4 Marks Questions

Question 14 Marks

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Answer

Given,
BE and CF are altitudes.
AC = AB To show,
BE = CF
In $\triangle\text{AEB}$ and $\triangle\text{AFC}$
$\angle \text{A}=\angle\text{A}$ (Common)
$\angle\text{AEB}=\angle \text{AFC}$ (Right angles)
AB = AC (Given)
Therefore, $\triangle\text{AEB}\cong \angle\text{AFC}$ by AAS congruence condition.
Thus, BE = CF (by CPCT).

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Question 24 Marks

In $\angle\text{B}<\angle\text{A}$ and $\angle\text{C}<\angle\text{D}.$ Show that AD < BC.

Answer

Given,
$\angle\text{B}<\angle\text{A}$ and $\angle\text{C}<\angle\text{D}$
Now,
AO < BO ...(i) (Side opposite to the smaller angle is smaller)
OD < OC ...(ii) (Side opposite to the smaller angle is smaller)
Adding (i) and (ii),
AO + OD < BO + OC
AD < BC.

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Question 34 Marks

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal Show that,

$\triangle\text{ABE}\cong \angle\text{ACF}$
AB = AC, i.e., ABC is an isosceles triangle.

Answer

Given,
BE = CF
In $\triangle\text{ABE}$ and $\triangle\text{ACF,}$
$\angle \text{A}=\angle\text{A}$ (Common)
$\angle \text{ABE}= \angle\text{ACF}$ (Right angles)
BE = CF (Given)
Therefore, $\triangle\text{ABE}\cong\triangle\text{ACF}$ by AAS congruence condition.
Thus, AB = AC by CPCT and therefore, ABC is an isosceles triangle.

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Question 44 Marks

ABCD is a quadrilateral in which AD = BC and . Prove that$\angle\text{ DAB }= \angle\text{CBA }.$

$\triangle\text{ABCD}\cong\triangle\text{BAC}$
BD = AC
$\angle\text{ DAB }= \angle\text{CBA }.$

Answer

Given,

AD = BC and $\angle\text{ DAB }= \angle\text{CBA }$

In $\triangle\text{ABD},\triangle\text{BAC},$

AB = BA (Common)

$\angle\text{ DAB }= \angle\text{CBA }$ (Given)

AD = BC (Given)

Therefore, $\triangle\text{ABCD}\cong\triangle\text{BAC}$ by SAS congruence condition.

Since, $\triangle\text{ABCD}\cong\triangle\text{BAC}$

Therefore BD = AC by CPCT

Since, $\triangle\text{ABCD}\cong\triangle\text{BAC}$

Therefore $\angle\text{ DAB }= \angle\text{CBA }$ by CPCT.

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Question 54 Marks

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of Show that:

$\triangle\text{APB}\cong\triangle\text{AQB}$
BP = BQ or B is equidistant from the arms of $\angle\text{A.}$

Answer

Given,

l is the bisector of an angle $\angle\text{A.}$

BP and BQ are perpendiculars.

In $\triangle\text{APB}$ and $\triangle\text{AQB},$

$\angle\text{ P} = \angle\text{ Q}$ (Right angles)

$\angle\text{BAP }= \angle\text{ BAQ}$ (l is bisector)

AB = AB (Common)

Therefore, $\triangle\text{APB}\cong\triangle\text{AQB}$ by AAS congruence condition.

BP = BQ by CPCT. Therefore, B is equidistant from the arms of $\angle\text{A.}$

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Question 64 Marks

ABC and DBC are two isosceles triangles on the same base BC . Show that $\angle\text{ABD}=\angle \text{ACD}.$

Answer

Given,
ABC and DBC are two isosceles triangles. To show,
$\angle\text{ABD}= \angle\text{ACD}$
In $\triangle\text{ABD}$ and $\triangle\text{ACD}$
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by SSS congruence condition.
Thus, $\angle\text{ABD}= \angle\text{ACD}$ (by CPCT).

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Question 74 Marks

In sides AB and AC of $\triangle\text{ABC}$ are extended to points P and Q respectively. Also, $\angle\text{PBC}<\angle\text{QCB}$ Show that AC > AB.

Answer

Given,
$\angle\text{PBC}<\angle\text{QCB}$
Now,
$\angle\text{ABC}+\angle\text{PBC}=180^{0}$
$\Rightarrow \angle\text{ABC}=180^{0}-\angle\text{PBC}$
Also,
$\angle\text{ABC}+\angle\text{QBC}=180^{0}$
$\Rightarrow \angle\text{ABC}=180^{0}-\angle\text{QBC}$
Since,
$\angle\text{PBC}<\angle\text{QCB}$ therefore, $\angle\text{ABC}>\angle\text{ACB}$
Thus, AC > AB as sides opposite to the larger angle is larger.

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Question 84 Marks

AD and BC are equal perpendiculars to a line segment AB . Show that CD bisects AB.

Answer

Given,
AD and BC are equal perprndiculars to AB.
To prove,
CD bisects AB
In $\triangle\text{AOD}$ and $\triangle\text{BOC}$
$\angle\text{A}=\angle\text{D}$(Perprndicular)
$\angle\text{ AOD} = \angle\text{ BOC}$ (Vertically opposite angle)
AD = BC (Given)
Therefore $\triangle \text{AOD}\cong\triangle \text{BOC}$ by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

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Question 94 Marks

ABC is an isosceles triangle with AB = AC. Draw $\text{AP}\perp\text{BC}$ to show that $\angle\text{B}=\angle\text{C}.$

Answer

In $\triangle \text{APB}$ and $\triangle \text{APC}$
$\angle\text{APB}=\angle\text{APC}$ (Each)
AB = AC (Given)
AP = AP (Common)
$\therefore \triangle\text{APB}\cong\triangle \text{APC}$ (by RHS)
AB = AC (Sides opposite to equal angles of a triangle are equal)
$\Rightarrow \angle\text{B}=\angle\text{C}$ (By using CPCT).

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Question 104 Marks

In $\triangle\text{ABC,}$, AD is the perpendicular bisector of BC. Show that D ABC is an isosceles triangle in which AB = AC.

Answer

Given,
AD is the perprndicular bisector of BC To Show,
AB = AC
In $\triangle\text{ADB}$ and $\triangle\text{ADC}$
AD = AD (Common)
$\angle\text{ADB}= \angle\text{ADC}$
BC = CD (AD is the perpendicular bisector)
$\therefore \angle\text{ADC}\cong \angle\text{ADC}$ by SAS congruence condition.
AB = AC (by CPCT).

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