5 Mark Question

Question 15 Marks

Check the congruence of triangles.

Answer

Draw $\triangle A B C$ of any measure on a card-sheet and cut it out.
Place it on a card-sheet. Make a copy of it by drawing its border.
Name it as $\Delta A_1 B_1 C_1$.
Now slide the $\triangle A B C$ which is the cut out of a triangle to some distance and make one more copy of it.
Name it $\Delta A _2 B_2 C _2$.
Then rotate the cut out of triangle ABC a little, as shown in the figure, and make another copy of it.
Name the copy as $\triangle A_3 B_3 C_3$.
Then flip the triangle $A B C$, place it on another card-sheet and make a new copy of it.
Name this copy as $\triangle A_4 B_4 C_4$.
Have you noticed that each of $\triangle A_1 B_1 C_1, \triangle A_2 B_2 C_2, \triangle A_3 B_3 C_3$ and $\triangle A_4 B_4 C_4$ is congruent with $\triangle A B C$ ?
Because each of them fits exactly with $\triangle A B C$. Let us verify for $\triangle A_3 B_3 C_3$.
If we place $\angle A$ upon $\angle A 3, \angle B$ upon $\angle B 3$ and $\angle C$ upon $\angle C 3$, then only they will fit each other and we can say that $\triangle A B C=\triangle A_3 B_3 C_3$.
We also have $A B=A_3 B_3, B C=B_3 C_3, C A=C_3 A_3$.
Note that, while examining the congruence of two triangles, we have to write their angles and sides in a specific order, that is with a specific one-to-one correspondence. If $\triangle A B C \cong \triangle P Q R$, then we get the following six equations:
$\angle A=\angle P, \angle B=\angle Q \angle C=\angle R$
and $A B=P Q B C=Q R, C A=R P$.
This means, with a one-to-one correspondence between the angles and the sides of two triangles, we get hree pairs of congruent angles and three pairs of congruent sides.

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Question 25 Marks

In the adjoining figure, seg $A D \perp$ seg $B C$. Seg $A E$ is the bisector of $\angle C A B$ and $B-E-C$. Prove that $\angle DAE =\frac{1}{2}(\angle C -\angle B )$.

Given: seg AD ⊥ seg BC
seg AE is the bisector of ∠CAB.
To prove: $\angle DAE =\frac{1}{2}(\angle C-\angle B)[\because AD \perp BC ]$
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED ….(ii)

Answer

Proof:
$\therefore \angle CAE =\frac{1}{2} \angle A \ldots$..(i) [seg AE is the bisector of $\angle CAB$ ]
In ∆DAE,
∠DAE + ∠ADE + ∠AED = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ ∠DAE + 90° + ∠AED = 180° [∵ AD ⊥ BC]
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED ….(ii)
In ∆ACE,
∴ ∠ACE + ∠CAE + ∠AEC = 180° [Sum of the measures of the angles of a triangle is 180°]
∠C + -∠A + ∠AED = 180° [From (i) and C-D-E]
$\therefore \angle A E D=180^{\circ}-\angle C-\frac{1}{2} \angle A \ldots \ldots \text {.iii) }$
$\therefore \angle D A E=90^{\circ}-180^{\circ}-\angle C+\frac{1}{2} \angle A[\text { Substituting (iii) in (ii)] }$
$\therefore \angle D A E=\angle C+\frac{1}{2} \angle A-90^{\circ} \ldots . \text { (iv) }$
$\text { In } \triangle A B C$
$\angle A+\angle B+\angle C=180^{\circ}$
$\begin{array}{ll}\therefore & \frac{1}{2} \angle A +\frac{1}{2} \angle B +\frac{1}{2} \angle C =90^{\circ} \quad \text { [Dividing both sides by } 2 \text { ] } \\ \therefore & \frac{1}{2} \angle A =90^{\circ}-\frac{1}{2} \angle C -\frac{1}{2} \angle B \quad \text { (v) } \\ \therefore \quad & \angle DAE \\ & =\angle C +\left(90^{\circ}-\frac{1}{2} \angle C -\frac{1}{2} \angle B \right)-90^{\circ} \quad \text { [Substituting (v) in (iv)] }\end{array}$
$\begin{array}{ll}\therefore \quad \angle DAE =\angle C -\frac{1}{2} \angle C -\frac{1}{2} \angle B \\ =\frac{1}{2} \angle C -\frac{1}{2} \angle B \\ \therefore \quad \angle DAE =\frac{1}{2}(\angle C -\angle B )\end{array}$

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Question 35 Marks

∆ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Given: In isosceles ∆ABC, AB = AC. seg BD and seg CE are the medians of ∆ABC.
To prove: BD = CE

Answer

Proof: $A E=\frac{1}{2} A B$.....(i) $[E$ is the midpoint of side $A B]$ $A D=\frac{1}{2} A C \ldots$...(ii) $[D$ is the midpoint of side $A C]$
Also, AB = AC ……(iii) [Given]
∴ AE = AD ….(iv) [From (i), (ii) and (iii)]
In ∆ADB and ∆AEC,

seg AB ≅ seg AC ∠BAD ≅ ∠CAE
seg AD ≅ seg AE
∴ ∆ADB ≅ ∆AEC
∴ seg BD ≅ seg CE
∴ BD = CE

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Question 45 Marks

Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

Answer

From the information shown in the figure,
In ∆ABC and ∆PQR,
∠ABC ≅ ∠PQR
seg BC ≅ seg QR
∠ACB ≅ ∠PRQ
∴ ∆ABC ≅ ∆PQR [ASA test]
∴ ∠BAC ≅ ∠QPR [Corresponding angles of congruent triangles]
seg AB ≅ segPQ and segAC ≅ seg PR [Corresponding sides of congruent triangles]
ii
.
From the information shown in the figure,
In ∆PTQ and ∆STR,
seg PT ≅ seg ST
∠PTQ ≅ ∠STR [Vertically opposite angles]
seg TQ ≅ seg TR
∴ ∆PTQ ≅ ∆STR [SAS test]
∴ ∠TPQ ≅ ∠TSR and ∠TQP ≅ ∠TRS [Corresponding angles of congruent triangles]
seg PQ ≅ seg SR [Corresponding sides of congruent triangles]

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Question 55 Marks

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

Answer

By SSS test
∆ABC ≅ ∆PQR
ii.

By SAS test
∆ XYZ ≅ ∆LMN
iii.

By ASA test
∆PRQ ≅ ∆STU
iv.

By hypotenuse side test
∆LMN ≅ ∆PTR

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Question 65 Marks

In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,

i. $\angle DEG =\frac{1}{2} \angle EDF$
ii. $EF = FG$

Answer

i. ∠DEG = ∠FEG = x° ….(i) [Ray EG bisects ∠DEF]
∠GFD = ∠GFM = y° …..(ii) [Ray FG bisects ∠DFM]

line DE || line GF and DF is their transversal. [Given]
∴ ∠EDF = ∠GFD [Alternate angles]
∴ ∠EDF = y° ….(iii) [From (ii)]
line DE || line GF and EM is their transversal. [Given]

∴ ∠DEF = ∠GFM [Corresponding angles]
∴ ∠DEG + ∠FEG = ∠GFM [Angle addition property]
∴ x°+ x° = y° [From (i) and (ii)]
∴ 2x° = y°
$\therefore x ^{\circ}=\frac{1}{2} y ^{\circ}$
$\therefore \angle DEG =\frac{1}{2} \angle EDF [\text { From (i) and (iii)] }$
ii. line DE || line GF and GE is their transversal. [Given]

∴ ∠DEG = ∠FGE …(iv) [Alternate angles]
∴ ∠FEG = ∠FGE ….(v) [From (i) and (iv)]
∴ In ∆FEG,
∠FEG = ∠FGE [From (v)]
∴ EF = FG [Converse of isosceles triangle theorem]

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Question 75 Marks

In ∆ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°, find the measure of ∠AOB.

Answer

∠OAB = ∠OAC = – ∠BAC ….(i) [Seg AO bisects ∠BAC]
∠OBA = ∠OBC = – ∠ABC …..(ii) [Seg RO bisects ∠ABC]
In AABC,
∠BAC + ∠ABC + ∠ACB = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ ∠BAC + ∠ABC + 70° = 180°
∴ ∠BAC + ∠ABC = 180°- 70°
∴ ∠BAC + ∠ABC = 110°
$\therefore \frac{1}{2}(\angle BAC )+\frac{1}{2}(\angle ABC )=\frac{1}{2} \times 110^{\circ}\left[\right.$ Multiplying both sides by $\left.\frac{1}{2}\right]$
∴ ∠OAB + ∠OBA = 55° ….(iii) [From (i) and (ii)]
In AOAB,
∠OAB + ∠OBA + ∠AOB = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 55° + ∠AOB = 180° [From (iii)]
∴ ∠AOB = 180°- 55°
∴ ∠AOB = 125°

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Maths 5 Mark Question

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