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 […]
LMN ~ PQR, 9 ×A(PQR ) = 16 ×A(LMN). If QR = 20 then find MN.
Given 9 ×A(PQR ) = 16 ×A(LMN)A(PQR )/ A(LMN) = 16/9…………(i)LMN ~ PQRA(PQR )/ A(LMN) = QR2/MN2 …..(ii)From (i) and (ii)QR2/MN2 = 16/9Given QR = 20202/MN2 = 16/9Taking square root on both sides20/MN = 4/3MN = 20×3/4MN = 15Hence the measure of MN is 15 units.
If ABC ~ PQR, A( ABC) = 80, A( PQR) = 125, then fill in the blanks.
Given A( ABC) = 80, A( PQR) = 125( ABC) / A( PQR) = 80/125 = 16/25( ABC) / A( PQR) = AB2/PQ2 [Theorem of areas of similar triangles]AB2/PQ2 = 16/25Taking square root on both sidesAB/PQ = 4/5Hence AB/PQ = 4/5
If ABC ~ PQR and AB: PQ = 2:3, then fill in the blanks.
A(ABC)/ A(PQR) = AB2/PQ2 = 22/32 = 4/9 [Theorem of areas of similar triangles]
The ratio of corresponding sides of similar triangles is 3:5; then find the ratio of their areas
When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.Given , the ratio of corresponding sides of the triangle is 3:5.Ratio of their areas = 32/52 [Theorem of areas of similar triangles]= 9/25Hence ratio of their areas = 9:25
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 [ […]