Triangles Test - 18

In ΔABC and ΔEDF

AB = DE (Given)

∠BAC = ∠EDF (Given)

AC = DF ( AB = AC and DE = DF)

∴ΔABC ≅ ΔEDF [SAS congruence rule]

So, ∠ACB = ∠DFE = 70° [CPCT]

⇒ ∠DEF = ∠DFE = 70 ( DE = DF)

Now, ∠BAC = 180° − (DFE + DEF)

= 180° − 140°

= 40°

The correct answer is B.

It is known that in any triangle, the side opposite to the larger angle is longer.

It can be observed that: ∠ABC > ∠ACB

∴ AC > AB

Thus, the relation given in alternative D is not correct.

The correct answer is D.

It is given that BE = FD

∴ BE + EF = FD + EF

⇒ BF = DE ... (1)

In ΔABF and ΔCDE:

AB = CD (Given)

∠ABF = ∠CDE = 90°

BF = DE [From (1)]

∴ ΔABF ≅ ΔCDE (By SAS congruency criterion)

Thus, ΔABF is congruent to ΔCDE by SAS congruency criterion.

The correct answer is A.

In ΔPQR, applying exterior angle property of triangles:

∠QRS = ∠RPQ + ∠PQR

⇒ 136° = 67° + ∠PQR

⇒ ∠PQR = 69°

Now, applying angle sum property of triangles in ΔPQR:

∠RPQ + ∠PQR + ∠PRQ = 180°

⇒ 67° + 69° + ∠PRQ = 180°

⇒ ∠PRQ = 44°

It can be observed that: ∠PQR > ∠RPQ > ∠PRQ

It is known that in any triangle, the side opposite to the larger angle is longer.

∴ PR > QR > PQ

Thus, the sides PQ, QR, PR of ΔPQR can be arranged in ascending order as PQ < QR < PR.

The correct answer is A.

It is given that AB = BC.

∴∠BAD = ∠BCD      (Angles apposite to equal sides are equal)

Consider ΔBAD and ΔBCD.

∠BDA = ∠BDC = 90°   (Given)

∠BAD = ∠BCD             (Shown above)

AB = BC                        (Given)

Therefore, by AAS congruency rule, ΔBADΔBCD

∴ AD = DC (Corresponding sides of congruent triangles are equal)

Applying Pythagoras theorem in ΔABC,
AB² + BC² = AC²

⇒ (10 cm)² + (10 cm)² = AC²

⇒ AC² = 100 cm² + 100 cm²

⇒ AC² = 200 cm²

Thus, the length of AD is

Hence, the correct answer is option B.

Since XYCD is a parallelogram, its opposite sides and angles are equal.

∴ XC = DY and CY = DX

Also, ∠ECF = ∠ADB … (1)

∴ ∠ADB = 60°

XC = DY

⇒ XC + XE = DY + YB [XE = YB]

⇒ CE = DB … (2)

Also, CY = DX

⇒ CY + YF = DX + AX [YF = AX]

⇒ CF = DA … (3)

From (1), (2), (3),

ΔADB ≅ ΔFCE [By SAS congruency rule]

⇒ ∠BAD = ∠CFE [By C.P.C.T]

Also, ∠ABD = ∠PBQ = 70° [Vertically opposite angles]

In ΔADB, by angle sum property,

∠BAD + ∠ADB + ∠ABD = 180°

⇒ ∠BAD + 60° + 70° = 180°

⇒ ∠BAD + 130° = 180°

⇒ ∠BAD = 180° − 130° = 50°

∴ ∠CFE = ∠BAD = 50°

Thus, the measure of ∠CFE is 50°.

The correct answer is B.

In ΔABD and ΔBAC,

AD = BC (opposite sides of a parallelogram are equal)

BD = AC (given that diagonals are equal)

AB = BA (common)

Thus, ΔABD is congruent to ΔBAC (by Side-Side-Side criterion)

Thus, ∠BAD = ∠ABC (corresponding angles of congruent triangles)

In a parallelogram, the sum of adjacent angles is 180º.

Thus, in parallelogram ABCD, ∠ABC + ∠BAD = 180°

∠ABC + ∠ABC = 180°

2∠ABC = 180°

∠ABC = 90°

Thus, the given parallelogram is a rectangle.

Note that all rectangles are parallelograms but not all parallelograms are rectangles.

The correct answer is D.

Given that DP = CQ

⇒ DP + PQ = CQ + PQ

⇒ DQ = CP

In ΔADQ and ΔBCP,

AQ = BP                                            (Given)

∠ADQ = ∠BCP                               (Both 90°)

DQ = CP                                            (Already proved)

∴ ΔADQ $\cong$ ΔBCP        (By Right angle-Hypotenuse-Side congruence criterion)
So, ∠DAQ = ∠CBP     (CPCT)

In ΔBCP, ∠BCP + ∠CPB + ∠CBP = 180º

90º + 60º + ∠CBP = 180º

∠CBP = 180° − 150°

∠CBP = 30°

Thus, ∠DAQ = ∠CBP = 30°           
Hence, the correct answer is option B.

In ΔABC and ΔDCB,

AC = BD                                            (Given)

∠ABC = ∠BCD = 90°                     (Given)

BC = BC                                            (Common side)

∴ ΔABC ≅ ΔDCB                              (RHS congruence rule)

So, ∠DBC = ∠ACB = 60°                (By CPCT)

Thus, ∠BDF =  ∠DCB + ∠DBC     (Exterior angle property)

= 90° + 60°

= 150°

Hence, the correct answer is option C.

Let BC = x cm

It is known that in a triangle, the sum of any two sides is always greater than the third side.

∴AB + AC ≥ BC

6 cm + 6 cm ≥ x cm

⇒ x ≤ 12 cm

∴ Perimeter of ΔABC = AB + BC + AC

= (6 + x + 6) cm

= (12 + x) cm

≤ (12 + 12) cm

= 24 cm

Thus, the perimeter of ΔABC must be less than or equal to 24 cm.

Thus, the length 26 cm cannot be the perimeter of ΔABC.

The correct answer is D.