Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 1 Similar Triangles are provided here with simple step-by-step explanations. These solutions for Similar Triangles are extremely popular among Class 9 students for Maths Similar Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 133:

#### Question 1:

In a triangle, a line is drawn parallel to one side and a small triangle is cut off.

Prove that the triangle cut off is similar to the original triangle.

#### Answer:

Given: Δ*ABC* and, Δ*AEF* which is* *cut off from Δ*ABC*

Also,* EF *is parallel to *BC*.

As *EF* is parallel to *BC*, ∠*AEF* = ∠*ABC* and ∠*AFE** *= ∠*ACB* (Corresponding angles)

Now in Δ*AEF* and Δ*ABC*:

∠*AEF* = ∠*ABC** *(Proved above)

∠*AFE *= ∠*ACB* (Proved above)

∠*EAF* = ∠*BAC* (Common)

As all the angles of the triangle are equal to the angles of the other triangle, both the triangles are similar.

Thus, the triangle cut off is similar to the original triangle.

#### Page No 133:

#### Question 2:

In the first problem, if the parallel line drawn in through the mid-point of one side, then how much of the area of the original triangle is the area of the small triangle?

#### Answer:

Given: Δ*ABC* and, Δ*AEF* which is* *cut off from Δ*ABC*

Also,* EF *is parallel to *BC*.

As *EF* is parallel to *BC*, ∠*AEF* = ∠*ABC* and ∠*AFE** *= ∠*ACB* (Corresponding angles)

Now in Δ*AEF* and Δ*ABC*:

∠*AEF* = ∠*ABC** *(Proved above)

∠*AFE** *= ∠*ACB* (Proved above)

∠*EAF* = ∠*BAC* (Common)

As all the angles of the triangle are equal to the angles of the other triangle, both the triangles are similar.

We know that if two triangles are similar then the corresponding sides of the triangles are proportional.

*E* and* F *are the midpoints of *AB* and *AC* respectively.

⇒ *AE* = *EB * and *AF *= *FC*

Let us draw a perpendicular from *A* to *EF* and *BC*, touching them at points *D* and *K* respectively.

Now in Δ*ADF* and Δ*AKC*:

∠*ADF** *= ∠*AKC* = 90° (By construction)

∠*AFD* = ∠*ACK** *(Corresponding angles)

∠ *DAF* = ∠*KAC** * (Common)

As all the angles of the triangle are equal to the angles of the other triangle, both the triangles are similar.

We know that if two triangles are similar then the corresponding sides of the triangles are proportional.

Also, *AF *= *FC* (As *F *is the midpoint of *AC*)

Area of Δ*ABC* =

Area of Δ*AEF* =

∴ Area of Δ*ABC** *= 4 × Area of Δ*AEF*

#### Page No 133:

#### Question 3:

In the picture below, the sides of the two triangles are parallel:

Prove that they are similar:

#### Answer:

Given: Δ*ABC* and Δ*DEF* with *AB* || *DE*, *AC* || *DF* and *EF* || *BC*

∠*ABC* = ∠*DEF*, ∠*BAC* = ∠*EDF** *and ∠*ACB* = ∠*DFE** *(Corresponding angles)

Now in Δ*ABC* and Δ*DEF*:

∠*ABC* = ∠*DEF* (Proved above)

∠*BAC* = ∠*EDF** *(Proved above)

∠*ACB* = ∠*DFE** *(Proved above)

#### Page No 139:

#### Question 1:

The circles shown below have the same centre O.

Prove that ΔOAB and ΔOPQ are similar.

#### Answer:

In Δ*OAB*:

*OA* = *OB* (Radii of the small circle)

⇒ ∠*OBA* = ∠*OAB** *(Angles opposite to equal sides are equal in measure)

Similarly, *OQ* = *OP* (Radii of the bigger circle)

⇒ ∠*OPQ* = ∠*OQP*

Using angle sum property in Δ*OPQ*:

∠*OQP* + ∠*QPO* + ∠*POQ* = 180°

⇒∠*POQ** *= 180° − ∠*OQP* − ∠*QPO* ... (1)

Using angle sum property in Δ*OAB*:

∠*OBA* + ∠*BAO* + ∠*AOB* = 180°

⇒∠*AOB** *= 180° − ∠*OBA* − ∠*BAO** * … (2)

As ∠*AOB* = ∠*POQ*, from (1) and (2), we have:

180° − ∠*OQP* − ∠*QPO* = 180° − ∠*OBA* − ∠*BAO** *

⇒∠*OQP* + ∠*QPO* = ∠*OBA* + ∠*BAO** *

As ∠*OBA* = ∠*OAB** *and ∠*OPQ* = ∠*OQP*

⇒ 2∠*OQP* = 2∠*OBA*

⇒ ∠*OQP* = ∠*OBA*

Similarly, ∠*OAB* = ∠*OPQ*

Now in Δ*OAB* and Δ*OPQ*:

∠*OAB* = ∠*OPQ** *(Proved above)

∠*OBA* = ∠*OQP* (Proved above)

∠*AOB* = ∠*POQ** * (Common)

#### Page No 139:

#### Question 2:

Prove that the four triangles got by joining the midpoints of a triangle are all congruent and that they are similar to the original triangle.

#### Answer:

Given: *D*, *E *and *F *are the midpoints of the sides *BC*, *CA* and *AB* respectively.

∴ *BD* = *DC *= , *AE* = *EC *= , *AF* = *FB* = …(1)

Since line segment *DE* passes through the midpoints of the side *BC* and *AC*, *DE* is parallel to *AB *and *DE* = .

⇒ *DE* = *AF* = *FB* … (2)

In Δ*CDE* and Δ*DBF*:

*DC* = *BD * (From equation (1))

∠*CDE* = ∠*DBF** * (Corresponding angles)

*BF* = *DE * (From equation (2))

As the side, angle, and side of one triangle is equal to the corresponding side, angle, and side of the other triangle, Δ*CDE* ≅ Δ*DBF*.

Similarly, Δ*FBD*, Δ*DEF* and Δ*AFE** *are all congruent to each other.

Now, in Δ*DCE** *and Δ*BCA*:

∠*CED* = ∠*CAB** * (Corresponding angles)

∠*CDE* = ∠*CBA* (Corresponding angles)

∠*ECD* = ∠*ACB** * (Common angles)

Similarly, Δ*BDF*, Δ*FEA* and Δ*EFD* are all similar to Δ*BCA*.

Thus, the four triangles got by joining the midpoints of the triangle are all congruent and they are similar to the original triangle.

#### Page No 139:

#### Question 3:

Prove that by joining the midpoints of any quadrilateral, we get a parallelogram.

#### Answer:

Given: A quadrilateral *ABCD *with *P*, *Q*, *R *and *S* as the respective midpoints of sides *DA*, *AB*, *BC* and *CD*

In Δ*ABD*, *P *and *Q* are the midpoints of sides *AD* and *AB* respectively.

⇒ *PQ *|| *DB* and *PQ* = ...(1)

Similarly, in Δ*CBD*, *R *and *S* are the midpoints of sides *CB* and *CD* respectively.

⇒ *RS *|| *BD* and *RS* = …(2)

From equation (1) and (2), we have:

*PQ* || *SR* and *PQ* = *SR*

Similarly, *PS * || *QR* and *PS* = *QR*

Thus, the opposite sides of the quadrilateral *PQRS* are equal and parallel.

Hence,* PQRS* is a parallelogram.

Hence, we can say that by joining the midpoints of the sides of any quadrilateral, we get a parallelogram.

#### Page No 139:

#### Question 4:

In the last problem, for what kind of quadrilateral do we get a rectangle by joining the mid-points?

#### Answer:

By joining the midpoints of the sides of a rhombus, we get a rectangle.

#### Page No 140:

#### Question 1:

In the figure below, G is the centroid of ΔABC.

Prove that the triangles AGB, AGC, BGC have the same area.

#### Answer:

Given: *G* is the centroid of Δ*ABC*.

Construction: Extend* AG *such that it intersects *BC* at point *D*.

In Δ*ABD* and Δ*ACD*:

*BD* = *DC* (As *AD* is the median of the triangle)

Also, the height of both the triangles is equal.

∴ Area of Δ*ABD* = × Height × *BD*

= × Height × *DC*

= Area of Δ*ACD*

Area of Δ*ABD* = Area of Δ*ACD* …(1)

Now, in ΔGBD and ΔGCD:

*BD* = *DC* (As *AD* is the median of the triangle)

Also, the height of both the triangles is equal.

∴ Area of Δ*GBD* = × Height × *BD*

= × Height × *DC*

= Area of Δ*GCD*

Area of Δ*GBD** *= Area of Δ*GCD* …(2)

Subtracting equation (2) from equation (1):

Area of Δ*ABD* − Area of Δ*GBD** *= Area of Δ*ACD* − Area of Δ*GCD*

⇒ Area of Δ*AGB* = Area of Δ*AGC*

Similarly, Area of Δ*AGB* = Area of Δ*GBC*

⇒ Area of Δ*AGB* = Area of Δ*GBC** *= Area of Δ*AGC*

#### Page No 140:

#### Question 2:

Prove that in a triangle, a line dividing two sides proportionally is parallel to the third side.

#### Answer:

Given,

⇒

⇒

⇒

⇒

⇒

In Δ*ABC* and Δ*ADE*:

∠*A* = ∠*A* (Common angle)

**If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.**

**⇒ **Δ*ABC* ∼ Δ*ADE*

∴ ∠*ABC* = ∠*ADE*

By converse of corresponding angles axiom, *BC* || *DE*

Thus, in a triangle, a line dividing two sides proportionally is parallel to the third side.

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