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 Meritnation’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.

#### 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.

Given: ΔABC and, ΔAEF which is cut off from ΔABC

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

Now in ΔAEF and ΔABC:

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.

#### 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?

Given: ΔABC and, ΔAEF which is cut off from ΔABC

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

Now in ΔAEF and ΔABC:

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. 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 ΔABC = Area of ΔAEF =  Area of ΔABC = 4 × Area of ΔAEF

#### Question 3:

In the picture below, the sides of the two triangles are parallel: Prove that they are similar:

Given: ΔABC and ΔDEF with AB || DE, AC || DF and EF || BC ABC = DEF, BAC = EDF and ACB = DFE (Corresponding angles)

Now in ΔABC and ΔDEF:

ABC = DEF (Proved above)

BAC = EDF (Proved above)

ACB = DFE (Proved above)

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

#### Question 1:

The circles shown below have the same centre O. Prove that ΔOAB and ΔOPQ are similar.

In ΔOAB:

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 ΔOPQ:

OQP + QPO + POQ = 180°

⇒∠POQ = 180° OQP QPO ... (1)

Using angle sum property in ΔOAB

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

2OQP = 2OBA

⇒ ∠OQP = OBA

Similarly, OAB = OPQ

Now in ΔOAB and ΔOPQ:

OAB = OPQ (Proved above)

OBA = OQP (Proved above)

AOB = POQ (Common)

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

#### 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. 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 ΔCDE and ΔDBF:

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, ΔCDE ΔDBF.

Similarly, ΔFBD, ΔDEF and ΔAFE are all congruent to each other.

Now, in ΔDCE and ΔBCA:

CED = CAB (Corresponding angles)

CDE = CBA (Corresponding angles)

ECD = ACB (Common angles)

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

Similarly, ΔBDF, ΔFEA and ΔEFD are all similar to ΔBCA.

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

#### Question 3:

Prove that by joining the midpoints of any quadrilateral, we get a parallelogram. Given: A quadrilateral ABCD with P, Q, R and S as the respective midpoints of sides DA, AB, BC and CD In ΔABD, P and Q are the midpoints of sides AD and AB respectively.

PQ || DB and PQ = ...(1)

Similarly, in ΔCBD, 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.

#### Question 4:

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

By joining the midpoints of the sides of a rhombus, we get a rectangle. #### Question 1:

In the figure below, G is the centroid of ΔABC. Prove that the triangles AGB, AGC, BGC have the same area.

Given: G is the centroid of ΔABC.

Construction: Extend AG such that it intersects BC at point D In ΔABD and ΔACD:

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

Also, the height of both the triangles is equal.

Area of ΔABD = × Height × BD

= × Height × DC

= Area of ΔACD

Area of ΔABD = Area of ΔACD …(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 ΔGBD = × Height × BD

= × Height × DC

= Area of ΔGCD

Area of ΔGBD = Area of ΔGCD …(2)

Subtracting equation (2) from equation (1):

Area of ΔABD Area of ΔGBD = Area of ΔACD Area of ΔGCD

Area of ΔAGB = Area of ΔAGC

Similarly, Area of ΔAGB = Area of ΔGBC

Area of ΔAGB = Area of ΔGBC = Area of ΔAGC

#### Question 2:

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