ML Aggarwal Mid Point Theorem Chapter Test Class 9 ICSE Maths Solutions Ch-11. Step by Step Answer of Ch-Test questions on Mid Point Theorem of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Mid Point Theorem Chapter Test Class 9 ICSE Maths Solutions Ch-11

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-11 | Mid Point Theorem |

Topics | Solution of Ch-Test Questions |

Academic Session | 2024-2025 |

### Ch-Test Questions on Mid Point Theorem

ML Aggarwal Class 9 ICSE Maths Solutions Ch-11

**Question 1. ****ABCD is a rhombus with P, Q and R as midpoints of AB, BC and CD respectively. Prove that PQ ⊥ QR.**

**Answer : **ABCD is a rhombus with P, Q and R as mid-points of AB, BC and CD

**To prove: **PQ ⊥QR, Join AC and BD

**Proof: **Diagonals of rhombus intersect at right angle

∠MON = 90° **…(1)**

In Δ BCD

Q and R are mid-points of BC and CD.

RQ || DB and RQ = ½ DB **…(2)**

RQ || DB

MQ || ON

∠MQN + ∠MON = 180°

∠MQN + 90° = 180°

⇒ ∠MQN = 180° – 90° = 90°

So, NQ ⊥MQ or PQ ⊥QR

Hence proved.

**Question 2. ****The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.**

**Answer : **ABCD is a quadrilateral in which diagonals AC and BD are perpendicular to each other

P, Q, R and S are mid-points of AB, BC, CD and DA

**To prove: **PQRS is a rectangle

**Proof: **P and Q are the mid-points of AB and BC

PQ || AC and PQ = ½ AC **…(1)**

S and R are mid-points of AD and DC

SR || AC and SR = ½ AC **…(2)**

Using both equations

PQ || SR and PQ = SR

So PQRS is a parallelogram

AC and BD intersect at right angles

SP || BD and BD ⊥AC

So SP ⊥ AC i.e. SP ⊥SR

∠RSP = 90°

⇒ ∠RSP = ∠SRQ = ∠RQS = ∠SPQ = 90°

Hence,

PQRS is a rectangle.

**Question 3. ****If D, E, F are mid-points of the sides BC, CA and AB respectively of a ∆ ABC, Prove that AD and FE bisect each other.**

**Answer : **D, E, F are mid-points of sides BC, CA and AB of a Δ ABC

**To prove: **AD and FE bisect each other

Construction: Join ED and FD

**Proof: **D and E are the midpoints of BC and AB

DE || AC and DE || AF **…(1)**

D and F are the midpoints of BC and AC

DF || AB and DF || AE **…(2)**

Using both equations

ADEF is a parallelogram

the diagonals of a parallelogram bisect each other

AD and EF bisect each other.

Hence proved.

**Question 4. ****In ∆ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.**

**Answer : **In Δ ABC, D and E are the mid points of sides AB and AC

DE is joined from E

EF || AB is drawn AB = 8 cm and BC = 9 cm

**To prove: **(i) BDEI is a parallelogram

(ii) Find the perimeter of BDEF

**Proof: **In Δ ABC

B and E are the mid-points of AB and AC

Here DE || BC and DE = ½ BC

So EF || AB

DEFB is a parallelogram

DE = BF

DE = ½ BC = ½ × 9 = 4.5 cm

EF = ½ AB = ½ × 8 = 4 cm

Perimeter of BDEF = 2 (DE + EF)

= 2 (4.5 + 4)

= 2 ×8.5

= 17 cm

**Question 5. ****In the given figure, ABCD is a parallelogram and E is mid-point of AD. DL EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.**

**Answer : **ABCD is a parallelogram

E is the mid-point of AD

DL || EB meets AB produced at F

**To prove: **EB = LF

B is the mid-point of AF

**Proof: **BC || AD and BE || LD

BEDL is a parallelogram

BE = LD and BL = AE

Here E is the mid-point of AD

L is the mid-point of BC

In Δ FAD

E is the mid-point of AD and BE || LD at FLD

So,

B is the mid-point of AF

EB = ½ FD = LF

**Question 6. ****In the given figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively. Show that CR = AC.**

**Answer : **ABCD is a parallelogram

P and Q are mid-points of CD and BC

**To prove: **CR = ¼ AC

**Construction: ** Join AC and BD

Proof: In parallelogram ABCD

Diagonals AC and BD bisect each other at O

AO = OC or OC = ½ AC **…(1)**

In Δ BCD

P and Q are mid points of CD and BC

PQ || BD

In Δ BCO,

Q is the mid-point of BC and PQ || OB

Is the mid-point of CO

CR = ½ OC = ½ (½ BC)

⇒ CR = ¼ BC

Hence proved.

— : End of ML Aggarwal Mid Point Theorem Chapter Test Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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