Question 1: In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form.

(i)

Solution:

In \(\Delta\)

\(\angle\)^{o}

\(\angle\)^{o}

\(\angle\)^{o}

Therefore, \(\Delta\)

Question 2: In the given figure, \(\Delta\)^{o} and \(\angle\)^{o}.

Find:

(i) \(\angle\)

(ii) \(\angle\)

(iii) \(\angle\)

(iv) \(\angle\)

Solution:

\(\Delta\)

\(\angle\)^{o} and \(\angle\)^{o}

(i) \(\angle\)^{o} – \(\angle\)^{o} – 115^{o}) = 65^{o}

(ii) \(\angle\)^{o} – \(\angle\)

Therefore, \(\angle\)^{o} – (70^{o} + 65^{o}) = 45^{o}

(iii) Now, \(\Delta\)

\(\angle\)^{o} [Vertically opposite angles]

\(\angle\)^{o}

\(\angle\)^{o} [Alternate angles]

So, \(\angle\)^{o}

(iv) \(\angle\)^{o}

Question 3: In the given figure, \(\Delta\)

Solution:

Given: \(\Delta\)

AB = 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5 cm

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

OA = \(\frac{3.5\times 8}{5}=5.6\)

And DO = \(\frac{6.4\times 5}{8}=4\)

OA = 5.6 cm and DO = 4 cm

Question 4: In the given figure, if \(\angle\)

Solution:

Given: \(\angle\)

AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm

Proof:

In \(\Delta\)

\(\angle\)

\(\angle\)

Therefore, \(\Delta\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

DE = \(\frac{3.8\times 4.2}{5.7}=2.8\)

Hence, DE = 2.8 cm

Question 5: The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm, find AB.

Solution:

Given: \(\Delta\)

We know that the ratio of perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Therefore, \(\frac{Perimeter of \Delta ABC}{Perimeter of \Delta PQR}=\frac{AB}{PQ}\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, AB = 17 cm

Question 6: In the given figure, \(\angle\)^{o} and AD \(\perp\)

Solution:

Given: AB = 100 cm, BC = 125 cm, AC = 75 cm

Proof:

In \(\Delta\)

\(\angle\)^{o}

\(\angle\)

\(\Delta\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, AD = 60 cm

Question 7: In the given figure, \(\angle\)^{o} and BD \(\perp\)

Solution:

Given that AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm

In \(\Delta\)

\(\angle\)^{o}

\(\angle\)

Therefore, \(\Delta\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, BC = 8.1 cm

Question 8: In the given figure, \(\angle\)^{o} and BD \(\perp\)

Solution:

Given that BD = 8 cm and AD = 4 cm

In \(\Delta\)

\(\angle\)^{o}

\(\angle\)^{o} – \(\angle\)

Therefore, \(\Delta\)

Therefore, \(\frac{BC}{CD}=\frac{AD}{BD}\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, CD = 16 cm

Question 9: P and Q are points on the sides AB and AC respectively of a \(\Delta\)

Solution:

Given: P is a point on AB.

Then, AB = AP + PB = (2 + 4) cm = 6 cm

Also Q is a point on AC.

Then, AC = AQ + QC = (3 + 6) cm = 9 cm

Therefore, \(\frac{AP}{AB}=\frac{2}{6}=\frac{1}{3}\)

and \(\frac{AQ}{AC}=\frac{3}{9}=\frac{1}{3}\)

Therefore, \(\frac{AP}{AB}=\frac{AQ}{AC}\)

Thus, in \(\Delta\)

\(\angle\)

and \(\frac{AP}{AB}=\frac{AQ}{AC}\)

Therefore, \(\Delta\)

\(\Rightarrow\)

Therefore, \(\frac{PQ}{BC}=\frac{AQ}{AC}\)

\(\Rightarrow\)

Therefore, BC = 3PQ [Hence proved]

Question 10: ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF x FB = EF x FD.

Solution:

Given: ABCD is a parallelogram and E is a point on BC.

Diagonal DB intersects AE at F.

To prove: AF x FB = EF x FD

Proof: In \(\Delta\)

\(\angle\)

\(\angle\)

Therefore, \(\Delta\)

Therefore, \(\frac{AF}{EF}=\frac{FD}{FB}\)

AF . FB = EF . FD

Hence proved

Question 11: In the given figure, DB \(\perp\)

Prove that \(\frac{BE}{DE}=\frac{AC}{BC}\)

Solution:

In the given figure: DB \(\perp\)

Therefore, \(\angle\)

In \(\Delta\)

\(\angle\)^{o}

\(\angle\)

\(\Delta\)

\(\Rightarrow\)

Hence proved

Question 12: A vertical pole of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

Solution:

Let AB be the vertical stick and let AC be its shadow.

Then, AB = 7.5 m and AC = 5 m

Let DE be the vertical tower and let DF be its shadow

Then, DF = 24 m, Let DE = x meters

Now, in \(\Delta\)

\(\Delta\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, height of the vertical tower is 36 m.

Question 13: In an isosceles \(\Delta\)^{2}.

Prove that \(\Delta\)

Solution:

In \(\Delta\)

CA = CB

\(\angle\)

\(\Rightarrow\)^{o} – \(\angle\)^{o} – \(\angle\)

\(\Rightarrow\)

Now, AP x BQ = AC^{2}

\(\Rightarrow\)

\(\Rightarrow\)

Thus, \(\angle\)

Therefore, \(\Delta\)

Question 14: In the given figure, \(\angle\)

Solution:

\(\angle\)

\(\frac{AC}{BD}=\frac{CB}{CE}\)

Also, \(\angle\)

Thus, \(\frac{AC}{CB}=\frac{BD}{CE}\)

Therefore, by SAS similarity criterion \(\Delta\)

Question 15: ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Solution:

Given: ABCD is a quadrilateral in which AD = BC. P, Q, R, S are the midpoints of AB, AC, CD and BD.

To prove: PQRS is a rhombus

Proof: In \(\Delta\)

Since P and Q are mid points of AB and AC

Therefore, PQ || BC and PQ = ½ AC = ½ DA (Mid-point theorem)

Similarly,

In \(\Delta\)

Since R and Q are mid points of CD and AC

Therefore, RQ || DA and RQ = ½ DA

In \(\Delta\)

Since S and P are mid points of BD and AB

Therefore, SP || DA and SP = ½ DA

In \(\Delta\)

Since S and R are mid points of BD and CD

Therefore, SR || BC and SR = ½ BC = ½ DA

Therefore, SP || RQ and PQ || SR and PQ = RQ = SP = SR

Hence, PQRS is a rhombus.

## EXERCISE – 4C

Question 1: \(\Delta\)^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.

Solution:

Given: \(\Delta\)

Area of \(\Delta\)^{2} and area of \(\Delta\)^{2}

We know that the ratio of the area of two similar triangles is equal to the ratios of the squares of their corresponding sides.

Therefore, \(\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{BC^{2}}{EF^{2}}\)

\(\Rightarrow\)

x^{2} = \(\frac{64}{121}\times (15.4)^{2}\)

x = \(\sqrt{\frac{64}{121}\times 15.4\times 15.4}=\left ( \frac{8}{11}\times 15.4 \right )=11.2cm\)

Hence, BC = 11.2 cm

Question 2: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

Solution:

Given: \(\Delta\)

Let, the area of \(\Delta\)^{2}. Then area of \(\Delta\)^{2}.

We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Therefore, \(\frac{area of \Delta ABC}{area of \Delta PQR}=\frac{9}{16}=\frac{BC^{2}}{QR^{2}}\)

\(\Rightarrow\)

\(\Rightarrow\)^{2} = (4.5)^{2} x \(\frac{16}{9}\)

\(\Rightarrow\)

\(\Rightarrow\)

Hence, QR = 6 cm

Question 3: \(\Delta\)

Solution:

\(\Delta\)

area of \(\Delta\)

Let area of \(\Delta\)

We know that the ratios of the areas of two similar triangle is equal to the ratio of the square of their corresponding sides.

Therefore, \(\frac{area of \Delta ABC}{area of \Delta PQR}=\frac{BC^{2}}{QR^{2}}\)

\(\frac{4x}{x}=\frac{(12)^{2}}{QR^{2}}\Rightarrow 4=\frac{(12)^{2}}{y^{2}}\)

\(\Rightarrow\)^{2} = 144

\(\Rightarrow\)

Hence, QR = 6 cm

Question 4: The areas of two similar triangles are 169 cm^{2} and 121 cm^{2} respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Solution:

Given: \(\Delta\)^{2} and ar(\(\Delta\)^{2}

We know that the ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.

Therefore, \(\frac{area of \Delta ABC}{area of \Delta DEF}=\frac{BC^{2}}{EF^{2}}\)

\(\Rightarrow\)

EF^{2} = (26)^{2} x \(\frac{121}{169}\)

EF = \(\sqrt{(26)^{2}\times \frac{121}{169}}=26\times \frac{11}{13}=22cm\)

Hence, the longest side of smallest triangle side is 22 cm.

Question 5: \(\Delta\)^{2} and 49 cm^{2}. If the altitude of \(\Delta\)

Solution:

Given: \(\Delta\)

ar(\(\Delta\)^{2} and ar(\(\Delta\)^{2}

Let AL and DM be the corresponding altitude of ABC and DEF respectively such that AL = 5 cm and let DM = x cm

We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes.

Therefore, \(\frac{area of \Delta ABC}{area of \Delta DEF}=\frac{AL^{2}}{DM^{2}}\)

\(\Rightarrow\)

\(\Rightarrow\)^{2} = (5)^{2} x \(\frac{49}{100}\)

\(\Rightarrow\)

Hence, DM = 3.5 cm

Therefore, the required altitude is 3.5 cm

Question 6: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratios of their areas.

Solution:

Given: \(\Delta\)

Let AL and DM be the corresponding altitudes of \(\Delta\)

We know that the ratio of squares of altitudes of two similar triangles is equal to the ratio of the corresponding areas.

\(\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{AL^{2}}{DM^{2}}\)

\(\frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{6^{2}}{9^{2}}=\frac{36}{81}=\frac{4}{9}=4:9\)

Hence, ratio of their areas = 4 : 9