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
In ABC, ray BD bisects ABC and ray CE bisects ACB. If seg AB seg AC then prove that ED || BC.
Given , In ABC ray BD bisects ABC .AB/BC = AD/CD ………..(i) [Angle bisector theorem]Since ray CE bisects ACBAC/BC = AE/BE …………….(ii) [Angle bisector theorem]Given seg AB = seg AC.Substitute AB in (ii)AB/BC = AE/BE…..(iii)From (i) AD/CD = AE/BE [in (i) AB/BC = AD/CD]EDBC [ converse of basic proportionality theorem]Hence proved.
In the figure 1.44, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.
In XDE, PQ || DE……. GivenXP/PD = XQ/QE…….. (I) (Basic proportionality theorem)In XEF, QR || EF……. GivenXR/RF = XQ/QE……. .(II) (Basic proportionality theorem)XP/PD = XR/RF ………. from (I) and (II)seg PR || seg DE ………. (converse of basic proportionality theorem)
In ABC, seg BD bisects ABC. If AB = x, BC = x+5, AD = x-2, DC = x+2, then find the value of x.
Given BD bisects ABC.Also AB = x, BC = x+5AD = x-2, DC = x+2Since BD bisects ABC, AB/BC = AD/DC [Angle bisector theorem]x/(x+5) = (x-2)/(x+2)Cross multiplying, we getx(x+2) = (x+5)(x-2)x2+2x = x2+5x-2x-10×2+2x = x2+3x-10x = 10the value of x is 10 .
In D LMN, ray MT bisects LMN. If LM = 6, MN = 10, TN = 8, then find LT.
Given ray MT bisects LMN.LM = 6MN = 10TN = 8Since ray MT bisects LMN , LM/MN = LT/TN [Angle bisector theorem]6/10 = LT/8LT = 6×8/10LT = 4.8Hence measure of LT is 4.8 units.
In figure 1.41, if AB || CD || FE then find x and AE.
From figure BD = 8, DF = 4, AC = 12 and CE = xGiven ABCDFEBD/DF = AC/CE [Property of three parallel lines and their transversals]8/4 = 12/xx = 12×4/8x = 6CE = 6AE = AC+CEAE = 12+6AE = 18Hence measure of x is 6 units and AE is 18 units.
Find QP using given information in the figure.
From figure MN = 25, NP = 40, MQ = 14Given NQ bisects MNP.MN/NP = MQ/QP [Angle bisector theorem]25/40 = 14/QPQP = 40×14/25QP = 22.5Hence measure of QP is 22.5 units.