## AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Optional Exercise Textbook Questions and Answers.

### 10th Class Maths 8th Lesson Coordinate Geometry Optional Exercise Textbook Questions and Answers

Question 1.

In the given figure, \(\frac{QT}{PR}\) = \(\frac{QR}{QS}\) and ∠1 = ∠2. Prove that △PQS ~ △TQR.

Answer:

Given: \(\frac{QT}{PR}\) = \(\frac{QR}{QS}\)

∠1 = ∠2

R.T.P : △PQS ~ △TQR

Proof: In △PQR; ∠1 = ∠2 Thus, PQ = PR

[∵ sides opp. to equal angles are equal]

\(\frac{QT}{PR}\) = \(\frac{QR}{QS}\) ⇒ \(\frac{QT}{PQ}\) = \(\frac{QR}{QS}\)

i.e., the line PS divides the two sides QT and QR of △TQR in the same ratio.

Hence, PS // TR.

[∵ If a line join of any two points on any two sides of triangle divides the two sides in the same ratio, then the line is parallel to the third side]

Hence, PS // TR (converse of B.P.T)

Now in △PQS and △TQR

∠QPS = ∠QTR

[∵ ∠P, ∠T are corresponding angles for PS // TR]

∠QSP = ∠QRT

[∵ ∠S, ∠R are corresponding angles for PS // TR]

∠Q = ∠Q (common)

∴ △PQS ~ △TQR (by AAA similarity)

Question 2.

Ravi is 1.82 m tall. He wants to find the height of a tree in his backyard. From the tree’s base he walked 12.20 m. along the tree’s shadow to a position where the end of his shadow exactly overlaps the end of the tree’s shadow. He is now 6.10 m from the end of the shadow. How tall is the tree?

Answer:

Given:

Height of Ravi ‘BC’ = 1.82 m.

Distance of Ravi from the foot of the tree BD = 12.2 m.

Length of the shadow of Ravi = AB = 6.10 m

Let DE represent the tree.

From the figure, △ABC ~ △ADE.

Thus, \(\frac{AB}{AD}\) = \(\frac{BC}{DE}\) = \(\frac{AC}{AE}\)

Ratio of corresponding sides of two similar triangles are equal]

\(\frac{6.10}{6.10+12.20}\) = \(\frac{1.82}{\mathrm{DE}}\)

DE = \(\frac{1.82 \times 18.30}{6.10}\) = 5.46 m

Thus the height of the tree = 5.46 m.

Question 3.

The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where ‘P’ is any point on side AB. Prove that CQ × PQ = QA × QD.

Answer:

Given: □ ABCD is a parallelogram.

P is a point on AB.

DP and AC intersect at Q.

R.T.P.: CQ . PQ = QA . QD.

Proof: In △CQD, △AQP

∠QCD = ∠QAP

∠CQD = ∠AQP

∴ ∠QDC = ∠QPA

(∵ Angle sum property of triangles)

Thus, △CQD ~ △AQP by AAA similarity condition.

\(\frac{CQ}{AQ}\) = \(\frac{QD}{QP}\) = \(\frac{CD}{AP}\)

[∵ Ratio of corresponding sides of similar triangles are equal]

\(\frac{CQ}{AQ}\) = \(\frac{QD}{QP}\)

CQ . PQ = QA . QD [Q.E.D]

Question 4.

△ABC and △AMP are two right triangles right angled at B and M respectively. Prove that (i) △ABC ~ △AMP (ii) \(\frac{CA}{PA}\) = \(\frac{BC}{MP}\)

Answer:

Given: △ABC; ∠B = 90°

AAMP; ∠M = 90°

R.T.P : i) △ABC ~ △AMP

Proof: In △ABC and △AMP

∠B = ∠M [each 90° given]

∠A = ∠A [common]

Hence, ∠C = ∠P

[∵ Angle sum property of triangles]

∴ △ABC ~ △AMP (by A.A.A. similarity)

ii) △ABC ~ △AMP (already proved)

\(\frac{AB}{AM}\) = \(\frac{BC}{MP}\) = \(\frac{CA}{PA}\)

[∵ Ratio of corresponding sides of similar triangles are equal]

\(\frac{CA}{PA}\) = \(\frac{BC}{MP}\)

Question 5.

An aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1\(\frac{1}{2}\) hour?

Answer:

Given: Speed of the first plane due north = 1000 kmph.

Speed of the second plane due west = 1200 kmph.

Distance = Speed × Time

Distance travelled by the first plane in

1\(\frac{1}{2}\) hrs = 1000 × 1\(\frac{1}{2}\) = 1000 × \(\frac{3}{2}\) = 1500 km.

Distance travelled by the second plane

in 1\(\frac{1}{2}\) hrs = 1200 × \(\frac{3}{2}\) = 1800 km.

From the figure, △ABC is a right triangle; ∠A = 90°.

AB^{2} + AC^{2} = BC^{2}

[Pythagoras theorem]

1500^{2} + 1800^{2} = BC^{2}

2250000 + 3240000 = BC^{2}

∴ BC = √5490000

= 100 × √549 m

≅ 100 × 23.43

≅ 2243km.

Question 6.

In a right triangle ABC right angled at C. P and Q are points on sides AC and CB respectively which divide these sides in the ratio of 2:1. Prove that

i) 9AQ^{2} = 9AC^{2} + 4BC^{2}

ii) 9BP^{2} = 9BC^{2} + 4AC^{2}

iii) 9(AQ^{2} + BP^{2}) = 13AB^{2}

Answer:

Given: In △ABC; ∠C = 90°

R.T.P.: i) 9AQ^{2} = 9AC^{2} + 4BC^{2}

Proof: In △ACQ; ∠C = 90°

AC^{2} + CQ^{2} = AQ^{2}

[side^{2} + side^{2} = hypotenuse^{2}]

AQ^{2} = AC^{2} + \(\left(\frac{2}{3} \mathrm{BC}\right)^{2}\)

[∵ Q divides CB in the ratio 2 : 1

CQ = \(\frac{2}{3}\)BC]

AQ^{2} = AC^{2} + \(\frac{4}{9} \mathrm{BC}^{2}\)

AQ^{2} = \(\frac{9 \mathrm{AC}^{2}+4 \mathrm{BC}^{2}}{9}\)

⇒ 9AQ^{2} = 9AC^{2} + 4BC^{2}

ii) 9BP^{2} = 9BC^{2} + 4AC^{2}

Proof: In △PCB,

PB^{2} = PC^{2} + BC^{2} [Pythagoras theorem]

⇒ PB^{2} = AC^{2} + 9BC^{2}

[!! If we take P on CA, in the ratio 2 : 1 then we get

BP^{2} = PC^{2} + BC^{2}

BP^{2} = \(\left(\frac{2}{3} \mathrm{A}\right)^{2}\) + BC^{2}

BP^{2} = \(\frac{4}{9} \mathrm{AC}^{2}\) + BC^{2}

BP^{2} = \(\frac{4 \mathrm{AC}^{2}+9 \mathrm{BC}^{2}}{9}\)

9BP^{2} = 4AC^{2} + 9BC^{2}

iii) 9 (AQ^{2} + BP^{2}) = 13 AB^{2}

Proof: In △ABC,

AC^{2} + BC^{2} = AB^{2}

[Pythagoras theorem]

Also, from (i) and (ii),