Question - 12 : -
If ABC and DEF are two triangles such that ∆ABC ≅ ∆FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then, which of the following is true?
(a) DF = 5 cm, ∠F = 60°
(b) DE = 5 cm, ∠E = 60°
(c) DF = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°

Answer - 12 : -

∵ ∆ABC ≅ ∆FDE,

AB = 5 cm, ∠A = 80°, ∠B = 40°

∴ DF = 5 cm, ∠F = 80°, ∠D = 40°

∴ ∠C =180°- (80° + 40°) = 180° – 120° = 60°

∴ ∠E = ∠C = 60°

∴ DF = 5 cm, ∠E = 60° (c)

Question - 13 : -
In the figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ∆ABD is congruent to
(a) ∆EFC
(b) ∆ECF
(c) ∆CEF
(d) ∆FEC

Answer - 13 : -

In the figure, AB ⊥ BE, FE ⊥ BE

BC = DE, AB = EF,

then CD + BC = CD + DE BD = CE

In ∆ABD and ∆CEF,

BD = CE (Prove)

AB = FE (Given)

∠B = ∠E (Each 90°)

∴ ∆ABD ≅ ∆FCE (b)

Question - 14 : -
In the figure, if AE || DC and AB = AC, the value of ∠ABD is
(a) 70°
(b) 110°
(c) 120°
(d) 130°

Answer - 14 : -

In the figure, AE || DC

∴ ∠1 = 70° (Vertically opposite angles)

∴ ∠1 = ∠2 (Alternate angles)

∠2 = ∠ABC (Base angles of isosceles triangle)

∴ ABC = 90°

But ∠ABC + ∠ABD = 180° (Linear pair)

⇒ 70° +∠ABD = 180°

⇒∠ABD = 180°-70°= 110°

∴ ∠ABD =110° (b)

Question - 15 : -
In the figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
(a) 52°
(b) 76°
(c) 156°
(d) 104°

Answer - 15 : -

In ∆ABC, AB = AC

AC is produced to E

CD || BA is drawn

∠ABC = 52°

∴ ∠ACB = 52° (∵ AB = AC)

∴ ∠BAC = 180°-(52° +52°)

= 180°-104° = 76°

∵ AB || CD

∴ ∠ACD = ∠BAC (Alternate angles)

= 76°

and ∠BCE + ∠DCB = 180° (Linear pair)

∠BCE + 52° = 180°

⇒∠BCE = 180°-52°= 128°

∠x + ∠ACD = 380°

⇒ x + 76° = 180°

∴ x= 180°-76°= 104° (d)

Question - 16 : -
In the figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm

Answer - 16 : -

In the figure, AC is the bisector of ∠BAD, AB = 3 cm, AC = 5 cm

In ∆ABC and ∆ADC,

AC = AC (Common)

∠B = ∠D (Each 90°)

∠BAC = ∠DAC (∵ AC is the bisector of ∠A)

∴ ∆ABC ≅ ∆ADC (AAS axiom)

∴ BC = CD and AB = AD (c.p.c.t.)

Now in right ∆ABC,

AC2 = AB2 + BC2

⇒ (5)2 = (3)2 + BC2

⇒25 = 9 + BC2

⇒ BC2 = 25 – 9 = 16 = (4)2

∴ BC = 4 cm

But CD = BC

∴ CD = 4 cm (c)

Question - 17 : -
D, E, F are the mid-point of the sides BC, CA and AB respectively of ∆ABC. Then ∆DEF is congruent to triangle
(a) ABC
(b) AEF
(c) BFD, CDE
(d) AFE, BFD, CDE

Answer - 17 : -

In ∆ABC, D, E, F are the mid-points of the sides BC, CA, AB respectively

DE, EF and FD are joined

∵ E and F are the mid-points

AC and AB,

∴ EF =

BC and EF || BC

Similarly,

DE =

AB and DE || AB

DF =

AC and DF || AC

∴ ∆DEF is congruent to each of the triangles so formed

∴ ∆DEF is congruent to triangle AFE, BFD, CDE (d)

Question - 18 : -
ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =
(a) 55°
(b) 70°
(c) 35°
(d) 110°

Answer - 18 : -

In ∆ABC, AB = AC

AD is median to BC

∴ BD = DC

In ∆ADB, ∠D = 90°, ∠B = 35°

But ∠B + BAD + ∠D = 180° (Sum of angles of a triangle)

⇒ 35° + ∠BAD + 90° = 180°

⇒∠BAD + 125°= 180°

⇒ ∠BAD = 180°- 125°

⇒∠BAD = 55° (a)

Question - 19 : -
In the figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm

Answer - 19 : -

In the figure, ABCD and AXYZ are squares

DY = 3 cm, AZ = 2 cm

DZ = DY + YZ

= DY + Z = 3 + 2 = 5 cm

In ∆ADZ, ∠2 = 90°

AD2 + AZ2 + DZ2 = 22 + 52 cm

= 4 + 25 = 29

In ∠ABX, ∠X = 90°

AB2 = AX2 + BX2

AD2 = AZ2 + BX2

(∵ AB = AD, AX = AZ sides of square)

29 = 22 + BX2

⇒ 29 = 4 + BX2

⇒ BX2 = 29 – 4 = 25 = (5)2

∴ BX = 5 cm (a)

Question - 20 : -
In the figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is
(a) 72°
(b) 73°
(c) 74°
(d) 95°

Answer - 20 : -

In the figure, ∠B = 2∠C, AD and BE are the bisectors of ∠A and ∠B respectively,

AB = CD

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