Ncert exemplar solutions mathematics class 9 Solutions for Chapter 7 triangles

Which of the following is not a criterion for congruence of triangles?

(A) SAS

(B) ASA

(C) SSA

(D) SSS

If AB = QR, BC = PR and CA = PQ, then

(A) Δ ABC ≅ ΔPQR

(B) ΔCBA ≅ ΔPRQ

(C) ΔBAC ≅ Δ RPQ

(D) Δ PQR ≅ Δ BCA

In Δ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to

(A) 40° (B) 50° (C) 80° (D) 130°

In Δ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to

(A) 80° (B) 40° (C) 50° (D) 100°

In Δ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is

(A) 4 cm (B) 5 cm (C) 2 cm (D) 2.5 cm

In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of Δ PQR should be equal to side AB of Δ ABC so that the two triangles are congruent? Give a reason for your answer.

In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of Δ PQR should be equal to side BC of Δ ABC so that the two triangles are congruent? Give reason for your answer.

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm, and 7 cm? Give reason for your answer.

It is given that Δ ABC ≅ Δ RPQ. Is it true to say that BC = QR? Why?

ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.

In Fig. D and E are points on side BC of a Δ ABC such that BD = CE and AD = AE. Show that Δ ABD ≅ Δ ACE.

CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that

Δ ADE ≅ Δ BCE.

In Fig. BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that Δ ABC ≅ Δ DEF.

Q is a point on the side SR of a Δ PSR such that PQ = PR. Prove that PS > PQ.

Find all the angles of an equilateral triangle.

The image of an object placed at point A before a plane mirror LM is seen at point B by an observer at D as shown in Fig. 7.12. Prove that the image is as far behind the mirror as the object is in front of the mirror.

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. ). To prove that ∠BAD = ∠CAD, a student proceeded as follows:

In Δ ABD and Δ ACD,

AB = AC (Given)

∠B = ∠C (because AB = AC)

And ∠ADB = ∠ADC

Therefore, Δ ABD Δ Δ ACD (AAS)

So, ∠BAD = ∠CAD (CPCT)

What is the defect in the above arguments?

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.

ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that Δ OCD is an isosceles triangle.

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.