Answer: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the...
Answer: Given, AB and CD are two chords ∠???????????? + ∠???????????? = 180o …(1) ∠???????????? + ∠???????????? = 180p …(2) Using (1) and (2), ∠???????????? = ∠???????????? ∠???? = ∠A...
Answer: Given, AB and CD are two chords In ∆ PAC and ∆ PDB ∠???????????? = ∠???????????? ∠???????????? = ∠???????????? By ???????? ????????????????????????????????????????...
Answer: In ∆ ABC, P and Q are mid points of AB and AC respectively. So, PQ || BC, and PQ = $\begin{array}{l} \frac{{1}}{{2}} \end{array}$ ???????? …(1) Similarly, in...
. Prove that ∆ ACB ~ ∆ DCE.Answer: Given, $\begin{array}{l} \frac{{AC}}{{BD}} = \frac{CB}{CE}\\ \frac{{AC}}{{CB}} = \frac{BD}{CE}\\ \frac{{AC}}{{CB}} = \frac{CD}{CE}\\ \end{array}$ ∠1 = ∠2 ∠???????????? =...
Answer: Given. ∆ABC is an isosceles triangle. CA = CB ∠???????????? = ∠???????????? 180o − ∠???????????? = 180o − ∠???????????? ∠???????????? = ∠???????????? AP × BQ = ????????2...
Answer: Let AB be the vertical stick and BC be its shadow. AB = 7.5 m, BC = 5 m Let PQ be the tower and QR be its shadow. QR = 24 m Let the length of PQ be x m. In ∆ ABC...
Answer: In ∆BED and ∆ACB, ∠???????????? = ∠???????????? = 90o ∵ ∠???? + ∠???? = 180o ∴ BD || AC ∠???????????? = ∠???????????? By AA similarity theorem, ∆ BED ~ ∆ ACB $\begin{array}{l}...
Answer: Given, ∠???????????? = ∠???????????? ∵ DA || BC ∴ ∠???????????? = ∠???????????? ∆ DAF ~ ∆ BEF $\begin{array}{l} \frac{{AF}}{{EF}} = \frac{FD}{FB}\\ \end{array}$...
Answer: Given, $\begin{array}{l} \frac{{AP}}{{AB}} = \frac{2}{6} = \frac{1}{3}\\ \frac{{AQ}}{{AC}} = \frac{3}{9} = \frac{1}{3}\\ \\ \frac{{AP}}{{AB}} = \frac{{AQ}}{{AC}} \end{array}$ By AA...
Answer: Given, ABC is a right-angled triangle BD is the altitude drawn from the right angle to the hypotenuse. In ∆ DBA and ∆ DCB, ∠???????????? = ∠????????????...
Answer: Given, ABC is a right-angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In ∆ BDC and ∆ ABC, ∠???????????? = ∠???????????? = 90o ∠???? = ∠????...
Answer: In ∆ BDA and ∆ BAC, ∠???????????? = ∠???????????? = 90o ∠???????????? = ∠???????????? By AA similarity theorem, ∆ BDA - ∆ BAC $\begin{array}{l} \frac{{AD}}{{AC}} =...
Answer: Given, ∆ ABC - ∆ DEF Their corresponding sides will be proportional. Ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides. $\begin{array}{l}...
Answer: Given, Triangles ABC and PQR are similar. $\begin{array}{l} \frac{{Perimeter( ABC)}}{{ Perimeter( PQR)}} = \frac{{AB}}{{PQ}}\\ \end{array}$ $\begin{array}{l} \frac{{32}}{{24}} =...
Answer: Given, ∠???????????? = ∠???????????? ???????????? ∠???? = ∠???? Let DE be X cm By AA similarity theorem, ∆ ADE - ∆ ABC $\begin{array}{l} \frac{{AD}}{{AB}} =...
Answers: (i) Let OA be X cm. ∵ ∆ OAB - ∆ OCD $\begin{array}{l} \frac{{OA}}{{OC}} = \frac{{AB}}{{CD}}\\ \end{array}$ $\begin{array}{l} \frac{{x}}{{3.5}} =...
Answers: (i) Given, ∆ ODC - ∆ OBA ∠???????????? = ∠???????????? = 45 o (ii) Given, ∆ ODC- ∆ OBA ∠???????????? = ∠???????????? = 70...
Answers: (i) Given, DB is a straight line. ∠???????????? + ∠???????????? = 180o ∠???????????? = 180o − 115o = 65o (ii) In ∆ DOC, ∠???????????? +...
Answer: In ∆ ABC ∠A + ∠???? + ∠???? = 1800 (Angle Sum Property) 800 + ∠???? + 700 = 180o ∠???? = 30o ∠???? = ∠???? ???????????? ∠???? = ∠???? By AA similarity, ∆ ABC - ∆ MNR...
Answers: (iii) Given, $\begin{array}{l} \frac{{CA}}{{QR}} = \frac{{8}}{{6}}\\ \end{array}$ $\begin{array}{l} \frac{{CA}}{{QR}} =...
Answers: (i) Given, ∠BAC = ∠PQR = 50o ∠ABC = ∠QPR = 60 o ∠ACB = ∠PRQ = 70 o By AAA similarity theorem, ∆ ABC – QPR (ii) Given, $\begin{array}{l}...