Math Class 9 Question Bank

Question 1 of 24

1. Question
1 point(s)
Can we have a cyclic quadrialateral $ABCD $ such that $\angle A=90^o,\angle B=70^o,\angle C=95^o $ and $\angle D=105^o $?
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Hint

False
$\angle A+\angle C=90+95=185\ne180\\ \\ \angle B+\angle D=70+105=175\ne180$
Sum of opposite angles is not equal to $180^o$ . So it is not a cyclic quadrialateral
Question 2 of 24

2. Question
1 point(s)
Two congruent circles with centres $O $ and $O’ $ intersect at two points $ A$ and $B $. Check whether $\angle AOB=\angle AO’B $ or not.
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Hint

In $\triangle AOB$ and $\triangle AO'B$
$OA=O'A$ (Radius)
$OB=BO'$ (Radius)
$AB=AB$ (common)
$\triangle AOB\cong\triangle AO'B$
$\angle AOB=\angle AO'B$ (cpct)
Question 3 of 24

3. Question
1 point(s)
In the figure, PQR is right angled at Q. Point S is taken on side PR such that PS = SR and QR = QS. Find the measure of $\angle $QSR
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Hint

$SP=SQ=SR$ (Radius)
But $QS=SR$ (given) and $PS=SR$ (given)
$QS=SR=QR$
So $\triangle QSR$ is an equlateral $\triangle$
So $\angle QRS=60^o$
Question 4 of 24

4. Question
1 point(s)
Show that diameter of a circle is the greatest chord.
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Hint

Take any chord in a circle with end points $AB$ . Let $O$ be the centre of the circle. Then segments $AO$ and $BO$ are ratio of the circle $AOB$ is a triangle.
$AB>AO+OB$ (Sum of two sides is greater then the third side)
$Ao+Bo=2r=d$
The maximum length of chord $AB$ is $d$
Question 5 of 24

5. Question
1 point(s)
In the figure, $\triangle $ABC is equilateral. Find $\angle $BDC and $\angle $BEC
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Hint

$\angle BAC=60^o$
$\angle BAC=\angle BDC=60^o$ (angle made by same arc are equal)
$\angle BDC+\angle BEC=180^o$ (Sum of opposite vertices of cyclic qaudrilateral is $180^o$ )
Question 6 of 24

6. Question
1 point(s)
In the figure, if $AOB $ is a diameter and $\angle ADC=120^o $, find $\angle CAB $
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Hint

$\angle ADC+\angle CBA=180^o\\ \\ \angle CBA=180-120=60^o$
In $\triangle ACB$
$\angle CAB+60+90=180^o\\ \\ \angle CAB=180-150=30^o$
Question 7 of 24

7. Question
1 point(s)
If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD
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Hint

Let $AXB$ and $CYD$ are arcs of cirlces whose centre and radius are $o$ and $r$ units. Hence ratio of $AB$ and $CD$ is $1:1$
Question 8 of 24

8. Question
1 point(s)
Show that two circles cannot intersect at more than two points.
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Hint

Since 2 distinct circles intersect at more than 2 points. These points are non-collinear points as 3 non collinear points determine one and only one circle, there should be only one circle. This contradicts assumption, so 2 circles cannot intersect each other at more than 2 points.
Question 9 of 24

9. Question
1 point(s)
Show that the altitudes of a triangle are concurrent.
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Hint

$RA=BC$ (Since RACB a parallelogram)
$AQ=BC$ (ABCD is a parallelogram)
$RA=AQ$ …(1)
$AE\bot RQ$ Since $AE$ is $\bot BC$
$AE$ is the perpendicular bisector of $RQ$
Similarly $BD$ and $CF$ is the perpendicular bisector of $RP$ and $QP$ respectively.
$AE,BD$ and $CF$ are concurrent as we know that the perpendicular bisector of the sides of a triangle are concurrent at a point called circumcenter.
Question 10 of 24

10. Question
1 point(s)
Prove that angle bisector opposite angle of a triangle and the perpendicular bisector of the opposite side if intersect, they will interect on the circumcircle of the triangle.
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Hint

$\angle BCD=\angle BAD$ (Angles in the same segment)
$\angle BCD=\angle BAD=\dfrac{1}{2}\angle A$ …(1)
( $\because\ AD$ is bisector of $\angle A$ )
Similarly $\angle DBC=\angle DAC=\dfrac{1}{2}\angle A$ …(2)
From (1) and (2)
$\angle DBC=\angle BCD\\ \\ BD=DC$
$\Rightarrow D$ lies on perpendicular bisector of $BC$
Question 11 of 24

11. Question
1 point(s)
ABC is an equilateral triangle inscribed in a circle and P be any point on the minor are BC which does not coincide with B or C. Prove that PA is angle bisector of $\angle $BPC.
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Hint

$AB=AC=BC$ ( $\because\ ABC$ is equalatiral $\triangle$ )
$\angle AOB=\angle AOC=\angle BOC$ (equal chord subtend equal angles at centre)
$\angle AOB=\angle AOC$ …(1)
$\angle APB=\dfrac{1}{2}\angle AOB$ …(2)
$\angle APC=\dfrac{1}{2}\angle AOC$ …(3)
So, $\angle APB=\angle APC$ (Proved)
Question 12 of 24

12. Question
1 point(s)
Prove that the angles in a segment greater than a semi–circle is less than a right angle.
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Hint

$\angle AOB=2\angle ACB$ (central angle is twice of inscribed angle)
$\angle AOB<180^o\\ \\ 2\angle ACB<180^o$
$\angle ACB<90^o$ (Proved)
Question 13 of 24

13. Question
1 point(s)
If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection. Prove that the chords are equal.
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Hint

In $\triangle OLE$ and $\triangle OME$
$\angle OLE=\angle OME$ (each $90$ )
$\angle DEL=\angle OEM$ (given)
$OM=OM$ (common)
$\triangle OLE\cong\triangle OME$ (AAS)
$OL=OM$ (cpct)
$AB=CD$ (chords equidistant from the centre are equal)
Question 14 of 24

14. Question
1 point(s)
AB and AC are two chords of a circle of radius r units. If AB = 2AC, and the length of the perpendicular from the centre on these chords are a and b respectively, prove that $4b^2=a^2+3r^2 $.
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Hint

$AM=MC=\dfrac{P}{2}\\ \\ AN=NB=P$
In $\triangle AOM$
$AO^2=AM^2=MO^2$
$=\left(\dfrac{P}{2}\right)^2+b$ …(1)
In $\triangle OAN$
$AO^2=(AN)^2+(NO)^2\\ \\ =p^2+a^2$ …(2)
From (1) and (2)
$\left(\dfrac{p}{2}\right)^2+b^2=p^2+a^2\\ \\ \dfrac{p^2}{4}+b^2=p^2+a^2\\ \\ p^2+4b^2=4p^2+4a^2\\ \\ 4b^2=3P^2+4a^2\\ \\ =a^2+3p^2+3a^2\\ \\ =a^2+3(p^2+a^2)\\ \\ 4b^2=a^2+3r^2\ (\because\ r^2=a^2+p^2)$
Question 15 of 24

15. Question
1 point(s)
P is the centre of the circle. Prove that $\angle XPZ=2\ [\angle XYZ+\angle XZY] $.
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Hint

$\angle YPX=2\angle XZY$ …(1) (central angle is twice of inscribed angle)
$\angle YPZ=2\angle YXZ$ …(2) (central angle is twice of inscribed angle)
Adding equation (1) and (2)
$\angle YPX+\angle YPZ=2\ (\angle XZY+YXZ)\\ \\ \angle XPZ=2\ (\angle XZY+\angle YXZ)$
$(\because\ \angle YPX+\angle YPZ=\angle XPZ)$ (Proved)
Question 16 of 24

16. Question
1 point(s)
Prove that the angle subtended by an arc at the centre is double the angles subtended by it at any point on the remaining part of the circle.
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Hint

In $\triangle OAQ$
$OA=OQ$ (Radius) $\Rightarrow \angle OAQ=\angle OQA$ …(1)
$\angle BOQ=\angle OAQ+\angle AQO$ (exterior angle is equal to the sum of interior opposite angle)
$\Rightarrow \angle BOQ=2\angle OAQ$ …(2)
Similarly $\angle BOP=2\angle OAP$ ….(3)
Adding (2) and (3)
$\angle BOP+\angle BOQ=2(\angle OAP+\angle OAQ)\\ \\ \angle POQ=2\angle PAQ$ …(4)
In case 3
Relax angle $POQ=2\angle PAQ$ (Proved)
Question 17 of 24

17. Question
1 point(s)
In the figure. $ABCD $ is a cyclic quadrilateral, $O $ is the centre of the circle. If $ \angle BOD=160^o$, find $\angle BPD $.
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Hint

$\angle BAD=\dfrac{160}{2}=80^o$
Since $ABCD$ is a cyclic quadrilatral
$\angle BCD=180-80=100^o$
$\angle BPD=100^o$ (Since angles on the same segment are equal)
Question 18 of 24

18. Question
1 point(s)
In the figure, $ABCD $ is a cyclic quadrilateral in which $AB $ is produced to $F $ and $BE||DC $. If $\angle FBE=20^o $ and $\angle DAB=95^o $, find $ \angle ADC$.
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Hint

$\angle DAB+\angle BCD=180^o$ (opposite angles of a cyclic quadrilateral)
$\angle BCD=180-95=85^o\\ \\ \angle CBE=\angle BCD=85^o$ (alt $\angle S$ )
$\angle CDF=\angle CBE+\angle FBE=85+20=105^o\\ \\ \angle ABC+\angle ADC=\angle ABC+\angle CBF=180^o\\ \\ \Rightarrow \angle ADC=\angle CBF=105^o$
Question 19 of 24

19. Question
1 point(s)
In the figure, B and E are points on the line segment AC and DF respectively. Show that AD || CF.
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Hint

$\angle CAD+\angle BED=180^o$ …(1) (opposite $\angle S$ of a cyclic quadrilateral)
$\angle BED+\angle BEF=180^o$ (Linear pair) …(2)
From (1) and (2)
$\angle CAD=\angle BEF\\ \\ \angle BEF+\angle FCA=180^o$
$\angle CAD+\angle FCA=180^o$ (Since the co-interior angles are supplementry)
$AD||CF$
Question 20 of 24

20. Question
1 point(s)
If O is the centre of a circle as shown in the given figure, then prove that x + y = z.
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Hint

In $\triangle ADF$
$\angle y=\angle 1+\angle 3$ …(1)
In $\triangle ACE$
$\angle 1+\angle x=\angle 4$ ….(2)
$\angle 1+\angle x+\angle y=\angle 1+\angle 3+\angle4\\ \\ \angle x+\angle y=\angle 3+\angle 4$
$=2\angle 3$ ( $\angle3=\angle 4$ angles in the same segment)
$\angle x+\angle y=\angle z\ (AOB=2\angle ADB)$
Question 21 of 24

21. Question
1 point(s)
Two equal chords AB and CD of a circle when produced intersect at the point P. Prove that PB = PD.
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Hint

$AB=CD$
$OL=OM$ (equal chords are equidistant from the centre)
In $\triangle OLP$ and $\triangle OMP$
$OL=OM$
$\angle OLP=\angle OMP$ (each 90)
$OP=OP$ (common)
$\triangle OLP\cong\triangle OMP$ (RHS)
$LP=MP$ (cpct)…(1)
$AB=CD\\ \\ \dfrac{1}{2}AB=\dfrac{1}{2}CD\\ \\ BL=DM$ …(2)
$LP-BL=MP-DM\\ \\ \Rightarrow PB=PD$
Question 22 of 24

22. Question
1 point(s)
In the figrue, find the values of a, b, c and d. Given $\angle BCD=43^o$ and $\angle BAE=62^o $.
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Hint

$\angle C=\angle BAE$ (exterior angle of a cyclic quadrilateral is equal to its interior opposite angle)
$\angle C=62$ …(1)
In $\triangle AEC$
$\angle ACE+\angle CAE+\angle d=180^o$
$\angle d=75$ …(2)
$\angle a+\angle d=180^o\\ \\ \Rightarrow \angle a=180-75=105$
In $\triangle FDE,\angle C+(180-\angle d)+\angle b=180^o$
$62+180-75+\angle b=180^o\\ \\ \angle b=13$
Question 23 of 24

23. Question
1 point(s)
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that $\angle $ACP = $\angle $QCD.
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Hint

$\angle 1=\angle 3$ (Angles on the same segment) …(1)
$\angle 2=\angle 4$ (Angles on the same segment) …(2)
$\angle 3=\angle 4$ (VOA)
So, $\angle 1=\angle 2$
$\angle ACP=\angle QCD$
Question 24 of 24

24. Question
1 point(s)
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base of the triangle.
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Hint

$\angle ADB=90^o$ (angle in a semicircle is 90)
$\angle ADB+\angle ADC=180^o$ (Linear pair)
$90^o+\angle ADC=180^o\\ \\ \angle ADC=90^o$
In $\triangle ABD$ and $\triangle ACD$
$AB=AC$ (equal sides of isosceles triangle)
$\angle ADB=\angle ADC$ (each 90)
$AD=AD$ (common)
$\triangle ABD\cong\triangle ACD$ (RHS)
$BD=DC$ (Proved)