2 Mark Question - Maths STD 9 Questions - Vidyadip

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2 Mark Question

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26 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Draw a triangle $\Delta A _1 B_1 C _1$ on a card-sheet and cut it out.
Measure $\angle A _1, \angle B_1, \angle C _1$
Draw two more triangles $AA _2 B_2 C _2$ and $AA _3 B_3 C _3$ such that
$\angle A _1=\angle A _2=\angle A _3, \angle B_1=\angle B _2=\angle B _3, \angle C _1=\angle C _2=\angle C _3$ and $B _1 C _1> B _2 C _2> B _3 C _3$.
Now cut these two triangles also.
Measure the lengths of the three triangles. Arrange the triangles in two ways as shown in the figure.

Check the ratios $\frac{A_1 B_1}{A_2 B_2}, \frac{B_1 C_1}{B_2 C_2}, \frac{A_1 C_1}{A_2 C_2}$.
You will notice that the ratios are equal.
Similarly, see whether the ratios $\frac{A_1 C_1}{A_3 C_3}, \frac{B_1 C_1}{B_3 C_3}, \frac{A_1 B_1}{A_3 B_3}$ are equal.
What do you observe?

Answer

From the activity we observe that, when corresponding angles of two triangles are equal, the ratios of their corresponding sides are also equal i.e., their corresponding sides are in the same proportion.

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Question 22 Marks

The measures of angles of a set square in your compass box are 30°, 60° and 90°. Verify the property of the sides of the set square.

Answer

Given:-

30°,60° and 90°

To Find:-

The property of the sides of the set square.

Solution:-

30+60+90=180

180=180

=1 ans

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Question 32 Marks

Draw a triangle of any measure on a thick paper. Take a point T on ray QR as shown in the figure given below. Cut two pieces of thick paper which will exactly fit the comers of ∠P and ∠Q. See that the same two jpieces fit exactly at the comer of ∠PRT as shown in the figure.

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Question 42 Marks

In the adjoining figure, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR.

Given: Seg PT is the bisector of ∠QPR.
To prove: PS = PR
Construction: Draw seg SR || seg PT.

Answer

Proof:
seg PT is the bisector of ∠QPR. [Given]
∴ ∠QPT = ∠RPT ….(i)
seg PT || seg SR [Construction]
and seg QS is their transversal.

∴ ∠QPT = ∠PSR …(ii) [Corresponding angles]
seg PT || seg SR [Construction]
and seg PR is their transversal.

∴ ∠RPT = ∠PRS …..(iii) [Alternate angles]
∴ ∠PRS = ∠PSR …(iv) [From (i), (ii) and (iii)]
In ∆PSR,
∠PRS = ∠PSR [From (iv)]
∴ PS = PR [Converse of isosceles triangle theorem]

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Question 52 Marks

In the adjoining figure, bisector of ∠B AC intersects side BC at point D. Prove that AB > BD.

Given: Bisector of ∠BAC intersects side BC at point D.
To prove: AB > BD

Answer

Proof:
∠BAD ≅ ∠DAC ….(i) [Seg AD bisects ∠BAC]
∠ADB is the exterior angle of ∆ADC.
∴ ∠ADB > ∠DAC ….(ii) [Property of exterior angle]
∴ ∠ADB > ∠BAD ….(iii) [From (i) and (ii)]
In AABD,
∠ADB > ∠BAD [From (iii)]
∴ AB > BD [Side opposite to greater angle is greater]

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Question 62 Marks

Prove the Theorem : If two sides of a triangle are unequal, then the angle opposite to the greater side is greater than angle opposite to the smaller side.

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Question 72 Marks

Prove the Theorem : In a right angled triangle, the length of the median of the hypotenuse is half the length of the hypotenuse.

Answer

Given : In $\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}$, seg $\mathrm{BD}$ is the median.
To prove : $\mathrm{BD}=\frac{1}{2} \mathrm{AC}$
Construction : Take point E on the ray BD such that B - D - E and $l(\mathrm{BD})=l(\mathrm{DE})$. Draw seg EC.
Proof : (Main steps are given. Write the steps in between with reasons and complete the proof.)
$\Delta \mathrm{ADB} \cong \triangle \mathrm{CDE}$ by S-A-S test
line $\mathrm{AB} \|$ line $\mathrm{EC}$ by test of alternate angles
$\Delta \mathrm{ABC} \cong \triangle \mathrm{ECB}$ by $\mathrm{S}-\mathrm{A}-\mathrm{S}$ test
$\mathrm{BD}=\frac{1}{2} \mathrm{AC}$

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Question 82 Marks

Prove the Theorem: The measure of an exterior angle of a triangle is equal to the sum of its remote interior angles.

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Question 92 Marks

In the adjoining figure, in AABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD = seg BE.

Answer

Proof:
In ∆ADB and ∆BEA,
seg BD ≅ seg AE [Given]
∠ADB ≅ ∠BEA = 90° [Given]
seg AB ≅ seg BA [Common side]
∴ ∆ADB ≅ ∆BEA [Hypotenuse-side test]
∴ seg AD ≅ seg BE [c. s. c. t.]

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Question 102 Marks

In ∆FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.

Answer

In ∆FAN,
∠F + ∠A + ∠N = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 80° + 40° + ∠N = 180°
∴ ∠N = 180° – 80° – 40°
∴∠N = 60°
Since, 80° > 60° > 40°
∴ ∠F > ∠N > ∠A
∴ AN > FA > FN [Side opposite to greater angle is greater]
∴ In ∆FAN, AN is the greatest side and FN is the smallest side.

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Question 112 Marks

In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.

Answer

In ∆PQR,
PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given]
Since, 12 > 10 > 8
∴ QR > PQ > PR
∴ ∠QPR > ∠PRQ > PQR [Angle opposite to greater side is greater]
∴ In ∆PQR, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

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Question 122 Marks

In the adjoining figure, ∠RST = 56°, seg PT ⊥ ray ST, seg PR ⊥ ray SR and seg PR ≅ seg PT. Find the measure of ∠RSP.
State the reason for your answer.

Answer

seg PT ⊥ ray ST, seg PR ⊥ ray SR [Given]
seg PR ≅ seg PT
∴ Point P lies on the bisector of ∠TSR [Any point equidistant from the sides of an angle is on the bisector of the angle]
∴ Ray SP is the bisector of ∠RST.
∠RSP = 56° [Given]
$\therefore \angle RSP =\frac{1}{2} \angle RST$
$=\frac{1}{2} \times 56^{\circ}$
$\therefore \angle RSP =28^{\circ}$

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Question 132 Marks

In the adjoining figure, point A is on the bisector of ∠XYZ. If AX = 2 cm, then find AZ.

Answer

AX = 2 cm [Given]
Point A lies on the bisector of ∠XYZ. [Given]
Point A is equidistant from the sides of ∠XYZ. [Every point on the bisector of an angle is equidistant from the sides of the angle]
∴ A Z = AX
∴ AZ = 2 cm

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Question 142 Marks

In the given figure, ∆ABC is a right angled triangle, seg BD is the median on hypotenuse. Measure the lengths of the following segments.
i. AD
ii. DC
iii. BD
From the measurements verify that $BD =\frac{1}{2} AC$.

Answer

AD = DC = BD= 1.9 cm
AC = AD + DC [A – D – C]
= 1.9 + 1.9
= 2 x 1.9 cm
∴ AC = 2 x BD
$\therefore B D=\frac{1}{2} A C$

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Question 152 Marks

Can the theorem of isosceles triangle be proved without doing any construction?

Answer

Yes
Proof:
In ∆ABC and ∆ACB,
seg AB ≅ seg AC [Given]
∠BAC ≅ ∠CAB [Common angle]
seg AC ≅ seg AB [Given]
∴ ∆ABC ≅ ∆ACB [SAS test]
∴ ∠ABC ≅ ∠ACB [c. a. c. t.]

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Question 162 Marks

Can the theorem of isosceles triangle be proved by doing a different construction?

Answer

Yes
Construction: Draw seg AD ⊥ seg BC.
Proof:
In ∆ABD and ∆ACD,
seg AB≅ seg AC [Given]
∠ADB ≅ ∠ADC [Each angle is of measure 90°]
seg AD ≅ seg AD [Common side]
∴ ∆ABD ≅ ∆ACD [Hypotenuse side test]
∴ ∠ABD ≅ ∠ACD [c.a.c.t.]
∴ ∠ABC ≅ ∠ACB [B-D-C]

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Question 172 Marks

In the given figure, point G is the point of concurrence of the medians of ∆PQR. If GT = 2.5, find the lengths of PG and PT.

Answer

i. In ∆PQR, G is the point of concurrence of the medians. [Given]
The centroid divides each median in the ratio 2 : 1.
PG : GT = 2 : 1
$\therefore \quad \frac{ PG }{ GT }=\frac{2}{1}$
$\therefore \quad \frac{ PG }{2.5}=\frac{2}{1}$
$\therefore PG = 2 x 2.5$
∴ PG = 5 units
ii. Now, PT = PG + GT [P – G – T]
= 5 + 2.5
∴ l(PG) = 5 units, l(PT) = 7.5 units

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Question 182 Marks

The length of hypotenuse of a right angled triangle is 15. Find the length of median on its hypotenuse.

Answer

Length of hypotenuse $=15$ [Given]
Length of median on the hypotenuse $=\frac{1}{2} x$ length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse] $=\frac{1}{2} \times 15=7.5$
$\therefore$ The length of the median on the hypotenuse is 7.5 units.

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Question 192 Marks

In the adjoining figure, ZP ≅ ZR, seg PQ ≅ seg RQ. Prove that APQT ≅ ARQS.

Answer

Proof:
In ∆PQT and ∆RQS,
∠P ≅ ∠R
seg PQ ≅ seg RQ [Given]
∠Q ≅ ∠Q [Common angle]
∴ ∆PQT ≅ ∆RQS [ASA test]

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Question 202 Marks

In the adjoining figure, seg AB ≅ seg CB and seg AD ≅ seg CD. Prove that ∆ABD ≅ ∆CBD.

Answer

proof:
In ∆ABD and ∆CBD,
seg AB ≅ seg CB
seg AD ≅ seg CD [Given]
seg BD ≅ seg BD [Common side]
∴ ∆ABD ≅ ∆CBD [SSS test]

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Question 212 Marks

As shown in the adjoining figure, in ∆LMN and ∆PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

Answer

In ∆LMN and ∆PNM,
seg LM ≅ seg PN
seg LN ≅ seg PM [Given]
seg MN ≅ seg NM [Common side]
∴ ∆LMN ≅ ∆PNM [SSS test]
∴ ∠LMN ≅ ∠PNM,
∴ ∠MLN ≅ ∠NPM, and ∠LNM ≅ ∠PMN [c.a.c.t.]

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Question 222 Marks

From the information shown in the figure, state the test assuring the congruence of ∆ABC and ∆PQR. Write the remaining congruent parts of the triangles.

Answer

In ∆BAC and ∆PQR,
seg BA ≅ seg PQ
seg BC ≅ seg PR
∠BAC ≅ ∠PQR = 90° [Given]
∴ ∆BAC ≅ ∆PQR [Hypotenuse side test]
∴ seg AC ≅ seg QR [c.s.c.t.]
∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP [c.a.c.t.]

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Question 232 Marks

Can you give an alternative proof of the above theorem by drawing a line through point R and parallel to seg PQ in the above figure?

Answer

Yes.
Construction: Draw line RM parallel to seg PQ through a point R.
Proof:
seg PQ || line RM and seg PR is their transversal. [Construction]
∴ ∠PRM = ∠QPR ……..(i) [Alternate angles]
seg PQ || line RM and seg QR is their transversal. [Construction]
∴ ∠SRM = ∠PQR ……..(ii) [Corresponding angles]
∴ ∠PRM + ∠SRM = ∠QPR + ∠PQR [Adding (i) and (ii)]
∴ ∠PRS = ∠PQR + ∠QPR [Angle addition property]

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Question 242 Marks

Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.

Answer

i. ∠c + 100° = 180° [Angles in a linear pair]
∴ ∠c = 180° – 100°
∴ ∠c = 80°

ii. ∠b = 70° [Vertically opposite angles]
iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°]
∠a + 70° + 80° = 1800
∴ ∠a = 180° – 70° – 80°
∴ ∠a = 30°
∴ ∠a = 30°, ∠b = 70°,∠ c = 80°

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Question 252 Marks

In ∆PQR, ∠P = 70°, ∠Q = 65°, then find ∠R.

Answer

∠P = 70°, ∠Q = 65° [Given]
In ∆PQR,
∠P + ∠Q + ∠R = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + 65° + ∠R = 180°
∴ ∠R = 180° – 70° – 65°
∴ ∠R = 45°

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Question 262 Marks

In the adjoining figure, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.

Answer

∠A = 70° , ∠B = 40° [Given]
∠ACD is an exterior angle of ∆ABC. [Given]
∴ ∠ACD = ∠A + ∠B
= 70° + 40°
∴ ∠ACD = 110°

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