In a △ PQR , N is a point on PR , such that QN ⊥ PR . If PN × NR = QN 2 , prove that ∠ PQR = 90 ° .

From the given, Given, ∆ PQR in which QN ⊥ PR and PN × NR = QN 2 Now, in ∆ QNP and ∆ QNR ∵ QN ⊥ PR ∴ ∠ 1 = ∠ 2 = 90 ° And QN 2 = NR × NP QN × QN = NR × NP QN NR = NP QN ∴ ∆ QNP ~ ∆ QNR (By SAS similarity criteria) ∴ ∠ P = ∠ RQN = x And ∠ R = ∠ PQN = y Now, in ∆ PQR , We have ∠ P + ∠ Q + ∠ R = 180 ° ⇒ x + ( x + y ) + y = 180 ° ⇒ 2 ( x + y ) = 180 ° ⇒ x + y = 90 ° ⇒ ∠ PQR = 90 ° Hence, it is proved.

In figure, ∠ D = ∠ E and A D D B = A E E C , prove that B A C is an isosceles triangle.

Given triangle B A C : Here, A D D B = A E E C and ∠ D = ∠ E . From the expression A D D B = A E E C it is obvious that the line D E divides the two sides A B and A C of △ B A C in the same ratio. Hence, D E ∥ B C (Converse of basic proportionality) So, ∠ D = ∠ B and ∠ E = ∠ C (Corresponding angles). From the given data ∠ B = ∠ C . So, A B = A C (Sides opposite to equal angles) Now, if we observe, then we find that the triangle B A C is that triangle which has two equal sides and angles opposite to equal sides are also equal. Hence, it has proved that △ B A C is an isosceles triangle.

In Fig 2, D E ∥ A C and D C ∥ A P . Prove that B E E C = B C C P

In the given figure, it is given that D E ∥ A C and D C ∥ A P We know that, if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. In ∆ B A C , as D E ∥ A C , According to the basic proportionality theorem as stated above, B D A D = B E E C .....(i) In ∆ B A P , as D C ∥ A P , According to the basic proportionality theorem, B D A D = B C C P .....(ii) From (i) and (ii), we get B E E C = B C C P Hence, proved.

In an equilateral △ A B C , D is a point on side B C , such that B D = 1 3 B C . Prove that 9 A D 2 = 7 A B 2 .

Given: An equilateral △ A B C , D is a point on B C such that B D = 1 3 B C To Prove: 9 A D 2 = 7 A B 2 Construction: Draw A P ⊥ B C . Proof: We have B D = 1 3 B C = 1 3 A B Since △ A B C is an equilateral and A P ⊥ B C B P = P C = 1 2 B C = 1 2 A B Using Pythagoras theorem in △ A B P , A P 2 + B P 2 = A B 2 A P 2 = A B 2 - B P 2 = A B 2 - A B 2 2 = 3 4 A B 2 --[1] Using Pythagoras theorem in △ A D P , A D 2 = A P 2 + D P 2 A D 2 = 3 4 A B 2 + B P - B D 2 A D 2 = 3 4 A B 2 + 1 2 A B - 1 3 A B 2 A D 2 = 3 4 A B 2 + 1 36 A B 2 A D 2 = 27 A B 2 + A B 2 36 A D 2 = 28 36 A B 2 A D 2 = 7 9 A B 2 9 A D 2 = 7 A B 2 Hence Proved.

Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

To prove: The area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal. Let us assume, a square A B C E in which ∠ A B C = 90 ° and A B = B C Δ A B D and Δ A C E are equilateral triangles. To prove: a r ( Δ A B D ) = 1 2 ar ( Δ C A E ) Let, A B = B C = x units Then, C A = A B 2 + B C 2 = x 2 + x 2 = x 2 units. [ By Pythagoras Theorem] In Δ A B D and Δ C A E , each angle is 60 o as they are Equilateral Triangle. ∴ Δ A B D ~ Δ C A E Since, the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sides. ∴ ar ( Δ A B D ) ar ( Δ C A E ) = A B 2 C E 2 = x 2 ( x 2 ) 2 = x 2 2 x 2 = 1 2 Hence, a r ( Δ A B D ) = 1 2 ar ( Δ C A E ) , proved.

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to its first side is a right angle.

Given, in △ A B C , A C 2 = A B 2 + B C 2 . Construction: Construct a triangle P Q R right-angled at Q such that, P Q = A B and Q R = B C . To prove: ∠ A B C = 90 ° Proof: Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides. In △ P Q R , P R 2 = P Q 2 + Q R 2 ⇒ P R 2 = A B 2 + B C 2 [Given] ⇒ P R 2 = A C 2 [Given] ⇒ P R = A C _____________ 1 In △ A B C and △ P Q R , A B = P Q [Given] B C = Q R [Given] A C = P R [From 1 ] ∴ △ A B C ≅ △ P Q R (By SSS congruency) Corresponding parts of congruent triangles are equal. ⇒ ∠ A B C = ∠ P Q R ⇒ ∠ A B C = 90 ° ∵ ∠ P Q R = 90 ° Hence proved.

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Let △ A B C be a right triangle and A D is a perpendicular from A on B C . Pythagoras theorem states that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. To prove: B C 2 = A B 2 + A C 2 . Proof: In △ A B C and △ A C D , ∠ A C B = ∠ A C D [common angle] ∠ C A B = ∠ A D C [each 90 ° ] So by A A similarity △ A B C ~ △ A C D ... ( 1 ) So, A C C D = A B A D = B C A C . Now, on taking first and third results, we get A C C D = B C A C ∴ A C 2 = B C × C D ... ( 2 ) Now, in △ A B C and △ A B D , ∠ A B C = ∠ A B D [common angle] ∠ C A B = ∠ A D B [each 90 ° ] So, by A A similarity △ A B C ~ △ A B D ... ( 3 ) Hence, A C A D = A B B D = B C A B . On taking second and third result, we get A B B D = B C A B . On cross multiplying, A B 2 = B C × B D ... ( 4 ) On adding equation number ( 2 ) a n d ( 4 ) , A C 2 + A B 2 = B C × C D + B C × B D ⇒ A C 2 + A B 2 = B C × C D + B D A C 2 + A B 2 = B C × B C ∵ C D + B D = B C A C 2 + A B 2 = B C 2 ... ( 5 ) Hence, proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Delhi - II)>Q 24

MEDIUM

10th CBSE

IMPORTANT

Earn 100

If a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points then prove that the other two sides are divided in the same ratio.

Important Questions on Triangles

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Delhi - III)>Q 25

In the given figure, $PQ ∥ AC$ . If $BP = 4 cm, AP = 2.4 cm$ and $BQ = 5 cm$ , then the length of $BC$ is

Question Image

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Delhi - III)>Q 26

In a $△ PQR$ , $N$ is a point on $PR$ , such that $QN ⊥ PR$ . If $PN \times NR = {QN}^{2}$ , prove that $∠ PQR = 90 °$ .

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Delhi - III)>Q 27

In the given figure, $△ ABC$ and $△ DBC$ are on the same base $BC$ . If $AD$ intersects $BC$ at $O$ . Prove that $\frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AO}{DO}$ .

Question Image

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Outside Delhi)>Q 28

If

△ P Q R ~ △ A B C

;

P Q = 6 cm

,

A B = 8 cm

and the perimeter of

△ A B C

is

36 cm

, then the perimeter of

△ P Q R

is

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Outside Delhi)>Q 29

In the given figure $D E ∥ B C$ , if $A D = 3 cm$ , $A B = 7 cm$ and $E C = 3 cm$ , then the length of $A E$ is

Question Image

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Outside Delhi)>Q 30

In the given figure, $A B C$ is a triangle in which $D E ∥ B C$ . If $A D = x, D B = x - 2, A E = x + 2$ and $E C = x - 1$ , then find the value of $x$ .

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EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Outside Delhi)>Q 31

Diagonals $B C$ and $B D$ of a trapezium $A B C D$ with $A B ∥ D C$ intersect each other at point $O$ . Show that $\frac{O A}{O C} = \frac{O B}{O D}$

Question Image

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 6 - Triangles>CBSE-2023 (Outside Delhi)>Q 32

In the given figure, $E$ is a point on the side $C B$ produced of an isosceles triangle $A B C$ with $A B = A C$ . If $A D ⊥ B C$ and $E F ⊥ A C$ , then prove that $△ A B D ~ △ E C F$

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