5 Marks Questions

Question 15 Marks

Find the missing angles. As per the convention that we have been following, all line segments marked with a single ‘|’ are equal to each other, and those marked with a double ‘|’ are equal to each other, etc.

Answer

In ∆CUR
∠CUR = ∠CRU = x (say)
(∵ CU = CR)
According to the angle sum property of a triangle
x + x + 90° = 180°
⇒ 2x = 180° – 90° = 90°
⇒ x = 45°
∴ ∠CUR = ∠CRU = 45°
In ∆VRN
∠VRN = ∠VNR = a (say)
[∵ Angles opposite to equal sides are equal]
∵ VR = VN
Since the sum of the angles of a triangle is 180°.
So, a + a + 68° = 180°
⇒ 2a = 180° – 68°
⇒ 2a = 112°
⇒ a = 56°
∠VRN = ∠VNR = 56°
In ∆AUP
∠UAP = ∠UPA (∵ they are equal)
∠UPA = 56°
The sum of the angles of a triangle is 180°.
So, 56° + 56° + ∠AUP = 180°
⇒ 112° + ∠AUP = 180°
⇒ ∠AUP = 180° – 112°
⇒ ∠AUP = 68°
∆BOF is an equilateral triangle as all sides are equal.
So, OB = OF = BF
∠FOB = ∠FBO = ∠OFB = 60°
∠RVN + ∠DVN = 180°
⇒ 68° + ∠DVN = 180°
⇒ ∠DVN = 180° – 68°
⇒ ∠DVN = 112°
∠VND + ∠VDN + ∠NVD = 180°
∵ VN = VD
∴ ∠VND = ∠VDN = c
∴ c + c + 112° = 180°
⇒ 2c = 180° – 112°
⇒ 2c = 68°
⇒ c = 34°
∠VND = ∠VDN = 34°
In ∆OLB
∠OBL = 90° – 60° = 30°
∠LOB = 60° [∵ LO || BF and BO is transversal]
In ∆OPN
∠OPN + ∠PON + ∠PNO = 180°
⇒ ∠OPN + 56° + 90° = 180°
⇒ ∠OPN + 146° = 180°
⇒ ∠OPN = 180° – 146°
⇒ ∠OPN = 34°
Now, ∠APK + ∠KPO + ∠OPN = 180° [∵ Straight angle is 180°]
⇒ 44° + ∠KPO + 34° = 180°
⇒ ∠KPO = 180° – 78°
⇒ ∠KPO = 102° In ∆KPO
∠KPO + ∠POK + ∠PKO = 180°
⇒ 102° + 30° + ∠PKO = 180°
⇒ 132° + ∠PKO = 180°
⇒ ∠PKO = 180° – 132° = 48°
∠KAP + ∠KPA + ∠AKP = 180° [∵ Sum of angles of a triangle is 180°]
⇒ 34° + 44° + ∠AKP = 180°
⇒ 78° + ∠AKP = 180°
⇒ ∠AKP = 180° – 78°
⇒ ∠AKP = 102° and ∠PKO = 48°
So, ∠AKP + ∠PKO + ∠OKL = 180°
⇒ 102° + 48° + ∠OKL = 180°
⇒ 150° + ∠OKL = 180°
⇒ ∠OKL = 180° – 150°
⇒ ∠OKL = 30°
In ∆KOL
∠OKL + ∠OLK + ∠KOL = 180°
⇒ 30° + 90° + ∠KOL = 180°
⇒ ∠KOL = 180° – 120°
⇒ ∠KOL = 60°
Also, ∆OKL ≅ ∆OBL
KL = LB
∠OLK ≅ ∠OLB = 90°
Side angle side condition.

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Question 25 Marks

ABCD is a square. Show that ∆ABC ≅ ∆ADC. Is ∆ABC also congruent to ∆CDA?

Give more examples of two triangles where one triangle is congruent to the other in two different ways, as in the case above. Can you give an example of two triangles where one is congruent to the other in six different ways?

Answer

To show ∆ABC ≅ ∆ADC (∵ ABCD is a square)
Now, AD = AB
CD = CB
AC = AC (Common side)
So, ∆ABC ≅ ∆ADC (Using SSS criterion)
For ∆ABC to be congruent to ∆CDA
AB = CD
BC = DA
AC = CA
Since ABCD is a square
Hence, ∆ABC is congruent to ∆CDA by side side-side-side condition.
Now, let us take two congruent triangles ∆HEN and ∆BIG.

There are six ways to write a congruence statement for two congruent triangles.
The other five ways are
(i) ∆HNE ≅ ∆BGI
(ii) ∆EHN ≅ ∆IBG
(iii) ∆ENH ≅ ∆IGB
(iv) ∆NHE ≅ ∆GBI
(v) ∆NEH ≅ ∆GIB
(vi) ∆HEN ≅ ∆BEN

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Question 35 Marks

Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent.
(a) AB = DE
BC = EF
CA = DF
(b) AB = EF
∠A = ∠E
AC = ED
(c) AB = DF
∠B = ∠D = 90°
AC = FE
(d) ∠A = ∠D
∠B = ∠E
AC = DF
(e) AB = DF
∠B = ∠F
AC = DE

Answer

(a) Here, AB = DE
BC = EF
CA = FD

All three corresponding sides are equal.
Thus, triangles are congruent by the side-side-side congruence.
Hence, ∆ABC ≅ ∆DEF.
(b) Given AB = EF
∠A = ∠E
AC = ED

Two corresponding sides and the included angle are equal.
Thus, triangles satisfy the SAS condition.
Hence, ∆ABC ≅ ∆EFD
(c) Here, AB = FD
∠B = ∠D = 90°
AC = FE

The triangles have equal right angles, equal hypotenuses, and one equal corresponding side.
Thus, triangles satisfy the RHS conditions.
Hence, ∆ABC ≅ ∆FDE.
(d) Here, ∠A = ∠D
∠B = ∠E
AC = DF

Clearly, two corresponding angles and one corresponding side are equal.
Thus, triangles satisfy the AAS conditions.
Hence, ∆ABC ≅ ∆DEF.
(e) Here, AB = DF
∠B = ∠F
AC = DE

Here, two corresponding sides and a non-included angle are equal.
Thus, the triangles satisfy the SSA condition, which is not a valid congruence rule.
Hence, ∆ABC need not be congruent to ∆DFE.

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