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## AP Board Class 10 Maths Chapter 8 Similar Triangles Ex 8.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 8 Similar Triangles Ex 8.2 Book Answers AP Board Class 10 Maths Chapter 8 Similar Triangles Ex 8.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 8 Similar Triangles Ex 8.2 Book Answers

## Andhra Pradesh State Board Class 10th Maths Chapter 8 Similar Triangles Ex 8.2 Books Solutions

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## AP Board Class 10th Maths Chapter 8 Similar Triangles Ex 8.2 Textbooks Solutions with Answer PDF Download

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### 10th Class Maths 8th Lesson Similar Triangles Ex 8.2 Textbook Questions and Answers

Question 1.
In the given figure, ∠ADE = ∠B
i) Show that △ABC ~ △ADE
ii) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE. i) Given: △ABC and ∠ADE = ∠B
∠A = ∠A [∵ Common]
∴ ∠C = ∠AED [∵ By Angle Sum property of triangles] △ABC ~ △ADE by AAA similarity condition.]

ii) AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.
To find DE; △ABC ~ △ADE
Hence,
𝐴𝐵𝐴𝐷 = 𝐵𝐶𝐷𝐸 = 𝐴𝐶𝐴𝐸
[∵ Ratios of corresponding sides are equal] Question 2.
The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle. Given: △ABC ~ △PQR
Perimeter of △ABC = 30 cm.
Perimeter of △PQR = 20 cm.
AB = 12 cm.
To find: PQ⎯⎯⎯⎯⎯⎯⎯⎯
Ratio of perimeters = 30 : 20 = 3 : 2
Let the length of the side corresponding to the side with length 12 cm be x.
Then 30 : 20 : : 12 : x
30x = 20 x 12
𝑥=20×1230 = 8 cm

Question 3.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp-post is 3.6 m above the ground, find the length of her shadow after 4 seconds. Given:
A lamp-post AB⎯⎯⎯⎯⎯⎯⎯⎯ of height = 3.6 m
= 360 cm.
Speed of the girl = 1.2 m/sec.
Distance travelled in 4 sec = Speed x Time = 1.2 × 4 = 4.8 m = 480 cm.
CD⎯⎯⎯⎯⎯⎯⎯⎯, height of the girl = 90 cm.
Let the length of the shadow at a distance of 4.8 m from the lamp post = x cm.
From the figure,
△ABE ~ △DCE
[∵ ∠B = ∠C = 90°
∠E = ∠C common
(A.A. similarity)]
Hence,
𝐴𝐵𝐷𝐶 = 𝐵𝐸𝐶𝐸 = 𝐴𝐸𝐷𝐸
∴ 36090 = 480+𝑥𝑥
⇒ 4 = 480+𝑥𝑥
⇒ 4x = 480 + x
⇒ 4x – x = 480
⇒ 3x = 480
⇒ x = 160 cm = 1.6 m
∴ Length of the shadow = 1.6 m

Question 4.
CM and RN are respectively the medians of similar triangles △ABC and △PQR. Prove that
i) △AMC ~ △PNR
ii) 𝐶𝑀𝑅𝑁 = 𝐴𝐵𝑃𝑄
iii) △CMB ~ △RNQ Given : △ABC ~ △PQR
CM is a median through C of △ABC.
RN is a median through R of △PQR.
R.T.P:
i) △AMC ~ △PNR.
Proof: In △AMC and △PNR,
𝐴𝐶𝑃𝑅 = 𝐴𝑀𝑃𝑁 and ∠A = ∠P [∵ In △ABC, △PQR and M, N are the mid-points of AB and PQ]
∴ △AMC ~ △PNR
[∵ SAS similarity condition]

ii) 𝐶𝑀𝑅𝑁 = 𝐴𝐵𝑃𝑄
Proof: From (i) we have
△AMC ~ △PNR
Hence 𝐴𝐶𝑃𝑄 = 𝐴𝑀𝑃𝑁 = 𝐶𝑀𝑅𝑁
[∵ Ratio of corresponding sides of two similar triangles are equal]
Thus, 𝐶𝑀𝑅𝑁 = 𝐴𝑀×2𝑃𝑁×2
[Multiplying both numerator and the denominator by 2]
𝐶𝑀𝑅𝑁 = 𝐴𝐵𝑃𝑄 [2AM = AB; 2PN = PQ]

iii) △CMB ~ △RNQ
Proof: In △CMB and △RNQ
∠B = ∠Q [Corresponding angles of △ABC and △PQR]
Also, 𝐵𝐶𝑅𝑄 = 𝐵𝑀𝑄𝑁 Thus, △CMB ~ △RNQ by S.A.S similarity condition.

Question 5.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point ‘O’. Using the criterion of similarity for two triangles, show that 𝑂𝐴𝑂𝐶 = 𝑂𝐵𝑂𝐷. Given : □ ABCD, AB || DC
The diagonals AC and BD intersect at ‘O’.
R.T.P: 𝑂𝐴𝑂𝐶 = 𝑂𝐵𝑂𝐷
Construction: Draw EF || AB, passing through ‘O’.
Proof: In △ACD, OE || CD [∵ Construction]
Hence 𝑂𝐴𝑂𝐶 = 𝐸𝐴𝐸𝐷 …….. (1)
(∵ Line drawn parallel to one side of a triangle divides other two sides in the same ratio – Basic proportionality theorem)
Also in △ABD, EO || AB [Construction] Hence,
𝐸𝐴𝐸𝐷 = 𝑂𝐵𝑂𝐷 ……… (2)
(∵ Basic proportionality theorem) From (1) and (2), we have
𝑂𝐴𝑂𝐶 = 𝑂𝐵𝑂𝐷
∴ Hence proved.

Question 6.
AB, CD, PQ are perpendicular to BD. AB = x, CD = y and PQ = z, prove that 1𝑥+1𝑦=1𝑧. Given ∠B = ∠Q = ∠D = 90°
Thus, AB || PQ || CD.
Now in △BQP, △BDC
∠B = ∠B (Common)
∠Q = ∠D (90°)
∠P = ∠C [∵ Angle Sum property of triangles]
∴ △BQP ~ △BDC
(by A.A.A similarity condition)
Hence 𝐵𝑄𝐵𝐷 = 𝑃𝑄𝐶𝐷
[∵ Ratio of corresponding sides is equal] Also in △DQP and △DBA
∠D = ∠D (Common)
∠Q = ∠B (90°)
∴ △DQP ~ △DBA (by A.A. similarity condition)
𝑄𝐷𝐵𝐷 = 𝑃𝑄𝐴𝐵
[ Ratio of corresponding sides is equal]
Adding (1) and (2), we get Question 7.
A flag pole 4 m tall casts a 6 m., shadow. At the same time, a nearby building casts a shadow of 24 m. How tall is the building?
Given: 4 m length flag pole casts a shadow 6 m.
Let x m length/tall building casts a shadow 24 m. Let AB be the length of flag pole = 4 m.
Shadow of AB = BC = 6 m.
PQ be the building = x m (say)
QR, the shadow of the building = 24 m
From the figure,
∠A = ∠P
∠B = ∠Q
∴ △ABC ~ △PQR by A.A. similarity condition
Hence 𝐴𝐵𝑃𝑄 = 𝐵𝐶𝑄𝑅
[∵ Ratio of corresponding angles is equal]
46 = 𝑥24
x = 24×46 = 16 m
∴ Height of the building = 16 m.

Question 8.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of △ABC and △FEG respectively. If △ABC ~ △FEG then show that
i) 𝐶𝐷𝐺𝐻 = 𝐴𝐶𝐹𝐺
ii) △DCB ~ △HGE
iii) △DCA ~ △HGF Given: △ABC ~ △FEG.
CD is the bisector of ∠C and GH is the bisector of ∠G.
R.T.P.:
i) 𝐶𝐷𝐺𝐻 = 𝐴𝐶𝐹𝐺
In △ACD and △FGH
∠A = ∠F
[∵ Corresponding angles of △ABC and △FEG]
∠ACD = ∠FGH
[∵ ∠C = ∠G ⇒ 12∠C = 12∠G ⇒ ∠ACD = ∠FGH]
∴ By A.A. similarity condition, △ACD ~ △FGH
𝐴𝐶𝐹𝐺 = 𝐶𝐷𝐺𝐻 = 𝐴𝐷𝐹𝐻
[∵ Ratio of the Corresponding angles is equal]
⇒ 𝐴𝐶𝐹𝐺 = 𝐶𝐷𝐺𝐻 [Q.E.D]

ii) △DCB ~ △HGE
In △DCB and △HGE,
∠B = ∠E
[∵ Corresponding angles of △ABC and △FEG]
∠DCB = ∠HGE
[∵ ∠C = ∠G ⇒ 12∠C = 12∠G ⇒ ∠DCB = ∠HGE]
∴ △DCB ~ △HGE . (by A.A. similarity condition)

iii) △DCA ~ △HGF
In △DCA and △HGF
∠A = ∠F
12∠C = 12∠G ⇒ ∠DCA = ∠HGF
[∵ Corresponding angles of the similar triangles]
∴ △DCA ~ △HGF
[ A.A. similarity condition]

Question 9.
AX and DY are altitudes of two similar triangles △ABC and △DEF. Prove that AX : DY = AB : DE. Given: △ABC ~ △DEF.
AX ⊥ BC and DY ⊥ EF.
R.T.P.: AX : DY = AB : DE.
Proof: In △ABX and △DEY ∠B = ∠E [∵ Corresponding angles of △ABC and △DEF]
∠AXB = ∠DYE [given]
∴ △ABX ~ △DEY
(by A.A. similarity condition)
Hence 𝐴𝐵𝐷𝐸 = 𝐵𝑋𝐸𝑌 = 𝐴𝑋𝐷𝑌
[∵ Ratios of corresponding sides of similar triangles are equal]
⇒ AX : DY = AB : DE [Q.E.D.]

Question 10.
Construct a triangle shadow similar to the given △ABC, with its sides equal to 53 of the corresponding sides of the triangle ABC. Steps of construction :

1. Draw a △ABC with certain measures.
2. Draw a ray BX−→− making an acute angle with BC on the side opposite to vertex A.
3. Locate 8 points (B1, B2, …., B8) on BX−→− such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7 = B7B8.
4. Join B5, C.
5. Draw a line parallel to B5C through which it intersects BC extended at C’.
6. Draw a line parallel to AC through ‘C’ which meets BA−→− produced at A’.
7. △A’BC’ is the required triangle.

Question 11.
Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it,whose sides are 2/3 of the corresponding sides of the first triangle. Steps of construction:

1. Draw △ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B1, B2 and B3 on BX−→− such that BB1 = B1B2 = B2B3.
4. Join B3, C.
5. Draw a line parallel to B3C through B2 meeting BC at C’.
6. Draw a line parallel to BA through C’ meeting BA at A’.
7. △A’BC’ is the required triangle.

Question 12.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another triangle whose sides are 112 times the corresponding sides of the isosceles triangle. Steps of construction:

1. Draw AABC in which BC = 8 cm and altitude AD = 4 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B1, B2 and B3 such that BB1 = B1B2 = B2B3.
4. Join B2C.
5. Draw a line parallel to B2C through B3 meeting BC produced C’.
6. Draw a line paral1e1 to AC through C’ meeting BA produced at A’.
7. △A’BC’ is the required triangle.

## Andhra Pradesh Board Class 10th Maths Chapter 8 Similar Triangles Ex 8.2 Textbooks for Exam Preparations

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