AREA AND VOLUME OF SIMILAR SHAPES EXAMPLES

If the corresponding sides of similar figures are in the ratio k, then :

Area of image  =  k2 x area of object

For example, for the similar triangles alongside :

If the corresponding sides of similar solids are in the ratio k, then :

Volume of image  =  k3 x volume of object.

For each pair of similar figures, find the unknown area.

Example 1 :

Solution :

Let A1 and A2 be the area of given figures.

A1  =  5.1 m2 and A =  ?

Let L1 and L2 be the length of sides of shape 1 and shape 2.

L1  =  3 and L2  =  5

k  =  L1/L ==>  3/5

k =  (3/5)2

Using the formula :

A1/A2  =  (l1/l2)2

A1/A2  =  (3/5)2

5.1/A2  =  (9/25)

A2  =  (25/9) ⋅ (5.1)

A2  =  14.16 m2

Example 2 :

Solution :

Let A1 and A2 be the area of given figures.

A1  =  20 mand A =  ?

Let L1 and L2 be the length of sides of shape 1 and shape 2.

L1  =  5 m and L2  =  7.2 m

k  =  L1/L ==>  5/7.2

k =  (5/7.2)2

Using the formula :

A1/A2  =  (5/7.2)2

20/A2  =  0.48

A2  =  20/(0.48)

A2  =  41.6 m2

Example 3 :

Solution :

Let A1 and A2 be the area of given figures.

A1  =  38.8 mand A =  ?

Let L1 and L2 be the length of sides of shape 1 and shape 2.

L1  =  6 m and L2  =  2.8 m

k  =  L1/L ==>  6/2.8

k =  (6/2.8)2.

Using the formula :

A1/A2  =  (6/2.8)2

38.8/A2  =  4.58

A2  =  38.8/4.58

A2  =  8.47 m2

Example 4 :

In the given figure, the area of triangle BCD is 6.4 cm2. Find the area  of

(a)  Triangle ACE

(b)  Quadrilateral ABDE

Solution :

In triangles BDC and AEC,

<BDC  =   <AEC

<BCD  =  <ACE

So, BDC is similar to AEC.

(a)  Side lengths are in the ratio 5:7

k  =  5/7

Area of triangle BCD  =  (5/7)2 area of triangle ACE

6.4  =  (25/49) Area of triangle ACE

Area of triangle ACE  =  6.4(49/25)

=  12.54 cm2

(b)  Area of quadrilateral AECD  =  Area of triangle ACE - Area of triangle BCD

=  12.54-6.4

=  6.14 cm2

Example 5 :

Find

(a)  The value of x

(b)  The area of triangle PQT is given quadrilateral QRST has the area 22 m2

Solution :

By comparing triangles PQT and PRS.

<QPT  =  <RPS   (A)

<PTQ  =  <PRS   (A)

So, triangle PTQ is similar to the triangle PRS.

PQ/PS  =  QT/RS  =  PT/PR

5/(x+6)  =  6/(5+4)

5/(x+6)  =  6/9

5/(x+6)  =  2/3

15  =  2x+12

2x  =  3

x  =  1.5 cm.

(b)  Ratio of side length PQT to PRS is 2/3.

k  =  2/3

Using the formula :

Area of Δ PQT / area of ΔPRS  =  (2/3)2

Let "A" be the area of triangle PQT.

A/(A+22)  =  4/9

9A  =  4(A+22)

9A-4A  =  88

5A  =  88

A  =  88/5

A  =  17.6 m2

So, area of triangle PQT is 17.6 m2.

Example 6 :

What will happen to the volume of 

(a) a sphere if the radius is doubled

(b)  A cylinder if the radius and the height are increased by 50% ?

Solution :

(a)  Ratio of radius of old sphere and new sphere  =  1 : 2

Volume of old sphere/Volume of new sphere  =  (1/2)3

Volume of new sphere  =  8(Volume of old sphere)

(b)  Radius of old cylinder  =  r1

Radius of new cylinder  =  1.50r1

Ratio of the radius  =  1 : 1.50

Volume of old cylinder / Volume of new cylinder  =  (1/1.50)3

Volume of new cylinder  =  (1.50)3 Volume of old cylinder

Volume of new cylinder  =  (3.375) Volume of old cylinder

So, it is multiplied by 3.375.

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