1 Marks Question

Question 11 Mark

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

Answer

In $\triangle$ABE and $\triangle$ACF,
$\angle $A= $\angle $ A [Common]
$\angle $AEB = $\angle $AFC = [90°]
AB = AC [Given]
$\therefore $ $\triangle$ABE ≅ $\triangle$ACF [By ASA congruency]
$\therefore $ BE = CF [By C.P.C.T.]
So Altitudes are equal.

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Question 21 Mark

Line l is the bisector of an angle $\angle$A and B is any point on l. BP and BQ are perpendicular from B to the arms of $\angle$A.

Show that:

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

Answer

In $\triangle$APB and $\triangle$AQB
$\angle$BAP = $\angle$BAQ ...[As l bisects $\angle$A]
$\angle$BPA = $\angle$BQA ...[Each 90°]
AB = AB ...[Common]
$\therefore$ $\triangle$APB $\cong$ $\triangle$AQB proved ...[AAS property] ...(1)
$\triangle$APB $\cong$ $\triangle$AQB ...[From (1)]
$\therefore$ BP = BQ ...[c.p.c.t.]

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Question 31 Mark

l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that $\Delta$ABC $\cong$ $\Delta$CDA.

Answer

Given : l || m and p || q
To prove : DABC $\cong$ DCDA
Proof : l || m and p || q . . . . [Given]
In DABC and DCDA
$\angle$BAC = $\angle$DCA . . . . . [Alternate interior angles as AB || DC]
Similarly, $\angle$ACB = $\angle$CAD . . . [Alternate interior angles as BC || DA]
AC = DA . . . [Common]
DABC $\cong$ DCDA [By ASA congruency]

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Question 41 Mark

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

Answer

In $\triangle$BOC and $\triangle$AOD,
$\angle$OBC = $\angle$OAD = 90° [Given]
$\angle$BOC = $\angle$AOD [Vertically Opposite angles]
BC = AD [Given]
$\therefore$ $\triangle$BOC $\cong$ $\triangle$AOD [By ASA congruency]
$\Rightarrow$ OB = OA and OC = OD [By C.P.C.T.]

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Question 51 Mark

D is a point on side BC of $\triangle$ABC such that AD = AC (see Fig.). Show that AB > AD.

Answer

In $\triangle$DAC, AD = AC (Given)
So, $\angle$ADC = $\angle$ ACD (Angles opposite to equal sides)
Now, $\angle$ ADC is an exterior angle for $\triangle$ABD.
So, $\angle$ ADC > $\angle$ ABD
or, $\angle$ACD > $\angle$ABD
or, $\angle$ACB > $\angle$ ABC
So, AB > AC (Side opposite to larger angle in $\triangle$ ABC)
or, AB > AD (AD = AC)

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Question 61 Mark

In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (see Fig.). Show that AD = AE

Answer

In $\triangle$ ABD and $\triangle$ACE,
AB = AC (Given) ...(1)
$\angle$ B = $\angle$C (Angles opposite to equal sides) ...(2)
Also, BE = CD
So, BE – DE = CD – DE
That is, BD = CE ...(3)
So, $\triangle$ ABD $\cong$ $\triangle$ ACE (Using (1), (2), (3) and SAS rule).
This gives AD = AE (CPCT)

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Question 71 Mark

E and F are respectively the mid-points of equal sides AB and AC of ∆ ABC (see Fig.). Show that BF = CE

Answer

In $\triangle$ ABF and $\triangle$ ACE,
AB = AC (Given)
$\angle$ A = $\angle$ A (Common)
AF = AE (Halves of equal sides)
So, $\triangle$ABF $\cong$ $\triangle$ACE (SAS rule)
Therefore, BF = CE (CPCT)

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Question 81 Mark

In $\triangle$ABC, the bisector AD of $\angle$A is perpendicular to side BC (See figure).Show that AB = AC and $\triangle$ABC is isosceles.

Answer

In $\triangle$ADB and $\triangle$ADC,
BD = CD [AD bisects BC]
$\angle$ADB = $\angle$ADC = 90° [AD $\bot$ BC]
AD = AD [Common]
$\therefore $ $\triangle$ABD $\cong$ $\triangle $ACD [By SAS congruency]
$\Rightarrow$ AB = AC [By C.P.C.T.]
Therefore, ABC is an isosceles triangle.

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Question 91 Mark

In Fig., line segment AB is parallel to another line segment CD. O is the mid-point of AD. Show that:

$\triangle$AOB $\cong$ $\triangle$DOC
O is also the mid-point of BC.

Answer

Consider $\triangle$ AOB and $\triangle$DOC.
$\angle$ABO = $\angle$ DCO (Alternate angles as AB || CD and BC is the transversal)
$\angle$ AOB = $\angle$ DOC (Vertically opposite angles)
OA = OD (Given)
Therefore, $\triangle$AOB $\cong$ $\triangle$DOC (AAS rule)
OB = OC (CPCT)
So, O is the mid-point of BC

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