## Given : In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Given side PQ side SR.Also AR = 5AP, AS = 5AQSQ is the transversal of parallel sides PQ and SR.QSR = PQS [ Alternate interior angles]ASR = AQP….(i) [ Alternate interior angles]Consider ASR and AQPASR = AQP From (i)SAR = QAP [ vertical opposite angles]ASR ~ AQP [ AA test of similarity]AS/AQ = SR/PQ [ […]

## In ABC, AP ⊥ BC, BQ ⊥ AC B- P-C, A-Q – C then prove that, CPA ~ CQB. If AP = 7, BQ = 8, BC = 12 then find AC.

Consider CPA and CQB,CPA CQB [ From figure, angle is equal to 90˚]PCA QCB [ Common angle]CPA ~ CQB, [AA test of similarity]Hence proved.AC/BC = AP/BQ [ corresponding sides of similar triangles]Given AP = 7, BQ = 8, BC = 12AC/12 = 7/8AC = 12×7/8AC = 10.5Hence measure of AC is 10.5 units.

## As shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time ?

Here PR and AC represents the smaller and bigger poles, and QR and BC represents their shadows respectively.Given PR = 4m, QR = 6m , AC = 8m, BC = xPRQ ~ ACB [ ∵Vertical poles and their shadows form similar figures]PR/AC = QR/BC [ Corresponding sides of similar triangles]4/8 = 6/xx = 6×8/4x = […]

## Are the triangles in figure 1.56 similar? If yes, by which test ?

Consider PQR and LMN ,PQ/LM = 6/3 = 2/1 …………(i)QR/MN = 8/4 = 2/1 ……….(ii)PR/LN = 10/5 = 2/1 ……….(iii)PQ/LM = QR/MN = PR/LNPQR ~ LMN [SSS test of similarity]

## In figure1.55, ABC = 75°, EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

Given ABC = 75˚ , EDC = 75˚Consider ABC and EDCABC = EDC [Given ABC = 75˚ , EDC = 75˚]ACB = DCE [Common angle]ABC ~ EDC [AA test of similarity]One to one correspondence is ABC↔ EDC