self

View full question & answer→

Question 204 Marks

In the given figure, $A C$ is the bisector of $\angle A$. If $A B=A C$, $AD = CD$ and $\angle ABC =75^{\circ}$, find the values of $x$ and $y$.

Answer

x=30, y=120

View full question & answer→

Question 214 Marks

In the given figure, $\triangle A B C$ is an equilateral triangle and $B C$ is produced to $D$ such that $B C=C D$. Prove that $A D \perp A B$.

Answer

self

View full question & answer→

Question 224 Marks

In the given figure, $\triangle ABC$ is an equilateral triangle whose base $B C$ is produced to $D$ such that $B C=C D$. Calculate :
(i) $\angle ACD$
(ii) $\angle ADC$.

Answer

(i) $\angle ACD =120^{\circ}$
(ii) $\angle ADC =30^{\circ}$

View full question & answer→

Question 234 Marks

Of all the line segments that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest.
Given: $\quad A$ straight line $A B$ and a point $P$ outside it; $PM \perp AB$ and N is any other point on AB .
To prove : $PM < PN$.
Proof.

Answer

Statement	Reason
1. $\angle PMN =90^{\circ}$	$PM \perp AB$. In a right $\Delta$, one angle measures $90^{\circ}$ and each one of the remaining two is acute. From 1 and 2. Side opp. the smaller angle is smaller.
2. $\angle PNM <90^{\circ}$
3. $\angle PNM <\angle PMN$
4. $PM < PN$$\therefore PM$ is the shortest of all line segments from $P$ to $A B$.

View full question & answer→

Question 244 Marks

The sum of any two sides of a triangle is greater than its third side.
Given : A $\triangle ABC$.
To prove :
(i) $AB + AC > BC$
(ii) $AB + BC > AC$
(iii) $BC + AC > AB$.
Construction : Produce BA to D such that $AD = AC$. Join CD. Proof.

Answer

Statement	Reason
1. $AC = AD$	By construction. Angles opp. to equal sides of a $\Delta$ are equal. Whole is greater than its part. Using 2 and 3. Greater angle has greater side opp. to it. BAD is a straight line, $BD = BA + AD$. $AD = AC$, by construction.
2. $\angle ACD =\angle ADC$
3. $\angle BCD >\angle ACD$
4. $\angle BCD >\angle ADC$
5. $BD > BC$
6. $BA + AD > BC$
7. $BA + AC > BC$ Thus, $AB + AC > BC$.
8. Similarly, $A B+B C>A C$ and $BC + AC > AB$.

View full question & answer→

Question 254 Marks

(Converse of Theorem 1) : If two angles of a triangle are unequal, then the greater angle has greater side opposite to it.
Given :
$
\text { A } \triangle ABC \text { in which } \angle ABC>\angle ACB \text {. }
$
To prove : $\quad AC > AB$.
Proof.

Answer

Statement

Reason

1. We may have three possibilities only :
(i) $AC = AB$
(ii) $AC < AB$
(iii) $AC > AB$
Out of these exactly one must be true.
Case I. $A C=A B$.
In this case, $\angle ABC =\angle ACB$.
This is contrary to what is given.
$\therefore AC = AB$ is not true.
Case II.$A C < A B$.
In this case, $AB > AC$.
$
\therefore \angle ACB>\angle ABC .
$
This is contrary to what is given.
$\therefore AC < AB$ is not true.
Thus, we are left with the only possibility $A C>A B$, which must be true
Hence, $A C>A B$.
$
\therefore \angle ABC>\angle ACB \Rightarrow AC>AB .$

By property of order relation in real numbers.
Angles opp.to equal sides of a $\Delta$ are equal.
Greater side has greater angle opp. to it.

View full question & answer→

Question 264 Marks

If two sides of a triangle are unequal, then the greater side has greater angle opposite to it.
Given: $\quad A \triangle A B C$ in which $A C > A B$.
To prove: $\quad \angle ABC >\angle ACB$.
Construction : Mark a point D on AC such that $AD = AB$. Join BD.
Proof.

Answer

Statement	Reason
1. $AB = AD$	By construction. Angles opp. to equal sides in a $\Delta$ are equal. $($ Ext. $\angle$ of $\triangle BCD )>($ Each of its Int. Opp. $\angle s)$. Using 2 in 3. $\angle ABD$ is a part of $\angle ABC$. Using 5 in 4. $ \angle DCB=\angle ACB . $
2. $\angle ABD =\angle BDA$
3. $\angle BDA >\angle DCB$
4. $\angle ABD >\angle DCB$
5. $\angle ABC >\angle ABD$
6. $\angle ABC >\angle DCB$
7. $\angle ABC >\angle ACB$ Thus, $AC > AB \Rightarrow \angle ABC >\angle ACB$.

View full question & answer→

MATHEMATICS [4 marks sum]

Test yourself on this topic