RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Mark the correct alternative in each of the following:
Question 1.
In ∆ABC ≅ ∆LKM, then side of ∆LKM equal to side AC of ∆ABC is
(a) LX
(b) KM
(c) LM
(d) None of these
Solution:
Side AC of ∆ABC = LM of ∆LKM (c)

Question 2.
In ∆ABC ≅ ∆ACB, then ∆ABC is isosceles with
(a) AB=AC
(b) AB = BC
(c) AC = BC
(d) None of these
Solution:
∵ ∆ABC ≅ ∆ACB
∴ AB = AC (a)

Question 3.
In ∆ABC ≅ ∆PQR, then ∆ABC is congruent to ∆RPQ, then which of the following is not true:
(a) BC = PQ
(b) AC = PR
(c) AB = PQ
(d) QR = BC
Solution:
∵ ∆ABC = ∆PQR
∴ AB = PQ, BC = QR and AC = PR
∴ BC = PQ is not true (a)

Question 4.
In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, ∠B = ∠P and BC = PR State which of the congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In two triangles ∆ABC and ∆PQR,
AB = QP, ∠B = ∠P and BC = PR
The condition apply : SAS (a)

Question 5.
In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:
(a) SAS
(b) ASA
(c) SSS
(d) RHS
Solution:
In ∆ABC and ∆PQR,
∠A = ∠R
∠B = ∠P
AB = RP
∴ ∆ABC ≅ ∆PQR (ASA axiom) (b)

Question 6.
If ∆PQR ≅ ∆EFD, then ED =
(a) PQ
(b) QR
(c) PR
(d) None of these
Solution:
∵ ∆PQR = ∆EFD
∴ ED = PR (c)

Question 7.
If ∆PQR ≅ ∆EFD, then ∠E =
(a) ∠P
(b) ∠Q
(c) ∠R
(d) None of these
Solution:
∵ ∆PQR ≅ ∆EFD
∴ ∠E = ∠P (a)

Question 8.
In a ∆ABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C
But Ext. ∠ACD = ∠A + ∠B

∠ACB + ∠ACD = 180° (Linear pair)
∴ ∠ACB + 100° = 180°
⇒ ∠ACB = 180°-100° = 80°
∴ ∠B = ∠ACD = 80°
But ∠A + ∠B 4- ∠C = 180°
∴ ∠A + 80° + 80° = 180°
⇒∠A+ 160°= 180°
∴ ∠A= 180°- 160° = 20° (a)

Question 9.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is
(a) 100°
(b) 120°
(c) 110°
(d) 130°
Solution:
In ∆ABC,
∠A = 2(∠B + ∠C)
= 2∠B + 2∠C

Adding 2∠A to both sides,
∠A + 2∠A = 2∠A + 2∠B + 2∠C
⇒ 3∠A = 2(∠A + ∠B + ∠C)
⇒ 3∠A = 2 x 180° (∵∠A + ∠B + ∠C = 180° )
⇒ 3∠A = 360°
⇒∠A = \(\frac { { 360 }^{ \circ } }{ 3 }\)  = 120°
∴ ∠A = 120° (b)

Question 10.
Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) SSA
(c) ASA
(d) SSS
Solution:
SSA is not the criterion of congruence of triangles. (b)

Question 11.
In the figure, the measure of ∠B’A’C’ is
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Solution:
In the figure,
∆ABC ≅ ∆A’B’C’
∴ ∠A = ∠A
⇒3x = 2x- + 20
⇒ 3x – 2x = 20
⇒ x = 20
∠B’A’C’ = 2x + 20 = 2 x 20 + 20
= 40 + 20 = 60° (b)

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°
Solution:
∵ ∆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

Solution:
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°

Solution:
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°

Solution:
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

Solution:
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
Solution:
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 = \(\frac { 1 }{ 2 }\) BC and EF || BC
Similarly,
DE = \(\frac { 1 }{ 2 }\) AB and DE || AB
DF = \(\frac { 1 }{ 2 }\) 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°

Solution:
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

Solution:
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°

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

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS are helpful to complete your math homework.

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