Math Class 9 Question Bank

Question 1 of 24

1. Question
1 point(s)
A triangle $ ABC$ can be constructed in which $\angle B=60^o,\angle C=45^o $ and $AB + BC + AC = 12\ cm$. Is this statement true ? Justify your answer
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Hint

$\angle A+\angle B+\angle C=180^o\\ \\ \angle B+\angle C=60+45=105<180$
$\triangle ABC$ can be constructed with the given conditions.
Question 2 of 24

2. Question
1 point(s)
Can we construct a triangle $ ABC$ in which $\angle A=105^o,\angle B=75^o $ and $ AB + BC + AC = 12\ cm$?
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Hint

$\angle A+\angle B+\angle C=180^o\\ \\ \angle A+\angle B=105+75=180$
I.e. $\angle C=0$ which is not possible
So $\triangle ABC$ cannot be constructed with the given conditions
Question 3 of 24

3. Question
1 point(s)
Can we construct a triangle $ABC $ in which $\angle B=105^o,\angle C=90^o $ and $ AB+BC+AC=10\ cm$?
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Hint

$\angle A+\angle B+\angle C=180^o\\ \\ \angle B+\angle C=105+90=195>180$
$\triangle ABC$ cannot be constructed with the given conditions
Question 4 of 24

4. Question
1 point(s)
Construct $\triangle ABC $ in which $BC=6.4\ cm,\angle B=45^o $ and $AB – AC = 2\ cm $.
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Hint

Step 1 : Draw a line segment BC of length 6.4 cm.
Step 2 : Draw an angle of 45° at point B on CB using protractor
Step 3 : Draw Ray BP
Step 4 : From B take BX = 2 cm
Step 5 : Join CX
Step 6 : Draw perpendicular bisector of CX intersect ray BP at A
Step 7 : Join A to C,
ABC is the required triangle.
Question 5 of 24

5. Question
1 point(s)
Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B respectively. Are these lines parallel ?
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Hint

To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction.
1. Draw a line segment AB = 4 cm.
2.Taking 4 as centre and radius more than $\dfrac{1}{2}$ AB (i.e., 2 cm) draw an arc say it intersect AB at E.
3.Taking E as centre and with same radius as above draw an arc which intersect previous arc at F.
Again, taking F as centre and with same radius as above draw an arc which intersect previous arc (obtained in step ii) at G.
5.Taking G and F are centres, draw arcs which intersect each other at H.
6.Join AH . Thus, AX is perpendicular to AB at A. Similarly, draw BY $\bot$ AB at B.
Now, we know that if two lines are parallel, then the angle between them will be 0°
or
180°.
Here, $\angle XAB = 90^o [\therefore XA \bot AB]$
and $\angle YBA = 90^o\ [ \therefore\ YB \bot AB]$
$\angle XAB+ \angle YBA = 90^o + 90^o = 180^o$
So, the lines XA and YS are parallel.
Question 6 of 24

6. Question
1 point(s)
Construct $\triangle ABC $ in which $BC=4.5\ cm,\angle B=45^o $ and $AB+AC=5.6\ cm $
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Hint

(1) Draw a line segment $BC=4.5\ cm$
(2) $B$ draw angle $45^o$
(3) Cut a point $y$ on line $Bx$ like $By=5.6 \ cm$
(4) Join $Cy$
(5) Bisect the line $Cy$ which cuts at point $A$ Join $AC$
$\triangle ABC$ is the required triangle
Question 7 of 24

7. Question
1 point(s)
Construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm. Bisect its smallest angle and measure each part.
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Hint

1. Draw a line segment BC of length 4.8 cm.
2. From B, point A is at a distance of 3.6 cm. So, having B as centre, draw an arc of radius 3.6 cm.
3. From C, point A is at a distance of 3 cm. So, having C as centre, draw an arc of radius 3 cm which intersect previous arc at A.
4. Join AB and AC. Thus, $\triangle$ ABC is the required triangle.
Flere, angle B is smallest, as AC is the smallest side. To direct angle B, we use the following steps.
1.Taking B as centre, we draw an are intersecting AB and BC at D and E, respectively.
2.Taking D and E as centres we draw arcs intersecting at P.
3. Joining BP, we obtain angle bisector of $\angle$ B.
4. Flere, $\mathrm{\angle ABC=39^o}$
Thus, $\mathrm{\angle ABD = \angle DBC = \dfrac{1}{2} \times 139^o = 19.5^o}$
Question 8 of 24

8. Question
1 point(s)
Construct a right angled triangle in which the base is 3 cm and the difference of hypotenuse and perpendicular is 1 cm.
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Hint

(1) Draw a line segment $BC=3\ cm$
(2) At $B$ , draw $\angle xBC=90^o$
(3) Produce $xB$ to $x'$ such that $XBx'$ is a line
(4) From ray $BX$ , cast off $BD=1\ cm$
(5) Join $CD$
(6) Draw the perpendicular bisector $PQ$ of $CD$ they intermates $xBx'$ at $A$
(7) Join $AC$
Question 9 of 24

9. Question
1 point(s)
Construct a triangle $XYZ $ in which $\angle Y=30^o,\angle Z=90^o $ and $XY + YZ + ZX = 11\ cm $
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Hint

(1) Draw a line segment $AB=11\ cm$
(2) Make $\angle y=30^o$ from point $A$ .
(3) Make a angle $\angle z=90^o$ from point $B$
(4) Bisect $\angle LAB$ and $\angle MBA$ . let these bisectors intersect at $X$ .
(5) Make perpendicular bisector of $XA$ . Let it intersect $AB$ at $y$ .
(6) Make perpendicular bisector $MXB$ . Let it intersect $AB$ at $2$ .
(7) Join $XY$ and $XZ$
$\triangle XYZ$ is the required $\triangle$
Question 10 of 24

10. Question
1 point(s)
Construct a triangle $ABC $ in which $BC=7\ cm,\angle B=75^o $ and $AB+AC=13\ cm $.
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Hint

(1) Draw base $BC=7\ cm$
(2) $\angle B=75^o$ Let the ray be $BX$
From $B$ as centre cut an arc on ray $BX$ . Let it intersect $BX$ at $D$
(4) Join $CD$ and draw perpendicular bisector of $CD$
(5) Mark point $A$ where perpendicular bisector intersects $BD$
(6) Join $AC$
$\triangle ABC$ is the required $\triangle$
Question 11 of 24

11. Question
1 point(s)
Construct a triangle $ ABC$ in which $BC=4\ cm $. $\angle B=30^o $ and $AB+AC=6\ cm $
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Hint

When $AB+AC=6\ cm$
then $AB=3AC=3\ cm$
Question 12 of 24

12. Question
1 point(s)
Construct a $\triangle $PQR with its perimeter = 10.4 cm and base angle of 45° and 120°
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Hint

Let $\angle B=45^o$ and let $\angle C=120^o$ .
Steps.
1. Draw a line segment say XY and equal to perimeter i.e., $AB+ BC + CA = 10.4\ cm$ .
2. Draw $\angle LXY = \angle B = 45^o$ and $\angle MYX = \angle C = 120^o$ .
3. Make the bisector $\angle LXY$ and $\angle MYX$ and let them meet at a know point $A$ .
4. Draw perpendicular bisectors $PQ$ and $RS$ of $AX$ and $AY$ , respectively.
5. Let $PQ$ intersect $XY$ at $B$ and $RS$ intersect $XY$ at $C$ .
Join $AB$ and $AC$ .
Question 13 of 24

13. Question
1 point(s)
Construct $\triangle XYZ $ in which $\angle Y=30^o\ \angle Z=90^o $ and perimeter is 11 cm.
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Hint

Given: In triangle $XYZ,\angle Y=30^o,\angle Z=90^o$ and $XY+YZ+ZX=11 \ cm$ .
Required: To construct the $\triangle XYZ$ .
Steps of Construction:
1. Draw a line segment $BC=XY+YZ+ZX-11\ cm$
2. Make $\angle LBC=\angle Y\ (=30^o)$ and $\angle MCB=\angle Z\ =90^o$
3. Bisect $\angle LBC$ and $\angle MCB$ . Let these bisectors meet at a point $X$ .
4. Draw perpendicular bisectors $DE$ of $XB$ and $FG$ of $XC$ .
5. Let $DE$ intersect $BC$ at $Y$ and $FG$ intersect $BC$ at $Z$ .
6. Join $XY$ and $XZ$ .
Then, $XYZ$ is the required triangle.
Question 14 of 24

14. Question
1 point(s)
c a right triangle whose base is 6 cm and the difference of its hypotenuse and the other side is 2 cm.
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Hint

Steps of Construction :
1. Draw the base BC = 6 cm and at point B make an angle of $90^o$ .
2. Cut a line segment BD equal to AC – AB = 2 cm from the line BX extended on opposite side of line segment BC.
3. Join DC and draw the perpendicular bisector, say PQ of DC.
4. Let PQ intersect BX at A. Join AC
Then, ABC is the required triangle.
Question 15 of 24

15. Question
1 point(s)
Construct $\triangle ABC $ such that $\angle B=60^o,\ \angle C=45^o $ and $AB + BC + CA = 10\ cm. $
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Hint

Steps of Construction :
1. Draw a line segment, say XY equal to BC + CA + AB = 10 cm.
2. Make angle LXY equal to $60^o$ and angle MYX equal to $45^o$ .
3. Bisect $\angle$ LXY and $\angle$ MYX. Let these bisectors intersect at point A.
4. Draw perpendicular bisector PQ of AX and RS of AY.
5. Let PQ intersect XY at B and RS intersect XY at C.
6. Join AB and AC.
ABC is the required triangle.
Question 16 of 24

16. Question
1 point(s)
Construct an equilateral triangle if its altitude is 4 cm. Give justification of your construction.
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Hint

Steps of Construction :
1. Draw any line segment $PQ$ .
2. Take a point $D$ on $PQ$ and at $D$ , construct perpendicular $DR$ to $PQ$ . From $DR$ , cut off $DA=4\ cm$ .
A $A$ , construct $\angle DAS=\angle DAT=\dfrac{1}{2}\times60^o=30^o$ on either side of $AD$ . Let $AS$ and $AT$ meet $PQ$ at points $B$ and $C$ respectively. Then, $ABC$ is the required equilateral triangle.
Question 17 of 24

17. Question
1 point(s)
Construct a triangle $ ABC$ in which $\angle A=45^o,\ \angle B=120^o $ and $AB+BC=10.4\ cm $
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Hint

Steps of Contruction:
1. Draw a ray DX and cut off a line segment DE= 10.4 cm from it.
2. At point D, construct $\angle YDE=\dfrac{1}{2}\times45^o=22\dfrac{1}{2}^o$ .
3. At point E, construct $\angle ZED=\dfrac{1}{2}\times120^o=60^o$ .
4. Extend the rays DX and EY to meet at a point and name it as C.
5. Draw the perpendicular bisector of CD intersecting DE at a point A.
6. Draw a perpendicular bisector of CE intersecting DE at a point B.
7. Join CA and CB. $\triangle$ ABC is the required triangle.
Question 18 of 24

18. Question
1 point(s)
Construct a right triangle in which one side is 3.5 cm and sum of other side and potenuse is 5.5 cm
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Hint

Let given right triangle be ABC.
Then, given BC = 3.5 cm, $\angle$ B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct $\triangle$ ABC use the following steps
1. Draw the base BC = 3.5 cm
2. Make an angle XBC = 90° at the point B of base BC.
3. Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB.
4. Join DC and make an $\angle$ DCY equal to $\angle$ BDC.
5. Let Y intersect BX at A.
Thus, $\triangle$ ABC is the required triangle.
Justification
Base BC and $\angle$ B are drawn as given.
In $\triangle$ ACD, $\angle$ ACD = $\angle$ ADC [by construction]
AD = AC …(i)
[sides opposite to equal angles are equal] Now, AB = BD – AD = BD – AC [from Eq. (i)]
$\Rightarrow$ BD = AB + AC
Thus, our construction is justified.
Question 19 of 24

19. Question
1 point(s)
Construct a triangle ABC, given that perimeter is $12.5\ cm,\ \angle B=60^o $ and $\angle C=75^o $
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Hint

Steps of construction
1. Draw a line segment and locate points $P,Q$ such that $PQ = 12.5\ cm$ .
2. Construct rays $P\vec R$ such that $\angle QPR=60^o$ and $Q\vec S$ such that $\angle PQS=75^o$ .
3. Draw the bisectors $PL$ and $QM$ of $\angle QPR$ and $\angle PQS$ respectively. Let these intersect at A.
4. Draw the perpendicular bisector of AP and AQ and let these intersect PQ at B and C respectively.
5. Join AB and AC.
Then ABC is the required triangle.
Question 20 of 24

20. Question
1 point(s)
Construct a rhombus whose diagonals are 4 cm and 6 cm in lengths.
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Hint

We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another.
So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps.
1. Draw the diagonal say AC = 4 cm
2. Taking A and C as centres and radius more than $\dfrac{1}{2}$ AC draw arcs on both sides of the line segment AC to intersect each other.
3. Cut both arcs intersect each other at P and Q, then join PQ.
4. Let PQ intersect AC at the point O. Thus, PQ is perpendicular bisector of AC.
5. Cut off 3 cm lengths from OP and OQ, then we get points B and D.
Now, join AB, BC, CD, and DA.
Thus, ABCD is the required rhombus.
Question 21 of 24

21. Question
1 point(s)
Construct a triangle ABC such that AB = BC = 6 cm and median AD = 4 cm.
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Hint

Steps of construction:
Step 1: Draw a line segment BC of 6 cm.
Step 2: Take midpoint D of BC.
Step 3: with center B and D and radii 6 cm and 4 cm draw two arcs which intersects each other A
Step 4: Join AB, AD and AC
Therefore $\triangle$ ABC is the required triangle
Question 22 of 24

22. Question
1 point(s)
Construct a triangle with perimeter 10 cm and base angle 60º and 45º
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Hint

Steps of Construction :
1. Draw a line segment, say XY equal to BC + CA + AB = 10 cm.
2. Make angle LXY equal to $60^o$ and angle MYX equal to $45^o$ .
3. Bisect $\angle$ LXY and $\angle$ MYX. Let these bisectors intersect at point A.
4. Draw perpendicular bisector PQ of AX and RS of AY.
5. Let PQ intersect XY at B and RS intersect XY at C.
6. Join AB and AC.
ABC is the required triangle.
Question 23 of 24

23. Question
1 point(s)
Construct a triangle $ABC $ in which $BC=7.5\ cm $. $\angle B=45^o $ and $ AC-AB=2.5\ cm$.
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Hint

Steps of Construction :
1. Draw the base BC 7.5 cm and at point B make an angle say XBC of 45 degree.
2. Cut line segment BD equal to AC – BC (2.5 cm) from line BX extended on opiate side of BC.
3. Join DC and draw perpendicular bisector, say PQ of DC.
4. Let PQ intersect BX at A. Join AC.
Thus, ABC is the required triangle.
Question 24 of 24

24. Question
1 point(s)
Construct a $\triangle ABC $ in which $\angle B=60^o $ and $\angle C=45^o $ and the perimeter of the triangle is 11 cm.
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Hint

Steps of Construction :
1. Draw a line segment PQ = 11 cm (= AB + BC + CA).
2. At P construct an angle of 60° and at Q, an angle of 45°.
3. Bisect these angles. Let the bisectors of these angles intersect at a point A.
4. Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.
5. Join AB and AC
ABC is the required triangle.