Question

easy method for the chapter SIMILAR TRIANGLES​

Answer 1

Answer:

Before knowing areas of a similar triangle, let us know first the similarity conditions. The two triangles are similar to each other if,

Corresponding angles of the triangles are equal

Corresponding sides of the triangles are in proportion

If there are two triangles say ΔABC and ΔPQR, then they are similar if,

i) ∠A=∠P, ∠B=∠Q and ∠C=∠R

ii) ABPQ = BCQR = ACPR

If we have two similar triangles, then not only their angles and sides share a relationship but also the ratio of their perimeter, altitudes, angle bisectors, areas and other aspects are in ratio.

In the upcoming discussion, the relation between the areas of two similar triangles is discussed.

Theorems on the Area of Similar Triangles

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

To prove this theorem, consider two similar triangles ΔABC and ΔPQR;

According to the stated theorem,

area of ΔABCarea of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2

As, Area of triangle = 12 × Base × Height

To find the area of ΔABC and ΔPQR, draw the altitudes AD and PE from the vertex A and P of ΔABC andΔPQR, respectively, as shown in the figure given below:

Theorems on area of similar triangles

Now, area of ΔABC = 12 × BC × AD

area of ΔPQR = 12 × QR × PE

The ratio of the areas of both the triangles can now be given as:

area of ΔABCarea of ΔPQR = 12×BC×AD12×QR×PE

⇒ area of ΔABCarea of ΔPQR = BC × ADQR × PE ……………. (1)

Now in ∆ABD and ∆PQE, it can be seen that:

∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)

∠ADB = ∠PEQ (Since both the angles are 90°)

From AA criterion of similarity ∆ADB ~ ∆PEQ

⇒ ADPE = ABPQ …………….(2)

Since it is known that ΔABC~ ΔPQR,

ABPQ = BCQR = ACPR …………….(3)

Substituting this value in equation (1), we get

area of ΔABCarea of ΔPQR = ABPQ × ADPE

Using equation (2), we can write

area of ΔABCarea of ΔPQR = ABPQ × ABPQ

⇒area of ΔABCarea of ΔPQR =(ABPQ)2

Also from equation (3),

area of ΔABCarea of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2

This proves that the ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

To have a better insight consider the following example.

Example

Example 1: In ΔABC andΔAPQ, the length of the sides are given as AP = 5 cm , PB = 10cm and BC = 20 cm. Find the ratio of the areas of ΔABC and ΔAPQ.

Area Of Similar Triangles example

Solution: In ΔABC and ΔAPQ , ∠PAQ is common and ∠APQ = ∠ABC (corresponding angles)

⇒ ΔABC ~ ΔAPQ (AA criterion for similar triangles)

Since both the triangles are similar, using the theorem for areas of similar triangles we have,

area of ΔABCarea of ΔAPQ = (ABAP)2 = (155)2 = 9

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Step-by-step explanation: