## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress

Question 1.

State, giving reasons, whether the following pairs of triangles are congruent or not:

Solution:

(i) In the given figure, using the SSS criterion triangles one congruent.

(ii) Triangles are congruent for the criterion ASA criterion.

(iii) Triangles are congruent for the criterion RHS.

(iv) In the first triangle, third angle = 180° – (70° + 50°) = 180° – 120° = 60°

Now triangles are congruent for ASA criterion.

(v) Not congruent as included angles of the given two sides are not equal.

(vi) Not congruent as the included sides are different.

Question 2.

Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not. In case of congruence, give reasons and write in symbolic form:

∆ABC | ∆PQR |

(i) AB = 4 cm, BC = 5 cm, ∠B = 70° | (i) QR = 4 cm, RP = 5 cm, ∠R = 70° |

(ii) AB = 4 cm, BC = 5 cm, ∠B = 80° | (ii) PQ = 4 cm, RP = 5 cm, ∠R = 80° |

(iii) BC = 6 cm, ∠A = 90°, ∠C = 50° | (iii) QR = 6 cm, ∠R = 50°, ZQ = 40° |

(iv) AB = 5 cm, ∠A = 90°, BC = 8 cm | (iv) PR = 5 cm, ∠P = 90°, QR = 8 cm |

Solution:

(i) In ∆ABC and ∆PQR

AB = QR = 4 cm

BC = RP = 5 cm

∠B = ∠R = 70°

∆ABC = ∆PQR (SAS criterion)

(ii) In ∆ABC and ∆PQR

AB = PQ = 4 cm

BC = RP = 5 cm not corresponding sides

∠B = ∠R = 80° not corresponding angles

Triangles are not congruent.

(iii) BC = QR = 6 cm

∠A = ∠P = 90° (Third angle)

∠C = ∠R = 50°

Triangles are congruent for ASA criterion.

(iv) AB = PR = 5 cm (Side)

∠A = ∠P = 90°

BC = QR = 8 cm (Hypotenuse)

Triangles are congruent for RHS criterion.

Question 3.

In the given figure, ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.

(i) State the three pairs of equal parts in ∆ADB and ∆ADC.

(zz) Is ∆ADB = ∆ADC? Give reasons.

(iii) Is ∠B = ∠C? Why?

(iv) Is BD = DC? Why?

Solution:

∆ABC is an isosceles triangle with AB = AC

and AD is one of the altitudes.

(i) In ∆ADB and ∆ADC

Side AD = AD (Common)

Hypotenuse, AB = AC (Given)

∠ADB = ∠ADC = 90° (∵ AB ⊥ BC)

∆ADB = ∆ADC

∠B = ∠C (c.p.c.t)

and BD = CD (c.p.c.t)

Question 4.

In the given figure, OA bisects ∠A and ∠ABO = ∠OCA. Prove that OB = OC.

Solution:

In ∆OAB and ∆OAC

∠OAB = ∠OAC (∵ OA bisects ∠A)

∠ABO = ∠ACO (Given)

OA = OA (common)

∆OAB = ∆OAC (AAS congruence rule)

OB = OC (Corresponding parts of congruent As)

Question 5.

In the given figure , prove that

(i) AB = FC

(ii) AF = BC.

Solution:

In ∆ABE and ∆DFC

∠B = ∠F (each 90°)

AE = DC (Given)

BE = DF (Given)

∆ABE = ∆DFC (RHS congruence rule)

(i) AB = FC (Corresponding parts of congruent ∆s)

(ii) As AB = FC (Proved above)

⇒ AF + FB = FB + BC

⇒ AF + FB – FB = BC

⇒ AF = BC