Page 193 - ICSE Math 7

Solution: (a) In ∆ ABC and ∆ BAD
∠C = ∠D (each 90°)

AC = BD (each 3 cm)
AB = BA (common hypotenuse)
∆ ABC  ∆ BAD (by RHS congruence criterion)

(b) In ∆ ABC and ∆ ADC
∠B = ∠D (each 90°)
AB = AD (each 4.1 cm)
AC = AC (common hypotenuse)
∆ ABC  ∆ ADC (by RHS congruence criterion)

Example 11: You have to show that ∆ AMP  ∆ MAQ. In the following table, supply the missing
reasons. P

A
Steps Reasons
(i) PM = AQ (i) _______________
(ii) ∠PMA = ∠QAM (ii) _______________
(iii) AM = AM (iii) _______________
(iv) ∆ AMP ≅ ∆ MAQ (iv) _______________ M
Q

Solution: Steps Reasons Try This
(i) PM = AQ (i) given In ∆ ABC, ∠A = 50°, ∠B = 20° and
(ii) ∠PMA = ∠QAM (ii) given ∠C = 110°. In ∆ PQR, ∠P = 50°,
(iii) AM = AM (iii) common ∠Q = 20° and ∠R = 110°. A student
says that ∆ ABC  ∆ PQR by AAA
(iv) ∆ AMP ≅ ∆ MAQ (iv) SAS criterion congruence criterion. Is he justified?
Why or why not?

Example 12: In a squared sheet, draw two triangles of equal areas such that:
(a) the triangles are congruent. (b) the triangles are not congruent.
What can you say about their perimeters?

A P

Solution: (a)

B C Q R

∆ ABC  ∆ PQR; area of ∆ ABC = area of ∆ PQR

R C
(b)

A T O W

∆ RAT ∆ COW; area of ∆ RAT = area of ∆ COW

In (a), since the triangles are congruent,

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