## In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find

Given DE AB.AD/DC = BE/EC [Basic proportionality theorem]AD = 5, DC = 3, BC = 6.4 [Given]BE = x , EC = 6.4-x [Given]5/3 = x/(6.4-x)Cross multiplying we get5×(6.4-x) = 3× x32-5x = 3×32 = 8xx = 32/8 = 4Hence BE = 4 units.

## MNT ~ QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio A( MNT) /A( QRS) .

Given MNT ~ QRSTMN SQR [corresponding angles of similar triangles]Construction:Draw altitude from T to MN meeting at L.Draw altitude from S to QR meeting at P.TLM = SPQ = 90˚In MLT and QPSTMN SQRTLM SPQMLT ~ QPS [AA test of similarity]MT/QS = TL/SPMT/QS = 5/9MNT ~ QRS [Given]A( MNT) /A( QRS) = MT2/QS2 [Theorem of areas […]

## In figure 1.74, PM = 10 cm A(PQS) = 100 sq.cm A(QRS) = 110 sq.cm then find NR.

Given PM = 10 cmA(PQS) = 100 sq.cmA(QRS) = 110 sq.cmPQS and QRS have common base QSA(PQS)/ A(QRS) = PM/NR100/110= 10/NRNR = 110×10/100NR = 11Hence NR = 11 cm.

## In figure 1.73, ABC = DCB = 90° AB = 6, DC = 8 then A( ABC) /A( DCB) = ?

Given ABC = DCB = 90° AB = 6, DC = 8BC is the common base of ABC and DCBA( ABC) /A( DCB) = AB/DC= 6/8= 3/4

## Ratio of areas of two triangles with equal heights is 2:3. If the base of the smaller triangle is 6cm then what is the corresponding base of the bigger triangle ?

Given ratio of two triangles with equal height is 2:3Let b1 be base of smaller triangle and b2 be base of bigger triangle.b1 = 6 cmLet a1 and a2 be areas of the triangles.Since triangles have equal height , a1/a2 = b1/b22/3 = 6/b2b2 = 3×6/2b2 = 9Hence base of bigger triangle is 9 cm.

## In ABC, B – D – C and BD = 7, BC = 20 then find following ratios.

Given BD = 7, BC = 20Construction:Draw a perpendicular from A to BC meeting at E.BC = BD+DC20 = 7+DCDC = 13(1)A( ABD) /A( ADC) = BD/DC [Triangles having same height]A( ABD) /A( ADC) = 7/13(2) A( ABD) /A( ABC) = BD/BC [Triangles having same height]A( ABD) /A( ABC) = 7/20(3) A( ADC) /A( ABC) […]

## Select the appropriate alternative.

(1) In ABC and PQR, in a one to one correspondence AB/QR = BC/ PR = CA/ PQ then(A) PQR ~ ABC(B) PQR ~ CAB(C) CBA ~ PQR(D) BCA ~ PQR Given AB/QR = BC/ PR = CA/ PQBy SSS test of similarity , PQR ~ CAB .Correct option is (B). (2) If in DEF and […]