## 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,

AC^{2} = AB^{2} + BC^{2}

⇒ (5)^{2} = (3)^{2} + BC^{2}

⇒25 = 9 + BC^{2}

⇒ BC^{2} = 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°

AD^{2} + AZ^{2} + DZ^{2} = 2^{2} + 5^{2} cm

= 4 + 25 = 29

In ∠ABX, ∠X = 90°

AB^{2} = AX^{2} + BX^{2}

AD^{2} = AZ^{2} + BX^{2}

(∵ AB = AD, AX = AZ sides of square)

29 = 2^{2} + BX^{2}

⇒ 29 = 4 + BX^{2}

⇒ BX^{2} = 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

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