AREA AND VOLUME OF SIMILAR FIGURES EXAMPLE PROBLEMS

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:

Example 1 :

Triangle DEC has area 4.2 cm².

a) Find the area of triangle ABC.

b) Find the area of quadrilateral ABED.

Solution :

In the triangle given above, ΔABC and ΔDEC.

<BCA  =  <ECD  (A)

ABC  =  <EDC  (A)

ΔABC and ΔDEC are similar. Bases of triangle ABC and EDC are BC and DC respectively.

(i)  Area of Δ ABC/Area of Δ DEC  =  (BC/DC)²

Area of Δ ABC/4.2  =  (6/4)²

Area of Δ ABC  =  (36/16) x 4.2

=  9.45 cm²

(ii)  Area of quadrilateral AEBD  =  Area of triangle ABC - Area of triangle DEC

=  9.45-4.2

=  5.25 cm²

So, area of quadrilateral is AEBD is 5.25 cm².

Example 2 :

If triangle ABC has area 15 cm²:

a) find the area of ΔCDE

b) find the area of PQED.

Solution :

(a)  In triangles ABC and CDE

AB is parallel to DE

Area of ABC/Area of CDE  =  (AB/DE)²

15/Area of CDE  =  (5/8)²

15/Area of CDE  =  25/64

Area of CDE  =  15(64)/25

=  38.4 cm²

In triangles CPQ and CDE

PQ is parallel to DE

Area of ΔCPQ/area of ΔCDE  =  (CQ/CE)²

CQ  =  x and CE  =  2x

Area of ΔCPQ/38.4  =  (x/2x)²

Area of ΔCPQ  =  38.4/4

Area of ΔCPQ  =  9.6

(b)  Area of quadrilateral DPQE

  =  Area of CDE - Area of CPQ

=  38.4-9.6

=  28.8 cm²

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

Volume of image  =  k³ x volume of object.

Example 3 :

What will happen to the volume of:

a) a sphere if the radius is doubled

b) a sphere if the radius is increased by 20%

Solution :

Volume of sphere  =  (4/3)πr³

(a)  Radius of the sphere  =  r

Radius of new sphere  =  2r

Volume of new sphere  =  (4/3)π(2r)³

 =  (4/3)π(8r³)

 =  8 x (4/3)πr³

So, volume of the new sphere will be 8 times volume of old sphere.

(b)  Radius of sphere is increased by 20%.

Radius of new sphere  =  1.20r

Volume of new sphere  =  (4/3)π(1.20r)³

 =  (4/3)π(1.728r³)

 =  1.728 x (4/3)πr³

So, volume of the new sphere will be 1.728 times volume of old sphere.

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