# Similar Triangles Questions

## Tick the correct answer and justify : In $$\triangle$$, AB = $$6\sqrt{3}$$ cm, AC = 12 cm and BC = 6 cm. The angle is : (i) 120 (ii) 60 (iii) 90 (iv) 45

Solution : In triangle ABC, we have : AB = $$6\sqrt{3}$$ cm , AC = 12 cm and BC = 6 cm Now, $${AB}^2$$ + $${BC}^2$$ = $$(6\sqrt{3})^2$$ + $$(6)^2$$ = $$36 \times 3$$ + 36 = 108 + 36 = 144 = $$(AC)^2$$ Thus, triangle ABC is a right angled triangle at B. $$\therefore$$  …

## In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one its altitudes.

Solution : Let ABC be and equilateral triangle and let AD $$\perp$$ BC. In $$\triangle$$ ADB and ADC, we have : AB = AC          (given) AD = AD           (common side of triangle) and $$\angle$$ ADB = $$\angle$$ ADB        (each 90) By RHS criteria of …

## In an equilateral triangle ABC, D is a point on the side BC such that BD = $$1\over 3$$ BC. Prove that $$9{AD}^2$$ = $$7{AB}^2$$.

Solution : Let ABC be an equilateral triangle and let D be a point on BC such that BD = $$1\over 3$$ BC. Draw AE $$\perp$$ BC. Join AD. In $$\triangle$$ AEB and AEC, we have : AB = AC            (ABC is equilateral) $$\angle$$ AEB = $$\angle$$ AEC and AE …

## The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that $$2{AB}^2$$ = $$2{AC}^2$$ + $${BC}^2$$

Solution : We have : DB = 3CD Now, BC = DB + CD i.e. BC = 3CD + CD          [because BD = 3CD] BC = 4CD $$\therefore$$   CD = $$1\over 4$$ BC  and  DB = 3CD = $$3\over 4$$ BC             ……….(1) Since triangle ABD is …

## D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $${AE}^2$$ + $${BC}^2$$ = $${AB}^2$$ + $${DE}^2$$.

Solution : From triangle ACE, $${AE}^2$$ = $${EC}^2$$ + $${AC}^2$$       ……….(1)          (By Pythagoras Theorem) From triangle DCB, $${BD}^2$$ = $${BC}^2$$ + $${DC}^2$$        ………(2) Adding (1) and (2), we get $${AE}^2$$ + $${BD}^2$$ = $${EC}^2$$ + $${AC}^2$$ + $${BC}^2$$ + $${DC}^2$$ By Pythagoras Theorem in right triangle …

## Two poles of height 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m. find the distance between their tops.

Solution : Let AB and CD be two poles. AB = 11 m, CD = 6 m. Draw a line CE || to BD. In $$\triangle$$ AEC, $${AC}^2$$ = $${CE}^2$$ + $${AE}^2$$ = $$(12)^2$$ + $$(11 – 6)^2$$ = 144 + 25 = 169 $$\implies$$ AC = 13 m Hence, distance between their tops is …

## An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after $$3\over 2$$ hours.

Solution : Let the first plane starts from O and goes upto A towards north. Where OA = ($$1000 \times {3\over 2}$$) km = 1500 km Let the second plane starts from O at the same time and goes upto B towards west, where OB = ($$1200 \times {3\over 2}$$) km = 1800 km According …

## A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

Solution : Let AB = 24 m be guy wire attached to a vertical pole BC of height 18 m. To keep the wire taut, let it be fixed to stake at A. Then, ABC is a right angled triangle at C. $$\therefore$$  $${AB}^2$$ = $${AC}^2$$ + $${BC}^2$$ So, $${24}^2$$ = $${AC}^2$$ + $${18}^2$$ $$\implies$$ …

## A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution : Let ladder be AB, B be the window and CB be the wall. Then, ABC is a right triangle, angled at C. $$\therefore$$  $${AB}^2$$ = $${AC}^2$$ + $${BC}^2$$ So,  $${10}^2$$ = $${AC}^2$$ + $$8^2$$ or  $${AC}^2$$ = 100 – 64 $$\implies$$ $${AC}^2$$ = 36 $$\implies$$  AC = 6 m

## In the figure, O is a point in the interior of a triangle ABC, OD $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ AB.

Question : In the figure, O is a point in the interior of a triangle ABC, OD $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ AB. Show that (i)  $${OA}^2$$ + $${OB}^2$$ + $${OC}^2$$ – $${OD}^2$$ – $${OE}^2$$ – $${OF}^2$$ = $${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$ (ii)  $${AF}^2$$ + $${BD}^2$$ + $${CE}^2$$ = $${AE}^2$$ + …