Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(C)

Question 1

(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.

(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.

Solution 1

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

 

(i) Required ratio = 

(ii) Required ratio = 

Question 2

A line PQ is drawn parallel to the base BC of ABC which meets sides AB and AC at points P and Q respectively. If AP =PB; find the value of:

(i) 

(ii) 

Solution 2

 

(i) AP =PB 

 

(ii) 

Question 3

The perimeters of two similar triangles are 30 cm and 24 cm. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.

Solution 3

Let 

Question 4

In the given figure, AX: XB = 3: 5.

Find:

(i) the length of BC, if the length of XY is 18 cm.

(ii) the ratio between the areas of trapezium XBCY and triangle ABC.

 

Solution 4

Given, 

(i)

(ii)

Question 5

ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB.

Solution 5

From the given information, we have:

Question 6

In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4.

Calculate the value of ratio:

(i) 

(ii) 

(iii) 

Solution 6

(i)

(ii) Since LMN and MNR have common vertex at M and their bases LN and NR are along the same straight line

 

(iii) Since LQM and LQN have common vertex at L and their bases QM and QN are along the same straight line

Question 7

The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel; PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ. Calculate:

(i) the length of AP,

(ii) the ratio of the areas of triangle APQ and triangle ABC.

 

Solution 7

(i)

(ii)

Question 8

In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP: PC = 1: 2. DP produced meets AB produces at Q. Given the area of triangle CPQ = 20 cm². Calculate:

(i) area of triangle CDP,

(ii) area of parallelogram ABCD.

 

Solution 8

 

Question 9

In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm², Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also, find the area of triangle BCD.

 

Solution 9

Question 10

The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersects at point P. If AP: CP = 3: 5, find:

(i) APB : CPB(ii) DPC : APB

(iii) ADP : APB(iv) APB : ADB

 

Solution 10

(i) Since APB and CPB have common vertex at B and their bases AP and PC are along the same straight line

(ii) Since DPC and BPA are similar

(iii) Since ADP and APB have common vertex at A and their bases DP and PB are along the same straight line

(iv) Since APB and ADB have common vertex at A and their bases BP and BD are along the same straight line

Question 11

In the given figure, ABC is a triangle. DE is parallel to BC and .

(i) Determine the ratios 

(ii) Prove that DEF is similar to CBF. Hence, find .

(iii) What is the ratio of the areas of DEF and BFC?

Solution 11

 

(i) Given, DE || BC and 

In ADE and ABC,

A = A(Corresponding Angles)

ADE = ABC(Corresponding Angles)

(By AA- similarity)

..........(1)

Now

Using (1), we get.........(2)

(ii) In DEF and CBF,

FDE = FCB(Alternate Angle)

DFE = BFC(Vertically Opposite Angle)

DEF CBF(By AA- similarity)

using (2)

.

(iii) Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, therefore

Question 12

In the given figure, B = E, ACD = BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ABC and DEC.

Solution 12