Triangles

### Prove that:

Solution: To Prove: $\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\frac{x}{2}$ Formula Used: 1) $\sin A=2 \times \sin \frac{A}{2} \times \cos \frac{A}{2}$ 2) $1+\cos A=2 \cos ^{2} \frac{A}{2}$...

### Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between their feet is 12m, find the distance between their tops. (a) 12m (b) 13m (c) 14m (d) 15m

Correct Answer: (b)13 m Explanation:         Let the poles be and CD AB = 6 m CD = 11 m Let AC be 12 m Draw a perpendicular from CD, meeting CD at E. BE = 12 m Applying...

### The areas of two similar triangles are 25 and 36 respectively. If the altitude of the first triangle is 3.5cm, then the corresponding altitude of the other triangle. (a) 5.6cm (b) 6.3cm (c) 4.2cm (d) 7cm

Correct Answer: (c) 4.2cm Explanation: The ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes. Let ɦ be the altitude of the other triangle....

### If ∆ABC~∆DEF such that 2AB = DE and BC = 6cm, find EF.

Answer: ∆ABC ~ ∆ DEF $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}}\\ \frac{1}{2} = \frac{6}{{EF}}\\ EF = 12cm \end{array}$

### In the given figure, DE║BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer:       DE || BC $\begin{array}{l} \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\\ \frac{x}{3x+4} = \frac{x+3}{{3x+19}}\\ \end{array}$ ???? (3???? + 19) = (???? + 3)(3???? + 4) 3????2 +...

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

Answer:       Let the ladder be AB and BC be the height of the window from the ground. Given, AB 10 m BC = 8 m Applying theorem in right-angled triangle ACB, ????????2 = ????????2 +...

### Find the length of the altitude of an equilateral triangle of side 2a cm.

Answer:     Let the triangle be ABC with AD as its altitude. D is the midpoint of BC. In right-angled triangle ABD, ????????2 = ????????2 + ????????2 ????????2 = ????????2 − ????????2...

### The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is (a) congruent to the original triangle (b) similar to the original triangle (c) an isosceles triangle (d) an equilateral triangle

Correct Answer: (b) similar to the original triangle Explanation:             The line segments joining the midpoint of the sides of a triangle form four triangles,...

### Two isosceles triangles have their corresponding angles equal and their areas are in the ratio 25: 36. The ratio of their corresponding heights is (a) 25 : 36 (b) 36 : 25 (c) 5 : 6 (d) 6: 5

Correct Answer: (c) 5:6 Explanation: Let x and y be the corresponding heights of the two triangles. The corresponding angles of the triangles are equal. The triangles are similar. (By AA criterion)...

### In an equilateral ∆ABC, D is the midpoint of AB and E is the midpoint of AC. Then, ar(∆ABC) : ar(∆ADE) = ? (a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 4

Correct Answer: (b) 4:1 Explanation:           In ∆ABC, D is the midpoint of AB and E is the midpoint of AC. By midpoint theorem and Basic Proportionality Theorem,...

### It is given that ∆ABC~∆PQR and , then

Correct Answer: (d)\frac{9}{4}$Explanation: Given, ∆ABC ~ ∆PQR$\begin{array}{l} \frac{{BC}}{{QR}} = \frac{2}{3}\\ \end{array}\begin{array}{l} \frac{{ar(\Delta PQR)}}{{ar(\Delta ABC)}} =...

### If in ∆ABC and ∆PQR, we have: , then (a) ∆PQR ~ ∆CAB (b) ∆PQR ~ ∆ABC (c) ∆CBA ~ ∆PQR (d) ∆BCA ~ ∆PQR

Correct Answer: (a) ∆PQR ~ ∆CAB Explanation: In ∆ABC and ∆PQR, $\begin{array}{l} \frac{{AB}}{{QR}} = \frac{{BC}}{{PR}} = \frac{{CA}}{{PQ}}\\ \end{array}$ ∆ABC ~ ∆QRP

### In ∆ABC and ∆DEF, it is given that ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are (a) congruent but not similar (b) similar but not congruent (c) neither congruent nor similar (d) similar as well as congruent

Correct Answer: (b) similar but not congruent Explanation: In ∆ABC and ∆DEF, ∠???? = ∠???? ∠???? = ∠???? Applying AA similarity theorem, ∆ABC - ∆DEF. AB = 3DE AB ≠ DE ∆ABC and ∆DEF are similar but...

### If ∆ABC~∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true? (a) BC.EF = AC.FD (b) AB.EF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD

Correct Answer: (c) BC. DE = AB. EF Explanation: ∆ABC ~ ∆EDF $\begin{array}{l} \frac{{AB}}{{DE}} = \frac{{AC}}{{EF}} = \frac{{BC}}{{DF}}\\ BC.DE \ne AB.EF \end{array}$

### In ∆ABC, DE ║ BC such that . If AC = 5.6cm, then AE = ? (a) 4.2cm (b) 3.1cm (c) 2.8cm (d) 2.1cm

Correct Answer: (d) 2.1 cm Explanation:       Given, DE || BC. Applying Thales’ theorem, $\begin{array}{l} \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}\\ \end{array}$ AE be x cm. EC = (5.6 –...

### In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is (a) right-angled (b) isosceles (c) scalene (d) obtuse-angled

Correct Answer: (b) Isosceles Explanation: In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

### A man goes 12m due south and then 35m due west. How far is he from the starting point?

Answer: In right-angled triangle SOW, Using Pythagoras theorem, ????????2 = ????????2 + ????????2 => 352 + 122 => 1225 + 144 => 1369 ???????? = 37 ???? Hence,...

### Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the base is 14 cm.

Answer: The altitude drawn from the vertex opposite to the non-equal side bisects the non-equal side. ABC is an isosceles triangle having equal sides AB and BC. The altitude drawn from the vertex...

### Two triangles DEF an GHK are such that and . If ∆DEF ~ ∆GHK then find the measures of ∠F.

Answer: If two triangles are similar then the corresponding angles of the two triangles are equal. ∆DEF ~ ∆GHK ∴ ∠???? = ∠???? = 570 In ∆ DEF Using the a????????????????...

### Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.

Answer: ABCD is a rhombus. The diagonals of a rhombus perpendicularly bisect each other. ∠???????????? = 900 ???????? = 12 ???????? ???????? = 5 ???????? In right triangle AOB,...

### In an equilateral triangle with side a, prove that area =

Answer: We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides. ABC is an...

### Find the length of the altitude of an equilateral triangle of side 2a cm.

Answer: The altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides. ABC is an equilateral...

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

Answer: Let AB be A ladder and B is the window at 8 m above the ground C. In right triangle ABC By using Pythagoras theorem, ????????2 = ????????2 + ????????2 102 = 82 +...

### In the given figure, DE║BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer: In ∆ADE and ∆ABC, ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ???????????????????????? ???????? ???????? ∥ ????????)...

### In the given figure, D is the midpoint of side BC and AE⊥BC. If BC = a, AC = b, AB = c, AD = p and AE = h, prove that (i) (ii)

Answers: (i) In right-angled triangle AEC, Applying Pythagoras theorem, ????????2 = ????????2 + ????C2 $b^{2}=h^{2}+\left(x+\frac{a}{2}\right)^{2}$ …(1) In right – angled...

### In the given figure, CD ⊥ AB Prove that

Answer: Given, ∠???????????? = 90o ???????? ⊥ ????B In ∆ ACB and ∆ CDB ∠???????????? = ∠???????????? = 90o ∠???????????? = ∠???????????? By AA similarity-criterion, ∆ ACB ~ ∆CDB...

### In ∆ABC, D is the midpoint of BC and AE⊥BC. If AC>AB, show that

Answer: In right-angled triangle AED, applying Pythagoras theorem, ????????2 = ????????2 + ????????2 … (????) In right-angled triangle AED, applying Pythagoras theorem,...

### Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.

Answer: Let ABCD be the rhombus with diagonals meeting at O. AC = 24 cm BD = 10 cm We know that the diagonals of a rhombus bisect each other at angles. Applying Pythagoras...

### Find the length of a diagonal of a rectangle whose adjacent sides are 30cm and 16cm.

Answer: Let ABCD be the rectangle with diagonals AC and BD meeting at O. AB = CD = 30 cm BC = AD = 16 cm Applying Pythagoras theorem in right-angled triangle ABC, we get:...

### Find the height of an equilateral triangle of side 12cm.

Answer: Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. D will be the midpoint of BC. Applying Pythagoras theorem in right-angled triangle...

### ∆ABC is an equilateral triangle of side 2a units. Find each of its altitudes.

Answer: Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, D, E and F are the midpoint of BC, AC and AB, In right-angled ∆ABD, AB = 2a BD = a...

### Find the length of altitude AD of an isosceles ∆ABC in which AB = AC = 2a units and BC = a units.

Answer: In isosceles ∆ ABC, AB = AC = 2a units BC = a unit Let AD be the altitude drawn from A that meets BC at D. D is the midpoint of BC. BD = BC = ????2...

### ∆ABC is an isosceles triangle with AB = AC = 13cm. The length of altitude from A on BC is 5cm. Find BC.

Answer: Given, ∆ ABC is an isosceles triangle. AB = AC = 13 cm The altitude from A on BC meets BC at D. D is the midpoint of BC. AD = 5 cm ∆ ????????????...

### In the given figure, O is a point inside a ∆PQR such that ∠PQR such that ∠POR = 90o, OP = 6cm and OR = 8cm. If PQ = 24cm and QR = 26cm, prove that ∆PQR is right-angled.

Answer: Applying Pythagoras theorem in right-angled triangle POR, ????????2 = ????????2 + ????????2 ????????2 = 62 + 82 =>36 + 64 =>100 ???????? = √100 =>10 ???????? In...

### A guy wire attached to a vertical pole of height 18 m is 24m 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?

Answer: Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A. In right triangle ABC By using Pythagoras theorem,...

### Two vertical poles of height 9m and 14m stand on a plane ground. If the distance between their feet is 12m, find the distance between their tops.

Answer: Let the two poles be DE and AB and the distance between their bases be BE. DE = 9 m AB = 14 m BE = 12 m Draw a line parallel to BE from D, meeting AB at C. DC = 12 m AC...

### A ladder is placed in such a way that its foot is at a distance of 15m from a wall and its top reaches a window 20m above the ground. Find the length of the ladder.

Answer: Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, AB = 20 m BC = 15 m Applying Pythagoras theorem in...

### A man goes 80m due east and then 150m due north. How far is he from the starting point?

Answer: Let the man starts from point A and goes 80 m due east to B. From B, he goes 150 m due north to c. In right- angled triangle ABC, $A C^{2}=A B^{2}+B C^{2}$...

### The sides of certain triangles are given below. Determine which of them right triangles are (a – 1) cm, 2√???? cm, (a + 1) cm

Answer: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. P = (a-1) cm, q = 2 √???? ???????? ???????????? ???? = (???? + 1)...

### The sides of certain triangles are given below. Determine which of them right triangles are. (i) 1.4cm, 4.8cm, 5cm (ii) 1.6cm, 3.8cm, 4cm

Answers: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. (i) a = 1.4 cm b= 4.8 cm c= 5 cm ????2 + ????2 = (1.4) 2 + (4.8)...

### The sides of certain triangles are given below. Determine which of them right triangles are. (i) 9cm, 16cm, 18cm (ii) 7cm, 24cm, 25cm

Answers: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. (i) a = 9 cm b = 16 cm c = 18 cm ????2 + ????2 = 92 + 162...

### In ∆ABC, D and E are the midpoints of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.

Answer: Given, D and E are midpoints of AB and AC. Applying midpoint theorem, DE ‖ BC. By B.P.T., $\frac{A D}{A B}=\frac{A E}{A C}$ ∠???? = ∠???? Applying SAS similarity...

### In the given figure, DE║BC and DE: BC = 3:5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

Answer: Given, DE || BC. ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ????????????????????????) ∠???????????? = ∠????????????...

### ∆ABC is right angled at A and AD⊥BC. If BC = 13cm and AC = 5cm, find the ratio of the areas of ∆ABC and ∆ADC.

Answer: In ∆ABC and ∆ADC, ∠???????????? = ∠???????????? = 900 ∠???????????? = ∠???????????? (????????????????????????) By AA similarity, ∆ BAC~ ∆ ADC. The ratio of the areas of...

### In the given figure, DE║BC. If DE = 3cm, BC = 6cm and ar(∆ADE) = 15cm2, find the area of ∆ABC.

Answer: Given, DE || BC ∴ ∠???????????? = ∠???????????? (???????????????????????????????????????????????????? ????????????????????????) ∠???????????? = ∠????????????...

### In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1cm, PB = 3cm, AQ = 1.5cm, QC = 4.5cm, prove that area of ∆APQ is 1/16 of the area of ∆ABC.

Answer: Given, $\frac{A P}{A B}=\frac{1}{1+3}=\frac{1}{4}$ $\frac{A Q}{A C}=\frac{1.5}{1.5+4.5}=\frac{1.5}{6}=\frac{1}{4}$ $\frac{A P}{A B}=\frac{A Q}{A C}$ ∠???? = ∠???? By SAS...

### The areas of two similar triangles are 64cm2 and 100cm2 respectively. If a median of the smaller triangle is 5.6cm, find the corresponding median of the other.

Answer: Let the two triangles be ABC and PQR with medians AM and PN, The ratio of areas of two similar triangles will be equal to the ratio of squares of their corresponding...

### The areas of two similar triangles are 81cm2 and 49cm2 respectively. If the altitude of the first triangle is 6.3cm, find the corresponding altitude of the other.

Answer: Given, Triangles are similar The areas of these triangles will be equal to the ratio of squares of their corresponding sides. The ratio of areas of two similar triangles...

### The corresponding altitudes of two similar triangles are 6cm and 9cm respectively. Find the ratio of their areas.

Answer: Let the two triangles be ABC and DEF with altitudes AP and DQ, Given, ∆ ABC ~ ∆ DEF. The ratio of areas of two similar triangles is equal to the ratio of squares of...

### ∆ABC ~ ∆DEF and their areas are respectively 100cm2 and 49cm2 . If the altitude of ∆ABC is 5cm, find the corresponding altitude of ∆DEF.

Answer: Given, ∆ABC ~ ∆DEF. The ration of the areas of these triangles will be equal to the ratio of squares of their corresponding sides. The ratio of areas of two similar...

### The areas of two similar triangles are 169cm2 and 121cm2 respectively. If the longest side of the larger triangle is 26cm, find the longest side of the smaller triangle.

Answer: Given, Triangles are similar. The ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides. Let the longest side of smaller triangle be X cm....

### ∆ABC~∆PQR and ar(∆ABC) = 4, ar(∆PQR). If BC = 12cm, find QR.

Answer: Given, ???????? ( ∆ ???????????? ) = 4???????? (∆ ???????????? ) $$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{4}{1}$$ ∵ ∆ABC...

### The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find the length of QR.

Answer: Given, ∆ ABC ~ ∆ PQR The ratio of the areas of triangles will be equal to the ratio of squares of their corresponding sides. \frac{\operatorname{ar}(\triangle A B...

### ABCD is parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD.

Answer: Given, ∠???????????? = ∠???????????? ∵ DA || BC ∴ ∠???????????? = ∠???????????? ∆ DAF ~ ∆ BEF $\begin{array}{l} \frac{{AF}}{{EF}} = \frac{FD}{FB}\\ \end{array}$...

### P and Q are points on the sides AB and AC respectively of a ∆ABC. If AP = 2cm, PB = 4cm, AQ = 3cm and QC = 6cm, show that BC = 3PQ

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...

### In the given figure, ∠ABC = 90o and BD⊥AC. If BD = 8cm, AD = 4cm, find CD.

Answer: Given, ABC is a right-angled triangle BD is the altitude drawn from the right angle to the hypotenuse. In ∆ DBA and ∆ DCB, ∠???????????? = ∠????????????...

### In the given figure, ∠ABC = 90o and BD⊥AC. If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm, find BC.

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 ∠???? = ∠????...