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In ΔTQR, 

∠TQR + ∠QTR + ∠TRQ = 180° 

40° + 90° + ∠TRQ = 180° 

130° + ∠TRQ = 180° 

∠ TRQ = 180° – 130° 

∠ TRQ = 50° 

∠ TRS = 50°

 [∠PRS is same as ∠TRS ] 

In Δ PRS, 

RS is produced to Q 

∴Exterior∠PSQ = ∠SPR + ∠PRS 

= 30° + 50° 

∠PSQ = 80°

Correct option is (B) side

Two squares are congruent if they have same side.

Correct option is  B) side

Two rectangles are congruent, if they have same and length and breadth.

False

If the areas of two rectangles are the same, they are congruent.

If the areas of two rectangles are same, they are congruent.

False

If the areas of two circles are the same, they are congruent.

True

Correct answer is (d) 40°, 132°, 18°

Correct answer is (b) 2 cm, 4 cm, 6cm

(c) 

In Δs RTQ and PSQ, 

QR = PQ     (Given) 

∠P = ∠R      (Given)

∠TQR (∠SQR + ∠SQT) = ∠PQS (∠TQP + ∠SQT) 

ΔRTQ ≅ ΔPSQ      (ASA)

Correct option is  A) S. A. S

C) ΔPEN ≅ ΔRIM

Correct option is  D) S.S.S

Correct option is  D) S.S.S

Correct answer is (c) SSS

A) ΔABC ≅ ΔRQP

If Δ ABC ≅ ΔPQR 

⇒ AB = PQ, BC = QR, AC = PR

Correct answer is (b) SAS

Yes, if ΔPQR ≅ ΔEDF, then it means that corresponding angles and their sides are equal because we know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corrosponding sides and angles of other triangle.

Here, ΔPQR ≅ ΔEDF

∴ PQ = ED, QR = DF and PR = EF

Hence, it is true to say that PR = EF.

In the given figure \(\overline{AB}|| \overline{CD} || \overline{EF}\) at equal distances.

\(\overline{AF}\) is a transversal and  \(\overline{GH}\) is perpendicular to \(\overline{AB}\)

So, GH = 4 cm, AB = 4.5 cm, FB = 8 cm 

To find area of ΔGDF, 

From ΔABF, D is mid point of \(\overline{BF}\)

Similarly ‘G’ is mid point of \(\overline{AF}\)

So BD ⊥ DF, AG = GF

Median divides a triangle into two equal triangles.

∴ Area of ΔABG = Area of ΔBGF

Similarly \(\overline{GD}\) is median of ΔBGF.

Area of ΔABG = 2 × area of ΔDGF

Area of ΔDFG = 1/2 area of ΔABG

Area of ΔABG = 1/2 × 4.5 × 4 = 9 sq.cm

Area of ΔDGE = 1/2 × 9 = 4.5 sq.cm

Data: In ∆ABC, ∠C = 90°, D and E are points on the sides CA and CB respectively 

To Prove: AE² + BD² = AB² + DE² 

In ⊥∆ACE, ∠C = 90° 

∴ AE² = AC² + CE² ………. (i) 

In ⊥∆DEB, ∠C = 90° 

∴ BD² = DC² + CB² ………… (ii) 

From adding equations (i) + (ii) 

AE² + BD² = AC² + CE² + DC² + CB² 

= AC² + CB² + DC² + CE² 

∴ AE² + BD² = AB² + DE² (Theorem 8).

Given AD is the bisector of ∠A in ΔABC. Let AC be x cm. 

We know that the angle bisector theorem states that the internal bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

⇒ \(\frac{AB}{AC}\) = \(\frac{DB}{DC}\)

⇒ \(\frac{6}{x}\) = \(\frac{3}{2}\)

⇒ x = \(\frac{6(2)}{3}\)

⇒ x = 4 cm 

∴ AC = 4 cm

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side. 

(i) \(\overline {AD}\) is the median

(ii) \(\overline {AE}\) is the Altitude

From the given figure ΔABC, DE || BC. 

Let EC = x cm. 

We know that basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Then \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)

⇒ \(\frac{6}9\) = \(\frac{8}x\)

⇒ x = \(\frac{8(9)}{6}\)

⇒ x = 12 cm = EC 

Here, AC = AE + EC 

⇒ AC = 8 + 12 = 20 cm 

∴ AC = 20 cm

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is equal to one half of the third side.

Basic Proportionality Theorem: 

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

Converse of Basic Proportionality Theorem: 

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

1. c)Pythagoras Theorem

2. a)50m

3. d)(21,20,28)

4. b)38m

5. c)82m

If a line is draw parallel to one side of a triangle intersect the other two sides, then it divides the other two sides in the same ratio.

1. c)100m

2. d)60m

3. b)40m

4. a)16m

5. d) 8m

1.(c) 100m

2.(d) 60m

3.(b) 40m

4.(a) 16m

5. (d) 8m

The two triangles are similar if and only if 

1. The corresponding sides are in proportion. 

2. The corresponding angles are equal.

(i) The areas of two similar triangles are in the ratio of the squares of corresponding altitudes.

(ii) The areas of two similar triangles are in the ratio of the squares of the corresponding medians.

(iii) The area of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.

The two triangles are similar if and only if 

1. The corresponding sides are in proportion. 

2. The corresponding angles are equal.

Need to prove: \(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\) = \(\frac{(a^2+b^2+c^2)}{2abc}\)

Left hand side

= \(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\)

 = \(\frac{b^2+c^2-a^2}{2abc}\) + \(\frac{c^2+a^2-b^2}{2abc}\) + \(\frac{a^2+b^2-c^2}{2abc}\)

= \(\frac{a^2+b^2+c^2}{2abc}\)

= Right hand side. [proved]

We need to know that BC = QR. Here we use A.S.A criterion.

Given:sin²A + sin²B = sin²C

To prove: The triangle is right-angled 

sin²A + sin²B = sin²C

We know, a/sinA = b/sinB = c/sinC = 2R

So,

sin²A + sin²B = sin²C

a²/4R² + b²/4R² = c²/4R²

a² + b² = c²

This is one of the properties of right angled triangle. And it is satisfied here. Hence, the triangle is right angled. [Proved]

To prove

a² (cos²B – cos²C) + b² (cos²C – cos²A) + c² (cos²A – cos²B) = 0

From left hand side, 

= a² (cos²B – cos²C) + b² (cos²C – cos²A) + c² (cos²A – cos²B) 

= a² ((1 - sin²B) – (1 - sin²C)) + b² ((1 - sin²C) – (1 - sin²A)) + c2 ((1 - sin²A) – (1 - sin²B)) 

= a² ( - sin²B + sin²C) + b² ( - sin²C + sin²A) + c² ( - sin²A + sin²B)

We know that, a/sinA = b/sinB = c/sinC = 2R where R is the circumradius.

Therefore,

a = 2R sinA.....(a)

similarity, b = 2R sinB and C = 2R sinC

So,

= 4R² [ sin²A( - sin²B + sin²C) + sin²B( - sin²C + sin²A) + sin²C( - sin²A + sin²B) 

= 4R² [ - sin²Asin²B + sin²Asin2C – sin²Bsin²C + sin²Asin²B – sin²Asin²C + sin²Bsin²C ] 

= 4R² [0] 

= 0 [Proved]

Correct option is (D) corresponding angles are equal and sides are in proportion

Two polygons with the same number of sides are similar if their corresponding angles are equal and corresponding sides are in proportion.

Correct option is: (D) corresponding angles are equal and sides are in proportion

In ∆ABC, ∠ABC = 90° 

∠A + ∠C = 90° 

∠ABC > ∠BAC and ∠ABC < ∠BCA 

∴ D is the largest side of ∠ABC. 

∴ AC is opposite side of larger angle. 

∴ AC hypoternuse is largest side of ∆ABC.

Data : ∠B < ∠A and ∠C < ∠D. 

To Prove: AD < BC 

Proof: ∠B < ∠A 

∴ OA < OB …………. (i) 

Similarly, ∠C < ∠D , 

∴ OD < OC …………. (ii) 

Adding (i) and (ii), we have 

OA + OD < OB + OC 

∴ AD < BC.

Data : AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively. 

∠PBC < ∠QCB. 

To Prove: AC > AB 

Proof: ∠PBC < ∠QCB 

Now, ∠PBC + ∠ABC = 180° 

∠ABC = 180 – ∠PBC ………. (i) 

Similarly, ∠QCB + ∠ACB = 180° 

∠ACB = 180 – ∠QCB …………. (ii) 

But, ∠PBC < ∠QCB (Data) 

∴ ∠ABC > ∠ACB Comparing (i) and (ii), 

AC > AB (∵ Angle opposite to larger side is larger)

From the figure, 

∠PBC = ∠A + ∠ACB 

∠QCB = ∠A + ∠ABC 

Given that ∠PBC < ∠QCB 

⇒∠A + ∠ACB < ∠A + ∠ABC

⇒ ∠ACB < ∠ABC 

⇒ AB < AC 

⇒ AC > AB 

Hence proved.